3.310 \(\int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=923 \[ -\frac {6 b \text {Li}_3\left (-i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {6 b \text {Li}_3\left (i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {3 a \text {Li}_3\left (-e^{2 i (c+d x)}\right ) f^3}{2 \left (a^2-b^2\right ) d^4}-\frac {6 b^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 i b (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {6 i b (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {3 i a (e+f x) \text {Li}_2\left (-e^{2 i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}+\frac {6 i b^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 a (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 b^2 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d} \]

[Out]

6*I*b*f^2*(f*x+e)*polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^3+I*b^2*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b
^2)^(1/2)))/(a^2-b^2)^(3/2)/d+3*a*f*(f*x+e)^2*ln(1+exp(2*I*(d*x+c)))/(a^2-b^2)/d^2-I*a*(f*x+e)^3/(a^2-b^2)/d-I
*b^2*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-3*I*a*f^2*(f*x+e)*polylog(2,-exp
(2*I*(d*x+c)))/(a^2-b^2)/d^3-6*I*b^2*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(
3/2)/d^3-6*I*b*f*(f*x+e)^2*arctan(exp(I*(d*x+c)))/(a^2-b^2)/d^2+3*b^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))
/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2-3*b^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/
(a^2-b^2)^(3/2)/d^2-6*b*f^3*polylog(3,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^4+6*b*f^3*polylog(3,I*exp(I*(d*x+c)))/(a^
2-b^2)/d^4+3/2*a*f^3*polylog(3,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^4-6*I*b*f^2*(f*x+e)*polylog(2,I*exp(I*(d*x+c)))/
(a^2-b^2)/d^3+6*I*b^2*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^3-6*b^2*
f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^4+6*b^2*f^3*polylog(4,I*b*exp(I*(d*x+c
))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^4-b*(f*x+e)^3*sec(d*x+c)/(a^2-b^2)/d+a*(f*x+e)^3*tan(d*x+c)/(a^2-b^2
)/d

________________________________________________________________________________________

Rubi [A]  time = 1.94, antiderivative size = 923, normalized size of antiderivative = 1.00, number of steps used = 29, number of rules used = 13, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {4533, 3323, 2264, 2190, 2531, 6609, 2282, 6589, 6742, 4184, 3719, 4409, 4181} \[ -\frac {6 b \text {PolyLog}\left (3,-i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {6 b \text {PolyLog}\left (3,i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {3 a \text {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) f^3}{2 \left (a^2-b^2\right ) d^4}-\frac {6 b^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 i b (e+f x) \text {PolyLog}\left (2,-i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {6 i b (e+f x) \text {PolyLog}\left (2,i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {3 i a (e+f x) \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}+\frac {6 i b^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 a (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 b^2 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

((-I)*a*(e + f*x)^3)/((a^2 - b^2)*d) - ((6*I)*b*f*(e + f*x)^2*ArcTan[E^(I*(c + d*x))])/((a^2 - b^2)*d^2) + (I*
b^2*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (I*b^2*(e + f*x)
^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + (3*a*f*(e + f*x)^2*Log[1 + E^
((2*I)*(c + d*x))])/((a^2 - b^2)*d^2) + ((6*I)*b*f^2*(e + f*x)*PolyLog[2, (-I)*E^(I*(c + d*x))])/((a^2 - b^2)*
d^3) - ((6*I)*b*f^2*(e + f*x)*PolyLog[2, I*E^(I*(c + d*x))])/((a^2 - b^2)*d^3) + (3*b^2*f*(e + f*x)^2*PolyLog[
2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (3*b^2*f*(e + f*x)^2*PolyLog[2, (I*
b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - ((3*I)*a*f^2*(e + f*x)*PolyLog[2, -E^((2*
I)*(c + d*x))])/((a^2 - b^2)*d^3) - (6*b*f^3*PolyLog[3, (-I)*E^(I*(c + d*x))])/((a^2 - b^2)*d^4) + (6*b*f^3*Po
lyLog[3, I*E^(I*(c + d*x))])/((a^2 - b^2)*d^4) + ((6*I)*b^2*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a
- Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^3) - ((6*I)*b^2*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a +
Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^3) + (3*a*f^3*PolyLog[3, -E^((2*I)*(c + d*x))])/(2*(a^2 - b^2)*d^4) -
(6*b^2*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^4) + (6*b^2*f^3*PolyL
og[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^4) - (b*(e + f*x)^3*Sec[c + d*x])/((a
^2 - b^2)*d) + (a*(e + f*x)^3*Tan[c + d*x])/((a^2 - b^2)*d)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4533

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> -Dist[b^2/(a^2 - b^2), Int[((e + f*x)^m*Sec[c + d*x]^(n - 2))/(a + b*Sin[c + d*x]), x], x] + Dist[1/(a^2
- b^2), Int[(e + f*x)^m*Sec[c + d*x]^n*(a - b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m
, 0] && NeQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac {\int \left (a (e+f x)^3 \sec ^2(c+d x)-b (e+f x)^3 \sec (c+d x) \tan (c+d x)\right ) \, dx}{a^2-b^2}-\frac {\left (2 b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}-\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {a \int (e+f x)^3 \sec ^2(c+d x) \, dx}{a^2-b^2}-\frac {b \int (e+f x)^3 \sec (c+d x) \tan (c+d x) \, dx}{a^2-b^2}\\ &=\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (3 i b^2 f\right ) \int (e+f x)^2 \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}+\frac {\left (3 i b^2 f\right ) \int (e+f x)^2 \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}-\frac {(3 a f) \int (e+f x)^2 \tan (c+d x) \, dx}{\left (a^2-b^2\right ) d}+\frac {(3 b f) \int (e+f x)^2 \sec (c+d x) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {(6 i a f) \int \frac {e^{2 i (c+d x)} (e+f x)^2}{1+e^{2 i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}-\frac {\left (6 b^2 f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^2}+\frac {\left (6 b^2 f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {\left (6 b f^2\right ) \int (e+f x) \log \left (1-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}+\frac {\left (6 b f^2\right ) \int (e+f x) \log \left (1+i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (6 a f^2\right ) \int (e+f x) \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}-\frac {\left (6 i b^2 f^3\right ) \int \text {Li}_3\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^3}+\frac {\left (6 i b^2 f^3\right ) \int \text {Li}_3\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {\left (6 i b f^3\right ) \int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^3}+\frac {\left (6 i b f^3\right ) \int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^3}\\ &=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i a f^2 (e+f x) \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (6 b^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {\left (6 b^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^4}-\frac {\left (6 b f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {\left (6 b f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {\left (3 i a f^3\right ) \int \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^3}\\ &=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i a f^2 (e+f x) \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 b f^3 \text {Li}_3\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 b f^3 \text {Li}_3\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {\left (3 a f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^4}\\ &=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i a f^2 (e+f x) \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 b f^3 \text {Li}_3\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 b f^3 \text {Li}_3\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}+\frac {3 a f^3 \text {Li}_3\left (-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^4}-\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 9.34, size = 1438, normalized size = 1.56 \[ \frac {b \sec (c) (e+f x)^3}{\left (b^2-a^2\right ) d}+\frac {f \left (\frac {2 i a (e+f x)^3}{f}+\frac {3 (a-b) \left (1+e^{2 i c}\right ) \log \left (1-i e^{-i (c+d x)}\right ) (e+f x)^2}{d}+\frac {3 (a+b) \left (1+e^{2 i c}\right ) \log \left (1+i e^{-i (c+d x)}\right ) (e+f x)^2}{d}+\frac {6 (a+b) \left (1+e^{2 i c}\right ) f \left (i d (e+f x) \text {Li}_2\left (-i e^{-i (c+d x)}\right )+f \text {Li}_3\left (-i e^{-i (c+d x)}\right )\right )}{d^3}+\frac {6 (a-b) \left (1+e^{2 i c}\right ) f \left (i d (e+f x) \text {Li}_2\left (i e^{-i (c+d x)}\right )+f \text {Li}_3\left (i e^{-i (c+d x)}\right )\right )}{d^3}\right )}{\left (a^2-b^2\right ) d \left (1+e^{2 i c}\right )}+\frac {b^2 \left (2 \sqrt {b^2-a^2} e^3 \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right ) d^3+\sqrt {a^2-b^2} f^3 x^3 \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3+3 \sqrt {a^2-b^2} e f^2 x^2 \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3+3 \sqrt {a^2-b^2} e^2 f x \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3-\sqrt {a^2-b^2} f^3 x^3 \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 \sqrt {a^2-b^2} e f^2 x^2 \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 \sqrt {a^2-b^2} e^2 f x \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 i \sqrt {a^2-b^2} f (e+f x)^2 \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^2+3 i \sqrt {a^2-b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d^2+6 \sqrt {a^2-b^2} e f^2 \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d+6 \sqrt {a^2-b^2} f^3 x \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d-6 \sqrt {a^2-b^2} e f^2 \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d-6 \sqrt {a^2-b^2} f^3 x \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d+6 i \sqrt {a^2-b^2} f^3 \text {Li}_4\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-6 i \sqrt {a^2-b^2} f^3 \text {Li}_4\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2} \left (b^2-a^2\right ) d^4}+\frac {\sin \left (\frac {d x}{2}\right ) e^3+3 f x \sin \left (\frac {d x}{2}\right ) e^2+3 f^2 x^2 \sin \left (\frac {d x}{2}\right ) e+f^3 x^3 \sin \left (\frac {d x}{2}\right )}{(a+b) d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\sin \left (\frac {d x}{2}\right ) e^3+3 f x \sin \left (\frac {d x}{2}\right ) e^2+3 f^2 x^2 \sin \left (\frac {d x}{2}\right ) e+f^3 x^3 \sin \left (\frac {d x}{2}\right )}{(a-b) d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(f*(((2*I)*a*(e + f*x)^3)/f + (3*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[1 - I/E^(I*(c + d*x))])/d + (3*(a +
b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x
)*PolyLog[2, (-I)/E^(I*(c + d*x))] + f*PolyLog[3, (-I)/E^(I*(c + d*x))]))/d^3 + (6*(a - b)*(1 + E^((2*I)*c))*f
*(I*d*(e + f*x)*PolyLog[2, I/E^(I*(c + d*x))] + f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3))/((a^2 - b^2)*d*(1 + E^
((2*I)*c))) + (b^2*(2*Sqrt[-a^2 + b^2]*d^3*e^3*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 3*Sqrt[a^2
- b^2]*d^3*e^2*f*x*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*
Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1 - (b*E^(I*(c + d*
x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2
+ b^2])] - 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - Sqrt[a^2 -
b^2]*d^3*f^3*x^3*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x
)^2*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*Poly
Log[2, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] + 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, (b*E^(I*(c + d*
x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a
^2 + b^2])] - 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - 6*Sqrt[a
^2 - b^2]*d*f^3*x*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] + (6*I)*Sqrt[a^2 - b^2]*f^3*Poly
Log[4, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - (6*I)*Sqrt[a^2 - b^2]*f^3*PolyLog[4, -((b*E^(I*(c +
d*x)))/(I*a + Sqrt[-a^2 + b^2]))]))/(Sqrt[-(a^2 - b^2)^2]*(-a^2 + b^2)*d^4) + (b*(e + f*x)^3*Sec[c])/((-a^2 +
b^2)*d) + (e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/((a +
b)*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d*x
)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/((a - b)*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2]))

________________________________________________________________________________________

fricas [C]  time = 1.05, size = 4140, normalized size = 4.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(4*(a^2*b - b^3)*d^3*f^3*x^3 + 12*(a^2*b - b^3)*d^3*e*f^2*x^2 - 12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d
*x + c)*polylog(4, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2))/b) + 12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(4, 1/2*(2*I*a*cos(d*x + c) - 2*a
*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*I*b^3*f^3*sqrt(-(a^2 - b
^2)/b^2)*cos(d*x + c)*polylog(4, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(4, 1/2*(-2*I*a*co
s(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*(a^2*b -
b^3)*d^3*e^2*f*x + 4*(a^2*b - b^3)*d^3*e^3 - 12*(a^3 - a^2*b - a*b^2 + b^3)*f^3*cos(d*x + c)*polylog(3, I*cos
(d*x + c) + sin(d*x + c)) - 12*(a^3 + a^2*b - a*b^2 - b^3)*f^3*cos(d*x + c)*polylog(3, I*cos(d*x + c) - sin(d*
x + c)) - 12*(a^3 - a^2*b - a*b^2 + b^3)*f^3*cos(d*x + c)*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) - 12*(a^3
+ a^2*b - a*b^2 - b^3)*f^3*cos(d*x + c)*polylog(3, -I*cos(d*x + c) - sin(d*x + c)) - 2*(-3*I*b^3*d^2*f^3*x^2
- 6*I*b^3*d^2*e*f^2*x - 3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(2*I*a*cos(d*x + c)
+ 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*(3*I*b^3*d
^2*f^3*x^2 + 6*I*b^3*d^2*e*f^2*x + 3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(2*I*a*co
s(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2
*(3*I*b^3*d^2*f^3*x^2 + 6*I*b^3*d^2*e*f^2*x + 3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/
2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b
)/b + 1) - 2*(-3*I*b^3*d^2*f^3*x^2 - 6*I*b^3*d^2*e*f^2*x - 3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x +
c)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^
2)/b^2) + 2*b)/b + 1) + 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*sqrt(-(a^2 - b^2
)/b^2)*cos(d*x + c)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*(b^3*d
^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(2*b*cos(
d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b
^3*c^2*d*e*f^2 - b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) +
2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*s
qrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2
*I*a) + 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 +
b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*
x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3
*d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/
2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)
/b) + 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b
^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x
+ c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*
d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2
*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)
/b) - 12*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, 1/2*(2*I*a*cos(d*x + c) -
2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*(b^3*d*f^3*x + b^3*d*
e*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*
x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*
cos(d*x + c)*polylog(3, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sq
rt(-(a^2 - b^2)/b^2))/b) + 12*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, 1/2*(
-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - (1
2*I*(a^3 - a^2*b - a*b^2 + b^3)*d*f^3*x + 12*I*(a^3 - a^2*b - a*b^2 + b^3)*d*e*f^2)*cos(d*x + c)*dilog(I*cos(d
*x + c) + sin(d*x + c)) - (-12*I*(a^3 + a^2*b - a*b^2 - b^3)*d*f^3*x - 12*I*(a^3 + a^2*b - a*b^2 - b^3)*d*e*f^
2)*cos(d*x + c)*dilog(I*cos(d*x + c) - sin(d*x + c)) - (-12*I*(a^3 - a^2*b - a*b^2 + b^3)*d*f^3*x - 12*I*(a^3
- a^2*b - a*b^2 + b^3)*d*e*f^2)*cos(d*x + c)*dilog(-I*cos(d*x + c) + sin(d*x + c)) - (12*I*(a^3 + a^2*b - a*b^
2 - b^3)*d*f^3*x + 12*I*(a^3 + a^2*b - a*b^2 - b^3)*d*e*f^2)*cos(d*x + c)*dilog(-I*cos(d*x + c) - sin(d*x + c)
) - 6*((a^3 + a^2*b - a*b^2 - b^3)*d^2*e^2*f - 2*(a^3 + a^2*b - a*b^2 - b^3)*c*d*e*f^2 + (a^3 + a^2*b - a*b^2
- b^3)*c^2*f^3)*cos(d*x + c)*log(cos(d*x + c) + I*sin(d*x + c) + I) - 6*((a^3 - a^2*b - a*b^2 + b^3)*d^2*e^2*f
- 2*(a^3 - a^2*b - a*b^2 + b^3)*c*d*e*f^2 + (a^3 - a^2*b - a*b^2 + b^3)*c^2*f^3)*cos(d*x + c)*log(cos(d*x + c
) - I*sin(d*x + c) + I) - 6*((a^3 + a^2*b - a*b^2 - b^3)*d^2*f^3*x^2 + 2*(a^3 + a^2*b - a*b^2 - b^3)*d^2*e*f^2
*x + 2*(a^3 + a^2*b - a*b^2 - b^3)*c*d*e*f^2 - (a^3 + a^2*b - a*b^2 - b^3)*c^2*f^3)*cos(d*x + c)*log(I*cos(d*x
+ c) + sin(d*x + c) + 1) - 6*((a^3 - a^2*b - a*b^2 + b^3)*d^2*f^3*x^2 + 2*(a^3 - a^2*b - a*b^2 + b^3)*d^2*e*f
^2*x + 2*(a^3 - a^2*b - a*b^2 + b^3)*c*d*e*f^2 - (a^3 - a^2*b - a*b^2 + b^3)*c^2*f^3)*cos(d*x + c)*log(I*cos(d
*x + c) - sin(d*x + c) + 1) - 6*((a^3 + a^2*b - a*b^2 - b^3)*d^2*f^3*x^2 + 2*(a^3 + a^2*b - a*b^2 - b^3)*d^2*e
*f^2*x + 2*(a^3 + a^2*b - a*b^2 - b^3)*c*d*e*f^2 - (a^3 + a^2*b - a*b^2 - b^3)*c^2*f^3)*cos(d*x + c)*log(-I*co
s(d*x + c) + sin(d*x + c) + 1) - 6*((a^3 - a^2*b - a*b^2 + b^3)*d^2*f^3*x^2 + 2*(a^3 - a^2*b - a*b^2 + b^3)*d^
2*e*f^2*x + 2*(a^3 - a^2*b - a*b^2 + b^3)*c*d*e*f^2 - (a^3 - a^2*b - a*b^2 + b^3)*c^2*f^3)*cos(d*x + c)*log(-I
*cos(d*x + c) - sin(d*x + c) + 1) - 6*((a^3 + a^2*b - a*b^2 - b^3)*d^2*e^2*f - 2*(a^3 + a^2*b - a*b^2 - b^3)*c
*d*e*f^2 + (a^3 + a^2*b - a*b^2 - b^3)*c^2*f^3)*cos(d*x + c)*log(-cos(d*x + c) + I*sin(d*x + c) + I) - 6*((a^3
- a^2*b - a*b^2 + b^3)*d^2*e^2*f - 2*(a^3 - a^2*b - a*b^2 + b^3)*c*d*e*f^2 + (a^3 - a^2*b - a*b^2 + b^3)*c^2*
f^3)*cos(d*x + c)*log(-cos(d*x + c) - I*sin(d*x + c) + I) - 4*((a^3 - a*b^2)*d^3*f^3*x^3 + 3*(a^3 - a*b^2)*d^3
*e*f^2*x^2 + 3*(a^3 - a*b^2)*d^3*e^2*f*x + (a^3 - a*b^2)*d^3*e^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d^4*c
os(d*x + c))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \sec \left (d x + c\right )^{2}}{b \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sec(d*x + c)^2/(b*sin(d*x + c) + a), x)

________________________________________________________________________________________

maple [F]  time = 6.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \left (\sec ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

int((f*x+e)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^3/(cos(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{3} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*sec(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)**3*sec(c + d*x)**2/(a + b*sin(c + d*x)), x)

________________________________________________________________________________________