3.4.10 \(\int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [310]

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

[Out]

-I*a*(f*x+e)^3/(a^2-b^2)/d+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+3*a*f*(f*x+e)^2*ln(1+exp(2*I*(d*x+c)))/(a^2-b^2)/d^2-6*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*I*a*f^2*(f*x+
e)*polylog(2,-exp(2*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+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*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*p
olylog(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*f^2*(f*x+e)*polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^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 = 1.28, 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 = {4629, 3404, 2296, 2221, 2611, 6744, 2320, 6724, 6874, 4269, 3800, 4494, 4266} \begin {gather*} -\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 \text {ArcTan}\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} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3404

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 3800

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 4266

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 4269

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 4494

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 4629

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 6724

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 6744

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 6874

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2241\) vs. \(2(923)=1846\).
time = 9.37, size = 2241, normalized size = 2.43 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*(e + f*x)^3*Sec[c])/((-a^2 + b^2)*d) - (I*b^2*((3*I)*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 + (b*(Cos[2*c + d*x]
+ I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])]*(Cos[c] + I*Sin[c])
+ (3*I)*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 + (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2
+ b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])]*(Cos[c] + I*Sin[c]) + I*Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1 + (b*(Co
s[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])]*(Cos[c]
+ I*Sin[c]) + 3*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLog[2, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Co
s[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) - 3*Sqrt[a^2 - b^2]*d^2*f*(e
+ f*x)^2*PolyLog[2, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Si
n[c])^2] + a*Sin[c])]*(Cos[c] + I*Sin[c]) + (6*I)*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, -((b*(Cos[2*c + d*x] + I*
Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) + (6
*I)*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^
2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) - 6*Sqrt[a^2 - b^2]*f^3*PolyLog[4, -((b*(Cos[2*c +
d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Si
n[c]) + 6*Sqrt[a^2 - b^2]*f^3*PolyLog[4, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 +
b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*(Cos[c] + I*Sin[c]) + 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 - (b*(Cos[
2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*((-I)*C
os[c] + Sin[c]) + 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 - (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[
c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*((-I)*Cos[c] + Sin[c]) + Sqrt[a^2 - b^2]*d^3*f^3*x^
3*Log[1 - (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] +
a*Sin[c])]*((-I)*Cos[c] + Sin[c]) + 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]
))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*((-I)*Cos[c] + Sin[c]) + 6*Sqrt[a^2
- b^2]*d*f^3*x*PolyLog[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] +
I*Sin[c])^2] + a*Sin[c])]*((-I)*Cos[c] + Sin[c]) - (2*I)*d^3*e^3*ArcTan[(b*Cos[c + d*x] + I*(a + b*Sin[c + d*
x]))/Sqrt[a^2 - b^2]]*Sqrt[(-a^2 + b^2)*(Cos[2*c] + I*Sin[2*c])]))/((a^2 - b^2)^(3/2)*d^4*Sqrt[(-a^2 + b^2)*(C
os[2*c] + I*Sin[2*c])]) + (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])) + (f*((-6*I)*a*d^3*e^2*x - (6*I)*a*d^3*e*f*x^2 - (2*I)*a*d^3*f^2*x^3 - (
12*I)*b*d^2*e^2*ArcTan[Cos[c + d*x] + I*Sin[c + d*x]] - (24*I)*b*d^2*e*f*x*ArcTan[Cos[c + d*x] + I*Sin[c + d*x
]] - (12*I)*b*d^2*f^2*x^2*ArcTan[Cos[c + d*x] + I*Sin[c + d*x]] + 6*a*d^2*e^2*Log[1 + Cos[2*(c + d*x)] + I*Sin
[2*(c + d*x)]] + 12*a*d^2*e*f*x*Log[1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]] + 6*a*d^2*f^2*x^2*Log[1 + Cos[2
*(c + d*x)] + I*Sin[2*(c + d*x)]] - (12*I)*b*d*f*(e + f*x)*PolyLog[2, I*Cos[c + d*x] - Sin[c + d*x]] + (12*I)*
b*d*f*(e + f*x)*PolyLog[2, (-I)*Cos[c + d*x] + Sin[c + d*x]] - (6*I)*a*d*e*f*PolyLog[2, -Cos[2*(c + d*x)] - I*
Sin[2*(c + d*x)]] - (6*I)*a*d*f^2*x*PolyLog[2, -Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)]] + 12*b*f^2*PolyLog[3, I
*Cos[c + d*x] - Sin[c + d*x]] - 12*b*f^2*PolyLog[3, (-I)*Cos[c + d*x] + Sin[c + d*x]] + 3*a*f^2*PolyLog[3, -Co
s[2*(c + d*x)] - I*Sin[2*(c + d*x)]] + 6*a*d^3*e^2*x*Tan[c] + 6*a*d^3*e*f*x^2*Tan[c] + 2*a*d^3*f^2*x^3*Tan[c])
)/(2*(a^2 - b^2)*d^4)

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Maple [F]
time = 0.16, 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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4118 vs. \(2 (824) = 1648\).
time = 0.82, size = 4118, normalized size = 4.46 \begin {gather*} \text {Too large to display} \end {gather*}

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]

-1/2*(2*(a^2*b - b^3)*d^3*f^3*x^3 + 6*(a^2*b - b^3)*d^3*f^2*x^2*e - 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x
+ c)*polylog(4, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b
^2))/b) + 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(4, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*
cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*
polylog(4, -(-I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/
b) - 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(4, -(-I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(
d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 6*(a^2*b - b^3)*d^3*f*x*e^2 - 6*(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)) - 6*(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)) - 6*(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)) - 6*(a^3 + a^2*b - a*b^2 - b^3)*f^3*cos(d*x + c)*polylog(3, -I*cos(d*x + c) - si
n(d*x + c)) + 2*(a^2*b - b^3)*d^3*e^3 + 3*(I*b^3*d^2*f^3*x^2 + 2*I*b^3*d^2*f^2*x*e + I*b^3*d^2*f*e^2)*sqrt(-(a
^2 - b^2)/b^2)*cos(d*x + c)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqr
t(-(a^2 - b^2)/b^2) - b)/b + 1) + 3*(-I*b^3*d^2*f^3*x^2 - 2*I*b^3*d^2*f^2*x*e - I*b^3*d^2*f*e^2)*sqrt(-(a^2 -
b^2)/b^2)*cos(d*x + c)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2) - b)/b + 1) + 3*(-I*b^3*d^2*f^3*x^2 - 2*I*b^3*d^2*f^2*x*e - I*b^3*d^2*f*e^2)*sqrt(-(a^2 - b^2)/
b^2)*cos(d*x + c)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 -
b^2)/b^2) - b)/b + 1) + 3*(I*b^3*d^2*f^3*x^2 + 2*I*b^3*d^2*f^2*x*e + I*b^3*d^2*f*e^2)*sqrt(-(a^2 - b^2)/b^2)*
cos(d*x + c)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) - b)/b + 1) - (b^3*c^3*f^3 - 3*b^3*c^2*d*f^2*e + 3*b^3*c*d^2*f*e^2 - b^3*d^3*e^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) - (b^3*c^3*f^3 -
3*b^3*c^2*d*f^2*e + 3*b^3*c*d^2*f*e^2 - b^3*d^3*e^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) + (b^3*c^3*f^3 - 3*b^3*c^2*d*f^2*e + 3*b^3*c*d^2*f
*e^2 - b^3*d^3*e^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) + (b^3*c^3*f^3 - 3*b^3*c^2*d*f^2*e + 3*b^3*c*d^2*f*e^2 - b^3*d^3*e^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) - (b^3
*d^3*f^3*x^3 + b^3*c^3*f^3 + 3*(b^3*d^3*f*x + b^3*c*d^2*f)*e^2 + 3*(b^3*d^3*f^2*x^2 - b^3*c^2*d*f^2)*e)*sqrt(-
(a^2 - b^2)/b^2)*cos(d*x + c)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sq
rt(-(a^2 - b^2)/b^2) - b)/b) + (b^3*d^3*f^3*x^3 + b^3*c^3*f^3 + 3*(b^3*d^3*f*x + b^3*c*d^2*f)*e^2 + 3*(b^3*d^3
*f^2*x^2 - b^3*c^2*d*f^2)*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*
cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - (b^3*d^3*f^3*x^3 + b^3*c^3*f^3 + 3*(b^3*d^3*
f*x + b^3*c*d^2*f)*e^2 + 3*(b^3*d^3*f^2*x^2 - b^3*c^2*d*f^2)*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-(-I*a
*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b^3*d^3
*f^3*x^3 + b^3*c^3*f^3 + 3*(b^3*d^3*f*x + b^3*c*d^2*f)*e^2 + 3*(b^3*d^3*f^2*x^2 - b^3*c^2*d*f^2)*e)*sqrt(-(a^2
- b^2)/b^2)*cos(d*x + c)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(
-(a^2 - b^2)/b^2) - b)/b) + 6*(b^3*d*f^3*x + b^3*d*f^2*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, -(I*a
*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 6*(b^3*d*f^3
*x + b^3*d*f^2*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(
d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 6*(b^3*d*f^3*x + b^3*d*f^2*e)*sqrt(-(a^2 - b^2)/b^2)
*cos(d*x + c)*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2
- b^2)/b^2))/b) - 6*(b^3*d*f^3*x + b^3*d*f^2*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, -(-I*a*cos(d*x
+ c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 6*(-I*(a^3 - a^2*b -
a*b^2 + b^3)*d*f^3*x - I*(a^3 - a^2*b - a*b^2 + b^3)*d*f^2*e)*cos(d*x + c)*dilog(I*cos(d*x + c) + sin(d*x + c
)) + 6*(I*(a^3 + a^2*b - a*b^2 - b^3)*d*f^3*x + I*(a^3 + a^2*b - a*b^2 - b^3)*d*f^2*e)*cos(d*x + c)*dilog(I*co
s(d*x + c) - sin(d*x + c)) + 6*(I*(a^3 - a^2*b - a*b^2 + b^3)*d*f^3*x + I*(a^3 - a^2*b - a*b^2 + b^3)*d*f^2*e)
*cos(d*x + c)*dilog(-I*cos(d*x + c) + sin(d*x + c)) + 6*(-I*(a^3 + a^2*b - a*b^2 - b^3)*d*f^3*x - I*(a^3 + a^2
*b - a*b^2 - b^3)*d*f^2*e)*cos(d*x + c)*dilog(-...

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

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

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}

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