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

Optimal. Leaf size=698 \[ -\frac{9 f^2 (e+f x) \text{PolyLog}\left (3,-i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 f^2 (e+f x) \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 i f (e+f x)^2 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{9 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{8 a d^2}+\frac{5 i f^3 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^4}-\frac{5 i f^3 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^4}-\frac{i f^3 \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac{9 i f^3 \text{PolyLog}\left (4,-i e^{i (c+d x)}\right )}{4 a d^4}+\frac{9 i f^3 \text{PolyLog}\left (4,i e^{i (c+d x)}\right )}{4 a d^4}+\frac{f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}+\frac{f^2 (e+f x) \tan (c+d x) \sec (c+d x)}{4 a d^3}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}+\frac{f (e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{4 a d^2}+\frac{f^3 \tan (c+d x)}{4 a d^4}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{(e+f x)^3 \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 (e+f x)^3 \tan (c+d x) \sec (c+d x)}{8 a d}-\frac{i f (e+f x)^2}{2 a d^2} \]

[Out]

((-I/2)*f*(e + f*x)^2)/(a*d^2) - ((5*I)*f^2*(e + f*x)*ArcTan[E^(I*(c + d*x))])/(a*d^3) - (((3*I)/4)*(e + f*x)^
3*ArcTan[E^(I*(c + d*x))])/(a*d) + (f^2*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/(a*d^3) + (((5*I)/2)*f^3*PolyL
og[2, (-I)*E^(I*(c + d*x))])/(a*d^4) + (((9*I)/8)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - ((
(5*I)/2)*f^3*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^4) - (((9*I)/8)*f*(e + f*x)^2*PolyLog[2, I*E^(I*(c + d*x))])/
(a*d^2) - ((I/2)*f^3*PolyLog[2, -E^((2*I)*(c + d*x))])/(a*d^4) - (9*f^2*(e + f*x)*PolyLog[3, (-I)*E^(I*(c + d*
x))])/(4*a*d^3) + (9*f^2*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/(4*a*d^3) - (((9*I)/4)*f^3*PolyLog[4, (-I)*E
^(I*(c + d*x))])/(a*d^4) + (((9*I)/4)*f^3*PolyLog[4, I*E^(I*(c + d*x))])/(a*d^4) - (f^3*Sec[c + d*x])/(4*a*d^4
) - (9*f*(e + f*x)^2*Sec[c + d*x])/(8*a*d^2) - (f^2*(e + f*x)*Sec[c + d*x]^2)/(4*a*d^3) - (f*(e + f*x)^2*Sec[c
+ d*x]^3)/(4*a*d^2) - ((e + f*x)^3*Sec[c + d*x]^4)/(4*a*d) + (f^3*Tan[c + d*x])/(4*a*d^4) + (f*(e + f*x)^2*Ta
n[c + d*x])/(2*a*d^2) + (f^2*(e + f*x)*Sec[c + d*x]*Tan[c + d*x])/(4*a*d^3) + (3*(e + f*x)^3*Sec[c + d*x]*Tan[
c + d*x])/(8*a*d) + (f*(e + f*x)^2*Sec[c + d*x]^2*Tan[c + d*x])/(4*a*d^2) + ((e + f*x)^3*Sec[c + d*x]^3*Tan[c
+ d*x])/(4*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.73537, antiderivative size = 698, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 16, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4531, 4186, 4185, 4181, 2279, 2391, 2531, 6609, 2282, 6589, 4409, 3767, 8, 4184, 3719, 2190} \[ -\frac{9 f^2 (e+f x) \text{PolyLog}\left (3,-i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 f^2 (e+f x) \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 i f (e+f x)^2 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{9 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{8 a d^2}+\frac{5 i f^3 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^4}-\frac{5 i f^3 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^4}-\frac{i f^3 \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac{9 i f^3 \text{PolyLog}\left (4,-i e^{i (c+d x)}\right )}{4 a d^4}+\frac{9 i f^3 \text{PolyLog}\left (4,i e^{i (c+d x)}\right )}{4 a d^4}+\frac{f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}+\frac{f^2 (e+f x) \tan (c+d x) \sec (c+d x)}{4 a d^3}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}+\frac{f (e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{4 a d^2}+\frac{f^3 \tan (c+d x)}{4 a d^4}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{(e+f x)^3 \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 (e+f x)^3 \tan (c+d x) \sec (c+d x)}{8 a d}-\frac{i f (e+f x)^2}{2 a d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*f*(e + f*x)^2)/(a*d^2) - ((5*I)*f^2*(e + f*x)*ArcTan[E^(I*(c + d*x))])/(a*d^3) - (((3*I)/4)*(e + f*x)^
3*ArcTan[E^(I*(c + d*x))])/(a*d) + (f^2*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/(a*d^3) + (((5*I)/2)*f^3*PolyL
og[2, (-I)*E^(I*(c + d*x))])/(a*d^4) + (((9*I)/8)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - ((
(5*I)/2)*f^3*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^4) - (((9*I)/8)*f*(e + f*x)^2*PolyLog[2, I*E^(I*(c + d*x))])/
(a*d^2) - ((I/2)*f^3*PolyLog[2, -E^((2*I)*(c + d*x))])/(a*d^4) - (9*f^2*(e + f*x)*PolyLog[3, (-I)*E^(I*(c + d*
x))])/(4*a*d^3) + (9*f^2*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/(4*a*d^3) - (((9*I)/4)*f^3*PolyLog[4, (-I)*E
^(I*(c + d*x))])/(a*d^4) + (((9*I)/4)*f^3*PolyLog[4, I*E^(I*(c + d*x))])/(a*d^4) - (f^3*Sec[c + d*x])/(4*a*d^4
) - (9*f*(e + f*x)^2*Sec[c + d*x])/(8*a*d^2) - (f^2*(e + f*x)*Sec[c + d*x]^2)/(4*a*d^3) - (f*(e + f*x)^2*Sec[c
+ d*x]^3)/(4*a*d^2) - ((e + f*x)^3*Sec[c + d*x]^4)/(4*a*d) + (f^3*Tan[c + d*x])/(4*a*d^4) + (f*(e + f*x)^2*Ta
n[c + d*x])/(2*a*d^2) + (f^2*(e + f*x)*Sec[c + d*x]*Tan[c + d*x])/(4*a*d^3) + (3*(e + f*x)^3*Sec[c + d*x]*Tan[
c + d*x])/(8*a*d) + (f*(e + f*x)^2*Sec[c + d*x]^2*Tan[c + d*x])/(4*a*d^2) + ((e + f*x)^3*Sec[c + d*x]^3*Tan[c
+ d*x])/(4*a*d)

Rule 4531

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

Rule 4186

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

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 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 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 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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^3 \sec ^5(c+d x) \, dx}{a}-\frac{\int (e+f x)^3 \sec ^4(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{3 \int (e+f x)^3 \sec ^3(c+d x) \, dx}{4 a}+\frac{(3 f) \int (e+f x)^2 \sec ^4(c+d x) \, dx}{4 a d}+\frac{f^2 \int (e+f x) \sec ^3(c+d x) \, dx}{2 a d^2}\\ &=-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{3 \int (e+f x)^3 \sec (c+d x) \, dx}{8 a}+\frac{f \int (e+f x)^2 \sec ^2(c+d x) \, dx}{2 a d}+\frac{f^2 \int (e+f x) \sec (c+d x) \, dx}{4 a d^2}+\frac{\left (9 f^2\right ) \int (e+f x) \sec (c+d x) \, dx}{4 a d^2}+\frac{f^3 \int \sec ^2(c+d x) \, dx}{4 a d^3}\\ &=-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{(9 f) \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right ) \, dx}{8 a d}+\frac{(9 f) \int (e+f x)^2 \log \left (1+i e^{i (c+d x)}\right ) \, dx}{8 a d}-\frac{f^2 \int (e+f x) \tan (c+d x) \, dx}{a d^2}-\frac{f^3 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{4 a d^4}-\frac{f^3 \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{4 a d^3}+\frac{f^3 \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{4 a d^3}-\frac{\left (9 f^3\right ) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{4 a d^3}+\frac{\left (9 f^3\right ) \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{4 a d^3}\\ &=-\frac{i f (e+f x)^2}{2 a d^2}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}+\frac{9 i f (e+f x)^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{9 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f^3 \tan (c+d x)}{4 a d^4}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\left (2 i f^2\right ) \int \frac{e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{a d^2}-\frac{\left (9 i f^2\right ) \int (e+f x) \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx}{4 a d^2}+\frac{\left (9 i f^2\right ) \int (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx}{4 a d^2}+\frac{\left (i f^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}-\frac{\left (i f^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}+\frac{\left (9 i f^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}-\frac{\left (9 i f^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}\\ &=-\frac{i f (e+f x)^2}{2 a d^2}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}+\frac{f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac{5 i f^3 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^4}+\frac{9 i f (e+f x)^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{5 i f^3 \text{Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^4}-\frac{9 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}-\frac{9 f^2 (e+f x) \text{Li}_3\left (-i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{4 a d^3}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f^3 \tan (c+d x)}{4 a d^4}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{f^3 \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{a d^3}+\frac{\left (9 f^3\right ) \int \text{Li}_3\left (-i e^{i (c+d x)}\right ) \, dx}{4 a d^3}-\frac{\left (9 f^3\right ) \int \text{Li}_3\left (i e^{i (c+d x)}\right ) \, dx}{4 a d^3}\\ &=-\frac{i f (e+f x)^2}{2 a d^2}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}+\frac{f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac{5 i f^3 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^4}+\frac{9 i f (e+f x)^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{5 i f^3 \text{Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^4}-\frac{9 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}-\frac{9 f^2 (e+f x) \text{Li}_3\left (-i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{4 a d^3}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f^3 \tan (c+d x)}{4 a d^4}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\left (i f^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a d^4}-\frac{\left (9 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}+\frac{\left (9 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{4 a d^4}\\ &=-\frac{i f (e+f x)^2}{2 a d^2}-\frac{5 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac{3 i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{4 a d}+\frac{f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac{5 i f^3 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^4}+\frac{9 i f (e+f x)^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{8 a d^2}-\frac{5 i f^3 \text{Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^4}-\frac{9 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{8 a d^2}-\frac{i f^3 \text{Li}_2\left (-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac{9 f^2 (e+f x) \text{Li}_3\left (-i e^{i (c+d x)}\right )}{4 a d^3}+\frac{9 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{4 a d^3}-\frac{9 i f^3 \text{Li}_4\left (-i e^{i (c+d x)}\right )}{4 a d^4}+\frac{9 i f^3 \text{Li}_4\left (i e^{i (c+d x)}\right )}{4 a d^4}-\frac{f^3 \sec (c+d x)}{4 a d^4}-\frac{9 f (e+f x)^2 \sec (c+d x)}{8 a d^2}-\frac{f^2 (e+f x) \sec ^2(c+d x)}{4 a d^3}-\frac{f (e+f x)^2 \sec ^3(c+d x)}{4 a d^2}-\frac{(e+f x)^3 \sec ^4(c+d x)}{4 a d}+\frac{f^3 \tan (c+d x)}{4 a d^4}+\frac{f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac{f^2 (e+f x) \sec (c+d x) \tan (c+d x)}{4 a d^3}+\frac{3 (e+f x)^3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{f (e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d^2}+\frac{(e+f x)^3 \sec ^3(c+d x) \tan (c+d x)}{4 a d}\\ \end{align*}

Mathematica [B]  time = 9.98643, size = 1901, normalized size = 2.72 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*d^2*e^3*x + 16*e*f^2*x + 6*d^2*e^2*f*x^2 + 8*f^3*x^2 + 4*d^2*e*f^2*x^3 + d^2*f^3*x^4 + (4*e*(d^2*e^2 +
4*f^2)*((-I)*d*x + Log[-Cos[c + d*x] - I*(-1 + Sin[c + d*x])])*(Cos[c] + I*(-1 + Sin[c])))/d + (4*f*(3*d^2*e^2
+ 4*f^2)*x*Log[1 - I*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] + I*(-1 + Sin[c])))/d + 12*d*e*f^2*x^2*Log[1 - I*Co
s[c + d*x] - Sin[c + d*x]]*(Cos[c] + I*(-1 + Sin[c])) + 4*d*f^3*x^3*Log[1 - I*Cos[c + d*x] - Sin[c + d*x]]*(Co
s[c] + I*(-1 + Sin[c])) + (24*e*f^2*(I*d*x*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x]] + PolyLog[3, I*Cos[c + d*
x] + Sin[c + d*x]])*(Cos[c] + I*(-1 + Sin[c])))/d + (12*f^3*(I*d^2*x^2*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x
]] + 2*d*x*PolyLog[3, I*Cos[c + d*x] + Sin[c + d*x]] - (2*I)*PolyLog[4, I*Cos[c + d*x] + Sin[c + d*x]])*(Cos[c
] + I*(-1 + Sin[c])))/d^2 + (4*f*(3*d^2*e^2 + 4*f^2)*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x]]*(1 + I*Cos[c] -
Sin[c]))/d^2))/(32*a*d^2*(Cos[c] + I*(-1 + Sin[c]))) - ((28*f^2 + 3*d^2*(e + f*x)^2)^2/f + 12*f*(9*d^2*e^2 +
28*f^2)*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(1 - I*Cos[c] + Sin[c]) + 216*d*e*f^2*(d*x*PolyLog[2, (-I
)*Cos[c + d*x] - Sin[c + d*x]] - I*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(1 - I*Cos[c] + Sin[c]) + 108
*f^3*(d^2*x^2*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]] - (2*I)*d*x*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c +
d*x]] - 2*PolyLog[4, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(1 - I*Cos[c] + Sin[c]) - 12*d*f*(9*d^2*e^2 + 28*f^2)*
x*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] + I*(1 + Sin[c])) - 108*d^3*e*f^2*x^2*Log[1 + I*Cos[c + d*x]
+ Sin[c + d*x]]*(Cos[c] + I*(1 + Sin[c])) - 36*d^3*f^3*x^3*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] + I*
(1 + Sin[c])) + (12*I)*d*e*(3*d^2*e^2 + 28*f^2)*(d*x + I*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])])*(Cos[c] + I
*(1 + Sin[c])))/(96*a*d^4*(Cos[c] + I*(1 + Sin[c]))) + ((3*e^3*x*Cos[c])/(4*a) + (((3*I)/4)*e^3*x*Sin[c])/a)/(
1 + Cos[2*c] + I*Sin[2*c]) + ((9*e^2*f*x^2*Cos[c])/(8*a) + (((9*I)/8)*e^2*f*x^2*Sin[c])/a)/(1 + Cos[2*c] + I*S
in[2*c]) + ((3*e*f^2*x^3*Cos[c])/(4*a) + (((3*I)/4)*e*f^2*x^3*Sin[c])/a)/(1 + Cos[2*c] + I*Sin[2*c]) + ((3*f^3
*x^4*Cos[c])/(16*a) + (((3*I)/16)*f^3*x^4*Sin[c])/a)/(1 + Cos[2*c] + I*Sin[2*c]) + (e^3 + 3*e^2*f*x + 3*e*f^2*
x^2 + f^3*x^3)/(8*a*d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) - (3*(e^2*f*Sin[(d*x)/2] + 2*e*f^2*x*Sin[(d
*x)/2] + f^3*x^2*Sin[(d*x)/2]))/(4*a*d^2*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (-
e^3 - 3*e^2*f*x - 3*e*f^2*x^2 - f^3*x^3)/(8*a*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) + (e^2*f*Sin[(d*x
)/2] + 2*e*f^2*x*Sin[(d*x)/2] + f^3*x^2*Sin[(d*x)/2])/(4*a*d^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin
[c/2 + (d*x)/2])^3) + (-2*d^2*e^3*Cos[c/2] - d*e^2*f*Cos[c/2] - 2*e*f^2*Cos[c/2] - 6*d^2*e^2*f*x*Cos[c/2] - 2*
d*e*f^2*x*Cos[c/2] - 2*f^3*x*Cos[c/2] - 6*d^2*e*f^2*x^2*Cos[c/2] - d*f^3*x^2*Cos[c/2] - 2*d^2*f^3*x^3*Cos[c/2]
- 2*d^2*e^3*Sin[c/2] + d*e^2*f*Sin[c/2] - 2*e*f^2*Sin[c/2] - 6*d^2*e^2*f*x*Sin[c/2] + 2*d*e*f^2*x*Sin[c/2] -
2*f^3*x*Sin[c/2] - 6*d^2*e*f^2*x^2*Sin[c/2] + d*f^3*x^2*Sin[c/2] - 2*d^2*f^3*x^3*Sin[c/2])/(8*a*d^3*(Cos[c/2]
+ Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (7*d^2*e^2*f*Sin[(d*x)/2] + 2*f^3*Sin[(d*x)/2] + 14
*d^2*e*f^2*x*Sin[(d*x)/2] + 7*d^2*f^3*x^2*Sin[(d*x)/2])/(4*a*d^4*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + S
in[c/2 + (d*x)/2]))

________________________________________________________________________________________

Maple [B]  time = 0.325, size = 2161, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/8/d/a*ln(1+I*exp(I*(d*x+c)))*e^2*f*x-9/8/d^2/a*ln(1+I*exp(I*(d*x+c)))*c*e^2*f-3/8/d^4/a*f^3*ln(1+I*exp(I*(d
*x+c)))*c^3-9/8/d^3/a*e*f^2*c^2*ln(exp(I*(d*x+c))-I)+9/8/d^2/a*e^2*f*c*ln(exp(I*(d*x+c))-I)-3/8/d/a*f^3*ln(1+I
*exp(I*(d*x+c)))*x^3-9/4*I*f^3*polylog(4,-I*exp(I*(d*x+c)))/a/d^4-3/8/d/a*e^3*ln(exp(I*(d*x+c))-I)+7/2/d^3/a*e
*f^2*ln(exp(I*(d*x+c))+I)-9/4/d^3/a*e*f^2*polylog(3,-I*exp(I*(d*x+c)))-2/d^3/a*e*f^2*ln(exp(I*(d*x+c)))+3/8/d^
4/a*f^3*c^3*ln(exp(I*(d*x+c))-I)+2/d^4/a*f^3*c*ln(exp(I*(d*x+c)))-7/2/d^4/a*f^3*c*ln(exp(I*(d*x+c))+I)-9/4/d^3
/a*f^3*polylog(3,-I*exp(I*(d*x+c)))*x+7/2/d^3/a*f^3*ln(1-I*exp(I*(d*x+c)))*x+7/2/d^4/a*f^3*ln(1-I*exp(I*(d*x+c
)))*c+9/8/d^3/a*ln(1+I*exp(I*(d*x+c)))*c^2*e*f^2-9/8/d/a*ln(1+I*exp(I*(d*x+c)))*e*f^2*x^2+3/8/a/d*ln(exp(I*(d*
x+c))+I)*e^3+3/2/d^4/a*f^3*c*ln(exp(I*(d*x+c))-I)-3/2/d^3/a*f^3*ln(1+I*exp(I*(d*x+c)))*x-3/2/d^3/a*e*f^2*ln(ex
p(I*(d*x+c))-I)-3/2/d^4/a*f^3*c*ln(1+I*exp(I*(d*x+c)))-I/d^4/a*f^3*c^2-I/d^2/a*f^3*x^2+3/2*I/d^4/a*f^3*polylog
(2,-I*exp(I*(d*x+c)))-7/2*I/d^4/a*f^3*polylog(2,I*exp(I*(d*x+c)))-3/8/a/d^4*f^3*c^3*ln(exp(I*(d*x+c))+I)+9/4/a
/d^3*e*f^2*polylog(3,I*exp(I*(d*x+c)))+9/4/a/d^3*f^3*polylog(3,I*exp(I*(d*x+c)))*x+3/8/a/d^4*f^3*c^3*ln(1-I*ex
p(I*(d*x+c)))+3/8/a/d*f^3*ln(1-I*exp(I*(d*x+c)))*x^3-1/4*I*(-9*I*d^2*e^2*f*exp(5*I*(d*x+c))-6*I*d^3*f^3*x^3*ex
p(2*I*(d*x+c))-8*I*d^2*f^3*x^2*exp(3*I*(d*x+c))-8*I*d^2*e^2*f*exp(3*I*(d*x+c))+6*d^3*e^2*f*x*exp(3*I*(d*x+c))+
44*d^2*e*f^2*x*exp(2*I*(d*x+c))+36*d^2*e*f^2*x*exp(4*I*(d*x+c))+9*d^3*e*f^2*x^2*exp(5*I*(d*x+c))+9*d^3*e^2*f*x
*exp(5*I*(d*x+c))+2*f^3+2*d*f^3*x*exp(I*(d*x+c))+2*d*e*f^2*exp(I*(d*x+c))+3*d^3*f^3*x^3*exp(I*(d*x+c))-4*I*f^3
*exp(3*I*(d*x+c))-2*I*f^3*exp(5*I*(d*x+c))+2*d^3*f^3*x^3*exp(3*I*(d*x+c))+4*d^2*e^2*f+3*d^3*e^3*exp(5*I*(d*x+c
))+2*d^3*e^3*exp(3*I*(d*x+c))+3*d^3*e^3*exp(I*(d*x+c))-2*I*f^3*exp(I*(d*x+c))-18*I*d^3*e*f^2*x^2*exp(2*I*(d*x+
c))-18*I*d^3*e^2*f*x*exp(2*I*(d*x+c))-16*I*d^2*e*f^2*x*exp(3*I*(d*x+c))-18*I*d^2*e*f^2*x*exp(5*I*(d*x+c))+4*d^
2*f^3*x^2+8*d^2*e*f^2*x+6*I*d^3*f^3*x^3*exp(4*I*(d*x+c))-9*I*d^2*f^3*x^2*exp(5*I*(d*x+c))+2*f^3*exp(4*I*(d*x+c
))+4*f^3*exp(2*I*(d*x+c))+2*I*d^2*e*f^2*x*exp(I*(d*x+c))+4*d*f^3*x*exp(3*I*(d*x+c))+4*d*e*f^2*exp(3*I*(d*x+c))
+18*I*d^3*e*f^2*x^2*exp(4*I*(d*x+c))+18*I*d^3*e^2*f*x*exp(4*I*(d*x+c))+22*d^2*e^2*f*exp(2*I*(d*x+c))+22*d^2*f^
3*x^2*exp(2*I*(d*x+c))+18*d^2*f^3*x^2*exp(4*I*(d*x+c))+18*d^2*e^2*f*exp(4*I*(d*x+c))+2*d*f^3*x*exp(5*I*(d*x+c)
)+2*d*e*f^2*exp(5*I*(d*x+c))+3*d^3*f^3*x^3*exp(5*I*(d*x+c))-6*I*d^3*e^3*exp(2*I*(d*x+c))+6*I*d^3*e^3*exp(4*I*(
d*x+c))+6*d^3*e*f^2*x^2*exp(3*I*(d*x+c))+9*d^3*e^2*f*x*exp(I*(d*x+c))+9*d^3*e*f^2*x^2*exp(I*(d*x+c))+I*d^2*f^3
*x^2*exp(I*(d*x+c))+I*d^2*e^2*f*exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)^4/d^4/(exp(I*(d*x+c))-I)^2/a+9/4*I/d^2/a*po
lylog(2,-I*exp(I*(d*x+c)))*e*f^2*x-9/4*I/d^2/a*polylog(2,I*exp(I*(d*x+c)))*e*f^2*x+9/4*I*f^3*polylog(4,I*exp(I
*(d*x+c)))/a/d^4+9/8/a/d*e*f^2*ln(1-I*exp(I*(d*x+c)))*x^2-9/8/a/d^3*e*f^2*c^2*ln(1-I*exp(I*(d*x+c)))+9/8/a/d*e
^2*f*ln(1-I*exp(I*(d*x+c)))*x+9/8/a/d^2*e^2*f*ln(1-I*exp(I*(d*x+c)))*c-9/8/a/d^2*e^2*f*c*ln(exp(I*(d*x+c))+I)+
9/8/a/d^3*e*f^2*c^2*ln(exp(I*(d*x+c))+I)-9/8*I/d^2/a*f^3*polylog(2,I*exp(I*(d*x+c)))*x^2+9/8*I/d^2/a*f^3*polyl
og(2,-I*exp(I*(d*x+c)))*x^2+9/8*I/d^2/a*e^2*f*polylog(2,-I*exp(I*(d*x+c)))-9/8*I/d^2/a*e^2*f*polylog(2,I*exp(I
*(d*x+c)))-2*I/d^3/a*f^3*c*x

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [C]  time = 4.69846, size = 6093, normalized size = 8.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*d^3*f^3*x^3 + 6*d^3*e*f^2*x^2 + 6*d^3*e^2*f*x + 2*d^3*e^3 - 4*(2*d^2*f^3*x^2 + 4*d^2*e*f^2*x + 2*d^2*e
^2*f + f^3)*cos(d*x + c)^3 - 2*(3*d^3*f^3*x^3 + 9*d^3*e*f^2*x^2 + 3*d^3*e^3 + 2*d*e*f^2 + (9*d^3*e^2*f + 2*d*f
^3)*x)*cos(d*x + c)^2 - 14*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*cos(d*x + c) + ((-9*I*d^2*f^3*x^2 - 18*I*
d^2*e*f^2*x - 9*I*d^2*e^2*f - 12*I*f^3)*cos(d*x + c)^2*sin(d*x + c) + (-9*I*d^2*f^3*x^2 - 18*I*d^2*e*f^2*x - 9
*I*d^2*e^2*f - 12*I*f^3)*cos(d*x + c)^2)*dilog(I*cos(d*x + c) + sin(d*x + c)) + ((-9*I*d^2*f^3*x^2 - 18*I*d^2*
e*f^2*x - 9*I*d^2*e^2*f - 28*I*f^3)*cos(d*x + c)^2*sin(d*x + c) + (-9*I*d^2*f^3*x^2 - 18*I*d^2*e*f^2*x - 9*I*d
^2*e^2*f - 28*I*f^3)*cos(d*x + c)^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) + ((9*I*d^2*f^3*x^2 + 18*I*d^2*e*f^2
*x + 9*I*d^2*e^2*f + 12*I*f^3)*cos(d*x + c)^2*sin(d*x + c) + (9*I*d^2*f^3*x^2 + 18*I*d^2*e*f^2*x + 9*I*d^2*e^2
*f + 12*I*f^3)*cos(d*x + c)^2)*dilog(-I*cos(d*x + c) + sin(d*x + c)) + ((9*I*d^2*f^3*x^2 + 18*I*d^2*e*f^2*x +
9*I*d^2*e^2*f + 28*I*f^3)*cos(d*x + c)^2*sin(d*x + c) + (9*I*d^2*f^3*x^2 + 18*I*d^2*e*f^2*x + 9*I*d^2*e^2*f +
28*I*f^3)*cos(d*x + c)^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + ((3*d^3*e^3 - 9*c*d^2*e^2*f + (9*c^2 + 28)*d
*e*f^2 - (3*c^3 + 28*c)*f^3)*cos(d*x + c)^2*sin(d*x + c) + (3*d^3*e^3 - 9*c*d^2*e^2*f + (9*c^2 + 28)*d*e*f^2 -
(3*c^3 + 28*c)*f^3)*cos(d*x + c)^2)*log(cos(d*x + c) + I*sin(d*x + c) + I) - 3*((d^3*e^3 - 3*c*d^2*e^2*f + (3
*c^2 + 4)*d*e*f^2 - (c^3 + 4*c)*f^3)*cos(d*x + c)^2*sin(d*x + c) + (d^3*e^3 - 3*c*d^2*e^2*f + (3*c^2 + 4)*d*e*
f^2 - (c^3 + 4*c)*f^3)*cos(d*x + c)^2)*log(cos(d*x + c) - I*sin(d*x + c) + I) + ((3*d^3*f^3*x^3 + 9*d^3*e*f^2*
x^2 + 9*c*d^2*e^2*f - 9*c^2*d*e*f^2 + (3*c^3 + 28*c)*f^3 + (9*d^3*e^2*f + 28*d*f^3)*x)*cos(d*x + c)^2*sin(d*x
+ c) + (3*d^3*f^3*x^3 + 9*d^3*e*f^2*x^2 + 9*c*d^2*e^2*f - 9*c^2*d*e*f^2 + (3*c^3 + 28*c)*f^3 + (9*d^3*e^2*f +
28*d*f^3)*x)*cos(d*x + c)^2)*log(I*cos(d*x + c) + sin(d*x + c) + 1) - 3*((d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*c*
d^2*e^2*f - 3*c^2*d*e*f^2 + (c^3 + 4*c)*f^3 + (3*d^3*e^2*f + 4*d*f^3)*x)*cos(d*x + c)^2*sin(d*x + c) + (d^3*f^
3*x^3 + 3*d^3*e*f^2*x^2 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + (c^3 + 4*c)*f^3 + (3*d^3*e^2*f + 4*d*f^3)*x)*cos(d*x
+ c)^2)*log(I*cos(d*x + c) - sin(d*x + c) + 1) + ((3*d^3*f^3*x^3 + 9*d^3*e*f^2*x^2 + 9*c*d^2*e^2*f - 9*c^2*d*
e*f^2 + (3*c^3 + 28*c)*f^3 + (9*d^3*e^2*f + 28*d*f^3)*x)*cos(d*x + c)^2*sin(d*x + c) + (3*d^3*f^3*x^3 + 9*d^3*
e*f^2*x^2 + 9*c*d^2*e^2*f - 9*c^2*d*e*f^2 + (3*c^3 + 28*c)*f^3 + (9*d^3*e^2*f + 28*d*f^3)*x)*cos(d*x + c)^2)*l
og(-I*cos(d*x + c) + sin(d*x + c) + 1) - 3*((d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + (
c^3 + 4*c)*f^3 + (3*d^3*e^2*f + 4*d*f^3)*x)*cos(d*x + c)^2*sin(d*x + c) + (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*c
*d^2*e^2*f - 3*c^2*d*e*f^2 + (c^3 + 4*c)*f^3 + (3*d^3*e^2*f + 4*d*f^3)*x)*cos(d*x + c)^2)*log(-I*cos(d*x + c)
- sin(d*x + c) + 1) + ((3*d^3*e^3 - 9*c*d^2*e^2*f + (9*c^2 + 28)*d*e*f^2 - (3*c^3 + 28*c)*f^3)*cos(d*x + c)^2*
sin(d*x + c) + (3*d^3*e^3 - 9*c*d^2*e^2*f + (9*c^2 + 28)*d*e*f^2 - (3*c^3 + 28*c)*f^3)*cos(d*x + c)^2)*log(-co
s(d*x + c) + I*sin(d*x + c) + I) - 3*((d^3*e^3 - 3*c*d^2*e^2*f + (3*c^2 + 4)*d*e*f^2 - (c^3 + 4*c)*f^3)*cos(d*
x + c)^2*sin(d*x + c) + (d^3*e^3 - 3*c*d^2*e^2*f + (3*c^2 + 4)*d*e*f^2 - (c^3 + 4*c)*f^3)*cos(d*x + c)^2)*log(
-cos(d*x + c) - I*sin(d*x + c) + I) + (18*I*f^3*cos(d*x + c)^2*sin(d*x + c) + 18*I*f^3*cos(d*x + c)^2)*polylog
(4, I*cos(d*x + c) + sin(d*x + c)) + (18*I*f^3*cos(d*x + c)^2*sin(d*x + c) + 18*I*f^3*cos(d*x + c)^2)*polylog(
4, I*cos(d*x + c) - sin(d*x + c)) + (-18*I*f^3*cos(d*x + c)^2*sin(d*x + c) - 18*I*f^3*cos(d*x + c)^2)*polylog(
4, -I*cos(d*x + c) + sin(d*x + c)) + (-18*I*f^3*cos(d*x + c)^2*sin(d*x + c) - 18*I*f^3*cos(d*x + c)^2)*polylog
(4, -I*cos(d*x + c) - sin(d*x + c)) - 18*((d*f^3*x + d*e*f^2)*cos(d*x + c)^2*sin(d*x + c) + (d*f^3*x + d*e*f^2
)*cos(d*x + c)^2)*polylog(3, I*cos(d*x + c) + sin(d*x + c)) + 18*((d*f^3*x + d*e*f^2)*cos(d*x + c)^2*sin(d*x +
c) + (d*f^3*x + d*e*f^2)*cos(d*x + c)^2)*polylog(3, I*cos(d*x + c) - sin(d*x + c)) - 18*((d*f^3*x + d*e*f^2)*
cos(d*x + c)^2*sin(d*x + c) + (d*f^3*x + d*e*f^2)*cos(d*x + c)^2)*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) +
18*((d*f^3*x + d*e*f^2)*cos(d*x + c)^2*sin(d*x + c) + (d*f^3*x + d*e*f^2)*cos(d*x + c)^2)*polylog(3, -I*cos(d
*x + c) - sin(d*x + c)) + 2*(3*d^3*f^3*x^3 + 9*d^3*e*f^2*x^2 + 9*d^3*e^2*f*x + 3*d^3*e^3 - 5*(d^2*f^3*x^2 + 2*
d^2*e*f^2*x + d^2*e^2*f)*cos(d*x + c))*sin(d*x + c))/(a*d^4*cos(d*x + c)^2*sin(d*x + c) + a*d^4*cos(d*x + c)^2
)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3} \sec \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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