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

Optimal. Leaf size=937 \[ -\frac {6 i a \text {Li}_4\left (-i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {6 i a \text {Li}_4\left (i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}-\frac {6 i b \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right ) d^4}-\frac {6 i b \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {3 i b \text {Li}_4\left (-e^{2 i (c+d x)}\right ) f^3}{4 \left (a^2-b^2\right ) d^4}-\frac {6 a (e+f x) \text {Li}_3\left (-i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}+\frac {6 a (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {6 b (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {6 b (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{\left (a^2-b^2\right ) d^3}+\frac {3 b (e+f x) \text {Li}_3\left (-e^{2 i (c+d x)}\right ) f^2}{2 \left (a^2-b^2\right ) d^3}+\frac {3 i a (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}-\frac {3 i a (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 i b (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 i b (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{\left (a^2-b^2\right ) d^2}-\frac {3 i b (e+f x)^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) f}{2 \left (a^2-b^2\right ) d^2}-\frac {2 i a (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x)^3 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.62, antiderivative size = 937, normalized size of antiderivative = 1.00, number of steps used = 29, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4533, 4519, 2190, 2531, 6609, 2282, 6589, 6742, 4181, 3719} \[ -\frac {6 i a \text {PolyLog}\left (4,-i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}+\frac {6 i a \text {PolyLog}\left (4,i e^{i (c+d x)}\right ) f^3}{\left (a^2-b^2\right ) d^4}-\frac {6 i b \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 ) d^4}-\frac {6 i b \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 ) d^4}+\frac {3 i b \text {PolyLog}\left (4,-e^{2 i (c+d x)}\right ) f^3}{4 \left (a^2-b^2\right ) d^4}-\frac {6 a (e+f x) \text {PolyLog}\left (3,-i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}+\frac {6 a (e+f x) \text {PolyLog}\left (3,i e^{i (c+d x)}\right ) f^2}{\left (a^2-b^2\right ) d^3}-\frac {6 b (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 ) d^3}-\frac {6 b (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 ) d^3}+\frac {3 b (e+f x) \text {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) f^2}{2 \left (a^2-b^2\right ) d^3}+\frac {3 i a (e+f x)^2 \text {PolyLog}\left (2,-i e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}-\frac {3 i a (e+f x)^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right ) f}{\left (a^2-b^2\right ) d^2}+\frac {3 i b (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 ) d^2}+\frac {3 i b (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 ) d^2}-\frac {3 i b (e+f x)^2 \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) f}{2 \left (a^2-b^2\right ) d^2}-\frac {2 i a (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (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 ) d}-\frac {b (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 ) d}+\frac {b (e+f x)^3 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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

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Mathematica [B]  time = 10.03, size = 2496, normalized size = 2.66 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*((I*b*(e + f*x)^4)/f - (2*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^3*Log[1 - I/E^(I*(c + d*x))])/d + (2*(a + b)
*(1 + E^((2*I)*c))*(e + f*x)^3*Log[1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d^2*(e + f*x)
^2*PolyLog[2, (-I)/E^(I*(c + d*x))] + 2*f*(d*(e + f*x)*PolyLog[3, (-I)/E^(I*(c + d*x))] - I*f*PolyLog[4, (-I)/
E^(I*(c + d*x))])))/d^4 - ((6*I)*(a - b)*(1 + E^((2*I)*c))*f*(d^2*(e + f*x)^2*PolyLog[2, I/E^(I*(c + d*x))] -
(2*I)*d*f*(e + f*x)*PolyLog[3, I/E^(I*(c + d*x))] - 2*f^2*PolyLog[4, I/E^(I*(c + d*x))]))/d^4))/((a^2 - b^2)*(
1 + E^((2*I)*c))) + (4*b*((-4*I)*d^4*e^3*E^((2*I)*c)*x - (6*I)*d^4*e^2*E^((2*I)*c)*f*x^2 - (4*I)*d^4*e*E^((2*I
)*c)*f^2*x^3 - I*d^4*E^((2*I)*c)*f^3*x^4 - (2*I)*d^3*e^3*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d
*x))))] + (2*I)*d^3*e^3*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] - d^3*e^3*Log
[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + d^3*e^3*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c + d
*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a
^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e^2*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 +
b^2)*E^((2*I)*c)])] - 6*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)
*c)])] + 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c
)])] - 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*d^3*E^(
(2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*f*x
*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e^2*E^((2*I)*c)*f*x*Log
[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(
2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*
c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*
E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*d^3*E^((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I
*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(2*c
+ d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)^2*PolyLog[
2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*e*f^2*PolyLog[3, (I*b*E^(I*
(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, (I*b*E^(I*(2
*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*f^3*x*PolyLog[3, (I*b*E^(I*(2*c + d*x)))/(a
*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, (I*b*E^(I*(2*c + d*x)))/(a*E
^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*e*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqr
t[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt
[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)
*E^((2*I)*c)]))] + 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*
E^((2*I)*c)]))] - (12*I)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])
] + (12*I)*E^((2*I)*c)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])]
- (12*I)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + (12*I)*E^((
2*I)*c)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))]))/((-a^2 + b^2
)*d^4*(-1 + E^((2*I)*c))) - (8*b*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*Csc[c]^3)/((a - b)*(a + b)*(Csc
[c/2] - Sec[c/2])*(Csc[c/2] + Sec[c/2])))/8

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fricas [C]  time = 0.75, size = 3081, normalized size = 3.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*I*b*f^3*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) + 6*I*b*f^3*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) - 6*I*b*f^3*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) - 6*I*b*f^3*polylog(4, 1/2*(-2*I*a*c
os(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) - 6*I*(a - b
)*f^3*polylog(4, I*cos(d*x + c) + sin(d*x + c)) - 6*I*(a + b)*f^3*polylog(4, I*cos(d*x + c) - sin(d*x + c)) +
6*I*(a - b)*f^3*polylog(4, -I*cos(d*x + c) + sin(d*x + c)) + 6*I*(a + b)*f^3*polylog(4, -I*cos(d*x + c) - sin(
d*x + c)) + (3*I*b*d^2*f^3*x^2 + 6*I*b*d^2*e*f^2*x + 3*I*b*d^2*e^2*f)*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) + (3*I*b*d^2*f^3*x^2 +
6*I*b*d^2*e*f^2*x + 3*I*b*d^2*e^2*f)*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) + (-3*I*b*d^2*f^3*x^2 - 6*I*b*d^2*e*f^2*x - 3*I*b*d^2*e^
2*f)*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) + (-3*I*b*d^2*f^3*x^2 - 6*I*b*d^2*e*f^2*x - 3*I*b*d^2*e^2*f)*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) + (3*I*(
a - b)*d^2*f^3*x^2 + 6*I*(a - b)*d^2*e*f^2*x + 3*I*(a - b)*d^2*e^2*f)*dilog(I*cos(d*x + c) + sin(d*x + c)) + (
3*I*(a + b)*d^2*f^3*x^2 + 6*I*(a + b)*d^2*e*f^2*x + 3*I*(a + b)*d^2*e^2*f)*dilog(I*cos(d*x + c) - sin(d*x + c)
) + (-3*I*(a - b)*d^2*f^3*x^2 - 6*I*(a - b)*d^2*e*f^2*x - 3*I*(a - b)*d^2*e^2*f)*dilog(-I*cos(d*x + c) + sin(d
*x + c)) + (-3*I*(a + b)*d^2*f^3*x^2 - 6*I*(a + b)*d^2*e*f^2*x - 3*I*(a + b)*d^2*e^2*f)*dilog(-I*cos(d*x + c)
- sin(d*x + c)) + (b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*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*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*
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*d^3*e^3 - 3*b*c*d^2*e^2*f
+ 3*b*c^2*d*e*f^2 - b*c^3*f^3)*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*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*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*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2*e^2*f -
3*b*c^2*d*e*f^2 + b*c^3*f^3)*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) + (b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2*e^2
*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*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) + (b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2
*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*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) + (b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*
c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*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) - ((a + b)*d^3*e^3 - 3*(a + b)*c*d^2*e^2*f + 3*(a + b)*c
^2*d*e*f^2 - (a + b)*c^3*f^3)*log(cos(d*x + c) + I*sin(d*x + c) + I) + ((a - b)*d^3*e^3 - 3*(a - b)*c*d^2*e^2*
f + 3*(a - b)*c^2*d*e*f^2 - (a - b)*c^3*f^3)*log(cos(d*x + c) - I*sin(d*x + c) + I) - ((a + b)*d^3*f^3*x^3 + 3
*(a + b)*d^3*e*f^2*x^2 + 3*(a + b)*d^3*e^2*f*x + 3*(a + b)*c*d^2*e^2*f - 3*(a + b)*c^2*d*e*f^2 + (a + b)*c^3*f
^3)*log(I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d^3*f^3*x^3 + 3*(a - b)*d^3*e*f^2*x^2 + 3*(a - b)*d^3*e^
2*f*x + 3*(a - b)*c*d^2*e^2*f - 3*(a - b)*c^2*d*e*f^2 + (a - b)*c^3*f^3)*log(I*cos(d*x + c) - sin(d*x + c) + 1
) - ((a + b)*d^3*f^3*x^3 + 3*(a + b)*d^3*e*f^2*x^2 + 3*(a + b)*d^3*e^2*f*x + 3*(a + b)*c*d^2*e^2*f - 3*(a + b)
*c^2*d*e*f^2 + (a + b)*c^3*f^3)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d^3*f^3*x^3 + 3*(a - b)*d^3
*e*f^2*x^2 + 3*(a - b)*d^3*e^2*f*x + 3*(a - b)*c*d^2*e^2*f - 3*(a - b)*c^2*d*e*f^2 + (a - b)*c^3*f^3)*log(-I*c
os(d*x + c) - sin(d*x + c) + 1) - ((a + b)*d^3*e^3 - 3*(a + b)*c*d^2*e^2*f + 3*(a + b)*c^2*d*e*f^2 - (a + b)*c
^3*f^3)*log(-cos(d*x + c) + I*sin(d*x + c) + I) + ((a - b)*d^3*e^3 - 3*(a - b)*c*d^2*e^2*f + 3*(a - b)*c^2*d*e
*f^2 - (a - b)*c^3*f^3)*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 6*(b*d*f^3*x + b*d*e*f^2)*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) + 6*(b*
d*f^3*x + b*d*e*f^2)*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) + 6*(b*d*f^3*x + b*d*e*f^2)*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) + 6*(b*d*f^3*x + b*d*e*f^2)*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
) + 6*((a - b)*d*f^3*x + (a - b)*d*e*f^2)*polylog(3, I*cos(d*x + c) + sin(d*x + c)) - 6*((a + b)*d*f^3*x + (a
+ b)*d*e*f^2)*polylog(3, I*cos(d*x + c) - sin(d*x + c)) + 6*((a - b)*d*f^3*x + (a - b)*d*e*f^2)*polylog(3, -I*
cos(d*x + c) + sin(d*x + c)) - 6*((a + b)*d*f^3*x + (a + b)*d*e*f^2)*polylog(3, -I*cos(d*x + c) - sin(d*x + c)
))/((a^2 - b^2)*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.66, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \sec \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(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?

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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)*(a + b*sin(c + d*x))),x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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