3.29 \(\int (c+d x)^3 (a+b \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=364 \[ -\frac{12 a b d^2 (c+d x) \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{6 i a b d (c+d x)^2 \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{12 i a b d^3 \text{PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac{12 i a b d^3 \text{PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4}-\frac{3 i b^2 d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}+\frac{3 b^2 d^3 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^4}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}-\frac{i b^2 (c+d x)^3}{f} \]

[Out]

((-I)*b^2*(c + d*x)^3)/f + (a^2*(c + d*x)^4)/(4*d) - ((4*I)*a*b*(c + d*x)^3*ArcTan[E^(I*(e + f*x))])/f + (3*b^
2*d*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f^2 + ((6*I)*a*b*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x))])
/f^2 - ((6*I)*a*b*d*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - ((3*I)*b^2*d^2*(c + d*x)*PolyLog[2, -E^((
2*I)*(e + f*x))])/f^3 - (12*a*b*d^2*(c + d*x)*PolyLog[3, (-I)*E^(I*(e + f*x))])/f^3 + (12*a*b*d^2*(c + d*x)*Po
lyLog[3, I*E^(I*(e + f*x))])/f^3 + (3*b^2*d^3*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^4) - ((12*I)*a*b*d^3*Poly
Log[4, (-I)*E^(I*(e + f*x))])/f^4 + ((12*I)*a*b*d^3*PolyLog[4, I*E^(I*(e + f*x))])/f^4 + (b^2*(c + d*x)^3*Tan[
e + f*x])/f

________________________________________________________________________________________

Rubi [A]  time = 0.454757, antiderivative size = 364, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {4190, 4181, 2531, 6609, 2282, 6589, 4184, 3719, 2190} \[ -\frac{12 a b d^2 (c+d x) \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{6 i a b d (c+d x)^2 \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{12 i a b d^3 \text{PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac{12 i a b d^3 \text{PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4}-\frac{3 i b^2 d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}+\frac{3 b^2 d^3 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^4}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}-\frac{i b^2 (c+d x)^3}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b^2*(c + d*x)^3)/f + (a^2*(c + d*x)^4)/(4*d) - ((4*I)*a*b*(c + d*x)^3*ArcTan[E^(I*(e + f*x))])/f + (3*b^
2*d*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f^2 + ((6*I)*a*b*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x))])
/f^2 - ((6*I)*a*b*d*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - ((3*I)*b^2*d^2*(c + d*x)*PolyLog[2, -E^((
2*I)*(e + f*x))])/f^3 - (12*a*b*d^2*(c + d*x)*PolyLog[3, (-I)*E^(I*(e + f*x))])/f^3 + (12*a*b*d^2*(c + d*x)*Po
lyLog[3, I*E^(I*(e + f*x))])/f^3 + (3*b^2*d^3*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^4) - ((12*I)*a*b*d^3*Poly
Log[4, (-I)*E^(I*(e + f*x))])/f^4 + ((12*I)*a*b*d^3*PolyLog[4, I*E^(I*(e + f*x))])/f^4 + (b^2*(c + d*x)^3*Tan[
e + f*x])/f

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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 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 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 (c+d x)^3 (a+b \sec (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)^3+2 a b (c+d x)^3 \sec (e+f x)+b^2 (c+d x)^3 \sec ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^4}{4 d}+(2 a b) \int (c+d x)^3 \sec (e+f x) \, dx+b^2 \int (c+d x)^3 \sec ^2(e+f x) \, dx\\ &=\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}-\frac{(6 a b d) \int (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac{(6 a b d) \int (c+d x)^2 \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}-\frac{\left (3 b^2 d\right ) \int (c+d x)^2 \tan (e+f x) \, dx}{f}\\ &=-\frac{i b^2 (c+d x)^3}{f}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}-\frac{\left (12 i a b d^2\right ) \int (c+d x) \text{Li}_2\left (-i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac{\left (12 i a b d^2\right ) \int (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac{\left (6 i b^2 d\right ) \int \frac{e^{2 i (e+f x)} (c+d x)^2}{1+e^{2 i (e+f x)}} \, dx}{f}\\ &=-\frac{i b^2 (c+d x)^3}{f}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{12 a b d^2 (c+d x) \text{Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}+\frac{\left (12 a b d^3\right ) \int \text{Li}_3\left (-i e^{i (e+f x)}\right ) \, dx}{f^3}-\frac{\left (12 a b d^3\right ) \int \text{Li}_3\left (i e^{i (e+f x)}\right ) \, dx}{f^3}-\frac{\left (6 b^2 d^2\right ) \int (c+d x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac{i b^2 (c+d x)^3}{f}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{3 i b^2 d^2 (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^3}-\frac{12 a b d^2 (c+d x) \text{Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}-\frac{\left (12 i a b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^4}+\frac{\left (12 i a b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^4}+\frac{\left (3 i b^2 d^3\right ) \int \text{Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^3}\\ &=-\frac{i b^2 (c+d x)^3}{f}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{3 i b^2 d^2 (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^3}-\frac{12 a b d^2 (c+d x) \text{Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3}-\frac{12 i a b d^3 \text{Li}_4\left (-i e^{i (e+f x)}\right )}{f^4}+\frac{12 i a b d^3 \text{Li}_4\left (i e^{i (e+f x)}\right )}{f^4}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}+\frac{\left (3 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^4}\\ &=-\frac{i b^2 (c+d x)^3}{f}+\frac{a^2 (c+d x)^4}{4 d}-\frac{4 i a b (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{3 b^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{6 i a b d (c+d x)^2 \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{3 i b^2 d^2 (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^3}-\frac{12 a b d^2 (c+d x) \text{Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac{12 a b d^2 (c+d x) \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3}+\frac{3 b^2 d^3 \text{Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^4}-\frac{12 i a b d^3 \text{Li}_4\left (-i e^{i (e+f x)}\right )}{f^4}+\frac{12 i a b d^3 \text{Li}_4\left (i e^{i (e+f x)}\right )}{f^4}+\frac{b^2 (c+d x)^3 \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 2.59635, size = 646, normalized size = 1.77 \[ \frac{-48 a b c d^2 f \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )+48 a b c d^2 f \text{PolyLog}\left (3,i e^{i (e+f x)}\right )+24 i a b d f^2 (c+d x)^2 \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )-24 i a b d f^2 (c+d x)^2 \text{PolyLog}\left (2,i e^{i (e+f x)}\right )-48 a b d^3 f x \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )+48 a b d^3 f x \text{PolyLog}\left (3,i e^{i (e+f x)}\right )-48 i a b d^3 \text{PolyLog}\left (4,-i e^{i (e+f x)}\right )+48 i a b d^3 \text{PolyLog}\left (4,i e^{i (e+f x)}\right )-12 i b^2 c d^2 f \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )-12 i b^2 d^3 f x \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )+6 b^2 d^3 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )+6 a^2 c^2 d f^4 x^2+4 a^2 c^3 f^4 x+4 a^2 c d^2 f^4 x^3+a^2 d^3 f^4 x^4-48 i a b c^2 d f^3 x \tan ^{-1}\left (e^{i (e+f x)}\right )+8 a b c^3 f^3 \tanh ^{-1}(\sin (e+f x))-48 i a b c d^2 f^3 x^2 \tan ^{-1}\left (e^{i (e+f x)}\right )-16 i a b d^3 f^3 x^3 \tan ^{-1}\left (e^{i (e+f x)}\right )+12 b^2 c^2 d f^3 x \tan (e+f x)+12 b^2 c^2 d f^2 \log (\cos (e+f x))+4 b^2 c^3 f^3 \tan (e+f x)+12 b^2 c d^2 f^3 x^2 \tan (e+f x)+24 b^2 c d^2 f^2 x \log \left (1+e^{2 i (e+f x)}\right )-12 i b^2 c d^2 f^3 x^2+12 b^2 d^3 f^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )+4 b^2 d^3 f^3 x^3 \tan (e+f x)-4 i b^2 d^3 f^3 x^3}{4 f^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a^2*c^3*f^4*x - (12*I)*b^2*c*d^2*f^3*x^2 + 6*a^2*c^2*d*f^4*x^2 - (4*I)*b^2*d^3*f^3*x^3 + 4*a^2*c*d^2*f^4*x^
3 + a^2*d^3*f^4*x^4 - (48*I)*a*b*c^2*d*f^3*x*ArcTan[E^(I*(e + f*x))] - (48*I)*a*b*c*d^2*f^3*x^2*ArcTan[E^(I*(e
+ f*x))] - (16*I)*a*b*d^3*f^3*x^3*ArcTan[E^(I*(e + f*x))] + 8*a*b*c^3*f^3*ArcTanh[Sin[e + f*x]] + 24*b^2*c*d^
2*f^2*x*Log[1 + E^((2*I)*(e + f*x))] + 12*b^2*d^3*f^2*x^2*Log[1 + E^((2*I)*(e + f*x))] + 12*b^2*c^2*d*f^2*Log[
Cos[e + f*x]] + (24*I)*a*b*d*f^2*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x))] - (24*I)*a*b*d*f^2*(c + d*x)^2*P
olyLog[2, I*E^(I*(e + f*x))] - (12*I)*b^2*c*d^2*f*PolyLog[2, -E^((2*I)*(e + f*x))] - (12*I)*b^2*d^3*f*x*PolyLo
g[2, -E^((2*I)*(e + f*x))] - 48*a*b*c*d^2*f*PolyLog[3, (-I)*E^(I*(e + f*x))] - 48*a*b*d^3*f*x*PolyLog[3, (-I)*
E^(I*(e + f*x))] + 48*a*b*c*d^2*f*PolyLog[3, I*E^(I*(e + f*x))] + 48*a*b*d^3*f*x*PolyLog[3, I*E^(I*(e + f*x))]
+ 6*b^2*d^3*PolyLog[3, -E^((2*I)*(e + f*x))] - (48*I)*a*b*d^3*PolyLog[4, (-I)*E^(I*(e + f*x))] + (48*I)*a*b*d
^3*PolyLog[4, I*E^(I*(e + f*x))] + 4*b^2*c^3*f^3*Tan[e + f*x] + 12*b^2*c^2*d*f^3*x*Tan[e + f*x] + 12*b^2*c*d^2
*f^3*x^2*Tan[e + f*x] + 4*b^2*d^3*f^3*x^3*Tan[e + f*x])/(4*f^4)

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Maple [B]  time = 0.297, size = 1478, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*c*d^2*x^3+3/2*a^2*c^2*d*x^2-4*I*b/f*a*c^3*arctan(exp(I*(f*x+e)))-6*I*b^2/f^4*d^3*polylog(2,I*exp(I*(f*x+e)
))*e-6*I*b^2/f^4*d^3*polylog(2,-I*exp(I*(f*x+e)))*e-6*I*b^2/f^3*d^3*polylog(2,-I*exp(I*(f*x+e)))*x+3*I*b^2/f^4
*d^3*e*polylog(2,-exp(2*I*(f*x+e)))-6*I*b^2/f^3*d^3*polylog(2,I*exp(I*(f*x+e)))*x-6*I*b^2/f*c*d^2*x^2-6*I*b^2/
f^3*c*d^2*e^2-3*I*b^2/f^3*c*d^2*polylog(2,-exp(2*I*(f*x+e)))+6*I*b^2/f^3*d^3*e^2*x-12*I*a*b*d^3*polylog(4,-I*e
xp(I*(f*x+e)))/f^4+12*I*b/f^2*a*c^2*d*e*arctan(exp(I*(f*x+e)))-12*I*b/f^3*a*c*d^2*e^2*arctan(exp(I*(f*x+e)))+1
2*I*b/f^2*a*c*d^2*polylog(2,-I*exp(I*(f*x+e)))*x-12*I*b/f^2*a*c*d^2*polylog(2,I*exp(I*(f*x+e)))*x+1/4*a^2*d^3*
x^4+a^2*c^3*x+6*b^2/f^4*d^3*polylog(3,-I*exp(I*(f*x+e)))+6*b^2/f^4*d^3*polylog(3,I*exp(I*(f*x+e)))+12*I*a*b*d^
3*polylog(4,I*exp(I*(f*x+e)))/f^4+6*b/f^3*a*c*d^2*e^2*ln(1+I*exp(I*(f*x+e)))-6*b/f*a*c^2*d*ln(1+I*exp(I*(f*x+e
)))*x-6*b/f^2*a*c^2*d*ln(1+I*exp(I*(f*x+e)))*e-6*b/f^3*a*c*d^2*e^2*ln(1-I*exp(I*(f*x+e)))+6*b/f^2*a*c^2*d*ln(1
-I*exp(I*(f*x+e)))*e+6*b/f*a*c^2*d*ln(1-I*exp(I*(f*x+e)))*x+6*b/f*a*c*d^2*ln(1-I*exp(I*(f*x+e)))*x^2-6*b/f*a*c
*d^2*ln(1+I*exp(I*(f*x+e)))*x^2+6*I*b/f^2*a*c^2*d*polylog(2,-I*exp(I*(f*x+e)))+6*I*b/f^2*a*d^3*polylog(2,-I*ex
p(I*(f*x+e)))*x^2-6*I*b/f^2*a*d^3*polylog(2,I*exp(I*(f*x+e)))*x^2-12*I*b^2/f^2*c*d^2*e*x-6*I*b/f^2*a*c^2*d*pol
ylog(2,I*exp(I*(f*x+e)))+4*I*b/f^4*a*d^3*e^3*arctan(exp(I*(f*x+e)))+2*I*b^2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3
)/f/(1+exp(2*I*(f*x+e)))-2*b/f*a*d^3*ln(1+I*exp(I*(f*x+e)))*x^3+12*b/f^3*a*d^3*polylog(3,I*exp(I*(f*x+e)))*x-6
*b^2/f^3*d^3*e*ln(1+exp(2*I*(f*x+e)))*x-12*b/f^3*a*d^3*polylog(3,-I*exp(I*(f*x+e)))*x+6*b^2/f^2*c*d^2*ln(1+exp
(2*I*(f*x+e)))*x+6*b^2/f^3*d^3*ln(1+I*exp(I*(f*x+e)))*e*x+6*b^2/f^3*d^3*ln(1-I*exp(I*(f*x+e)))*e*x+12*b^2/f^3*
c*d^2*e*ln(exp(I*(f*x+e)))+2*b/f*a*d^3*ln(1-I*exp(I*(f*x+e)))*x^3-2*b/f^4*a*d^3*e^3*ln(1+I*exp(I*(f*x+e)))-12*
b/f^3*a*c*d^2*polylog(3,-I*exp(I*(f*x+e)))+12*b/f^3*a*c*d^2*polylog(3,I*exp(I*(f*x+e)))+2*b/f^4*a*d^3*e^3*ln(1
-I*exp(I*(f*x+e)))-2*I*b^2/f*d^3*x^3+4*I*b^2/f^4*d^3*e^3+3*b^2/f^2*d^3*ln(1-I*exp(I*(f*x+e)))*x^2+3*b^2/f^4*d^
3*ln(1+I*exp(I*(f*x+e)))*e^2-3*b^2/f^4*d^3*e^2*ln(1+exp(2*I*(f*x+e)))-6*b^2/f^4*d^3*e^2*ln(exp(I*(f*x+e)))+3*b
^2/f^2*c^2*d*ln(1+exp(2*I*(f*x+e)))-6*b^2/f^2*c^2*d*ln(exp(I*(f*x+e)))+3*b^2/f^2*d^3*ln(1+I*exp(I*(f*x+e)))*x^
2+3*b^2/f^4*d^3*ln(1-I*exp(I*(f*x+e)))*e^2

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Maxima [B]  time = 3.6196, size = 4415, normalized size = 12.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(f*x + e)*a^2*c^3 + (f*x + e)^4*a^2*d^3/f^3 - 4*(f*x + e)^3*a^2*d^3*e/f^3 + 6*(f*x + e)^2*a^2*d^3*e^2/f
^3 - 4*(f*x + e)*a^2*d^3*e^3/f^3 + 4*(f*x + e)^3*a^2*c*d^2/f^2 - 12*(f*x + e)^2*a^2*c*d^2*e/f^2 + 12*(f*x + e)
*a^2*c*d^2*e^2/f^2 + 6*(f*x + e)^2*a^2*c^2*d/f - 12*(f*x + e)*a^2*c^2*d*e/f + 8*a*b*c^3*log(sec(f*x + e) + tan
(f*x + e)) - 8*a*b*d^3*e^3*log(sec(f*x + e) + tan(f*x + e))/f^3 + 24*a*b*c*d^2*e^2*log(sec(f*x + e) + tan(f*x
+ e))/f^2 - 24*a*b*c^2*d*e*log(sec(f*x + e) + tan(f*x + e))/f - 4*(4*b^2*d^3*e^3 - 12*b^2*c*d^2*e^2*f + 12*b^2
*c^2*d*e*f^2 - 4*b^2*c^3*f^3 + (4*(f*x + e)^3*a*b*d^3 - 12*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e)^2 + 12*(a*b*d^3
*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f*x + e) + 4*((f*x + e)^3*a*b*d^3 - 3*(a*b*d^3*e - a*b*c*d^2*f)*(f*x
+ e)^2 + 3*(a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) - (-4*I*(f*x + e)^3*a*b
*d^3 + (12*I*a*b*d^3*e - 12*I*a*b*c*d^2*f)*(f*x + e)^2 + (-12*I*a*b*d^3*e^2 + 24*I*a*b*c*d^2*e*f - 12*I*a*b*c^
2*d*f^2)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(cos(f*x + e), sin(f*x + e) + 1) + (4*(f*x + e)^3*a*b*d^3 - 12*(a
*b*d^3*e - a*b*c*d^2*f)*(f*x + e)^2 + 12*(a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f*x + e) + 4*((f*x +
e)^3*a*b*d^3 - 3*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e)^2 + 3*(a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f
*x + e))*cos(2*f*x + 2*e) - (-4*I*(f*x + e)^3*a*b*d^3 + (12*I*a*b*d^3*e - 12*I*a*b*c*d^2*f)*(f*x + e)^2 + (-12
*I*a*b*d^3*e^2 + 24*I*a*b*c*d^2*e*f - 12*I*a*b*c^2*d*f^2)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(cos(f*x + e), -
sin(f*x + e) + 1) - (6*(f*x + e)^2*b^2*d^3 + 6*b^2*d^3*e^2 - 12*b^2*c*d^2*e*f + 6*b^2*c^2*d*f^2 - 12*(b^2*d^3*
e - b^2*c*d^2*f)*(f*x + e) + 6*((f*x + e)^2*b^2*d^3 + b^2*d^3*e^2 - 2*b^2*c*d^2*e*f + b^2*c^2*d*f^2 - 2*(b^2*d
^3*e - b^2*c*d^2*f)*(f*x + e))*cos(2*f*x + 2*e) + (6*I*(f*x + e)^2*b^2*d^3 + 6*I*b^2*d^3*e^2 - 12*I*b^2*c*d^2*
e*f + 6*I*b^2*c^2*d*f^2 + (-12*I*b^2*d^3*e + 12*I*b^2*c*d^2*f)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e) + 1) + 4*((f*x + e)^3*b^2*d^3 - 3*(b^2*d^3*e - b^2*c*d^2*f)*(f*x + e)^2 + 3*(b^2*d^3*
e^2 - 2*b^2*c*d^2*e*f + b^2*c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) + (6*(f*x + e)*b^2*d^3 - 6*b^2*d^3*e + 6*b^
2*c*d^2*f + 6*((f*x + e)*b^2*d^3 - b^2*d^3*e + b^2*c*d^2*f)*cos(2*f*x + 2*e) - (-6*I*(f*x + e)*b^2*d^3 + 6*I*b
^2*d^3*e - 6*I*b^2*c*d^2*f)*sin(2*f*x + 2*e))*dilog(-e^(2*I*f*x + 2*I*e)) + (12*(f*x + e)^2*a*b*d^3 + 12*a*b*d
^3*e^2 - 24*a*b*c*d^2*e*f + 12*a*b*c^2*d*f^2 - 24*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e) + 12*((f*x + e)^2*a*b*d^
3 + a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2 - 2*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e))*cos(2*f*x + 2*e) -
(-12*I*(f*x + e)^2*a*b*d^3 - 12*I*a*b*d^3*e^2 + 24*I*a*b*c*d^2*e*f - 12*I*a*b*c^2*d*f^2 + (24*I*a*b*d^3*e - 24
*I*a*b*c*d^2*f)*(f*x + e))*sin(2*f*x + 2*e))*dilog(I*e^(I*f*x + I*e)) - (12*(f*x + e)^2*a*b*d^3 + 12*a*b*d^3*e
^2 - 24*a*b*c*d^2*e*f + 12*a*b*c^2*d*f^2 - 24*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e) + 12*((f*x + e)^2*a*b*d^3 +
a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2 - 2*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e))*cos(2*f*x + 2*e) + (12*
I*(f*x + e)^2*a*b*d^3 + 12*I*a*b*d^3*e^2 - 24*I*a*b*c*d^2*e*f + 12*I*a*b*c^2*d*f^2 + (-24*I*a*b*d^3*e + 24*I*a
*b*c*d^2*f)*(f*x + e))*sin(2*f*x + 2*e))*dilog(-I*e^(I*f*x + I*e)) - (-3*I*(f*x + e)^2*b^2*d^3 - 3*I*b^2*d^3*e
^2 + 6*I*b^2*c*d^2*e*f - 3*I*b^2*c^2*d*f^2 + (6*I*b^2*d^3*e - 6*I*b^2*c*d^2*f)*(f*x + e) + (-3*I*(f*x + e)^2*b
^2*d^3 - 3*I*b^2*d^3*e^2 + 6*I*b^2*c*d^2*e*f - 3*I*b^2*c^2*d*f^2 + (6*I*b^2*d^3*e - 6*I*b^2*c*d^2*f)*(f*x + e)
)*cos(2*f*x + 2*e) + 3*((f*x + e)^2*b^2*d^3 + b^2*d^3*e^2 - 2*b^2*c*d^2*e*f + b^2*c^2*d*f^2 - 2*(b^2*d^3*e - b
^2*c*d^2*f)*(f*x + e))*sin(2*f*x + 2*e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)
- (-2*I*(f*x + e)^3*a*b*d^3 + (6*I*a*b*d^3*e - 6*I*a*b*c*d^2*f)*(f*x + e)^2 + (-6*I*a*b*d^3*e^2 + 12*I*a*b*c*
d^2*e*f - 6*I*a*b*c^2*d*f^2)*(f*x + e) + (-2*I*(f*x + e)^3*a*b*d^3 + (6*I*a*b*d^3*e - 6*I*a*b*c*d^2*f)*(f*x +
e)^2 + (-6*I*a*b*d^3*e^2 + 12*I*a*b*c*d^2*e*f - 6*I*a*b*c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) + 2*((f*x + e)^
3*a*b*d^3 - 3*(a*b*d^3*e - a*b*c*d^2*f)*(f*x + e)^2 + 3*(a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f*x +
e))*sin(2*f*x + 2*e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - (2*I*(f*x + e)^3*a*b*d^3 +
(-6*I*a*b*d^3*e + 6*I*a*b*c*d^2*f)*(f*x + e)^2 + (6*I*a*b*d^3*e^2 - 12*I*a*b*c*d^2*e*f + 6*I*a*b*c^2*d*f^2)*(f
*x + e) + (2*I*(f*x + e)^3*a*b*d^3 + (-6*I*a*b*d^3*e + 6*I*a*b*c*d^2*f)*(f*x + e)^2 + (6*I*a*b*d^3*e^2 - 12*I*
a*b*c*d^2*e*f + 6*I*a*b*c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) - 2*((f*x + e)^3*a*b*d^3 - 3*(a*b*d^3*e - a*b*c
*d^2*f)*(f*x + e)^2 + 3*(a*b*d^3*e^2 - 2*a*b*c*d^2*e*f + a*b*c^2*d*f^2)*(f*x + e))*sin(2*f*x + 2*e))*log(cos(f
*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - (24*a*b*d^3*cos(2*f*x + 2*e) + 24*I*a*b*d^3*sin(2*f*x + 2*e
) + 24*a*b*d^3)*polylog(4, I*e^(I*f*x + I*e)) + (24*a*b*d^3*cos(2*f*x + 2*e) + 24*I*a*b*d^3*sin(2*f*x + 2*e) +
24*a*b*d^3)*polylog(4, -I*e^(I*f*x + I*e)) - (-3*I*b^2*d^3*cos(2*f*x + 2*e) + 3*b^2*d^3*sin(2*f*x + 2*e) - 3*
I*b^2*d^3)*polylog(3, -e^(2*I*f*x + 2*I*e)) - (-24*I*(f*x + e)*a*b*d^3 + 24*I*a*b*d^3*e - 24*I*a*b*c*d^2*f + (
-24*I*(f*x + e)*a*b*d^3 + 24*I*a*b*d^3*e - 24*I*a*b*c*d^2*f)*cos(2*f*x + 2*e) + 24*((f*x + e)*a*b*d^3 - a*b*d^
3*e + a*b*c*d^2*f)*sin(2*f*x + 2*e))*polylog(3, I*e^(I*f*x + I*e)) - (24*I*(f*x + e)*a*b*d^3 - 24*I*a*b*d^3*e
+ 24*I*a*b*c*d^2*f + (24*I*(f*x + e)*a*b*d^3 - 24*I*a*b*d^3*e + 24*I*a*b*c*d^2*f)*cos(2*f*x + 2*e) - 24*((f*x
+ e)*a*b*d^3 - a*b*d^3*e + a*b*c*d^2*f)*sin(2*f*x + 2*e))*polylog(3, -I*e^(I*f*x + I*e)) - (-4*I*(f*x + e)^3*b
^2*d^3 + (12*I*b^2*d^3*e - 12*I*b^2*c*d^2*f)*(f*x + e)^2 + (-12*I*b^2*d^3*e^2 + 24*I*b^2*c*d^2*e*f - 12*I*b^2*
c^2*d*f^2)*(f*x + e))*sin(2*f*x + 2*e))/(-2*I*f^3*cos(2*f*x + 2*e) + 2*f^3*sin(2*f*x + 2*e) - 2*I*f^3))/f

________________________________________________________________________________________

Fricas [C]  time = 3.2186, size = 4317, normalized size = 11.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(24*I*a*b*d^3*cos(f*x + e)*polylog(4, I*cos(f*x + e) + sin(f*x + e)) + 24*I*a*b*d^3*cos(f*x + e)*polylog(4
, I*cos(f*x + e) - sin(f*x + e)) - 24*I*a*b*d^3*cos(f*x + e)*polylog(4, -I*cos(f*x + e) + sin(f*x + e)) - 24*I
*a*b*d^3*cos(f*x + e)*polylog(4, -I*cos(f*x + e) - sin(f*x + e)) + (-12*I*a*b*d^3*f^2*x^2 - 12*I*a*b*c^2*d*f^2
+ 12*I*b^2*c*d^2*f - 12*I*(2*a*b*c*d^2*f^2 - b^2*d^3*f)*x)*cos(f*x + e)*dilog(I*cos(f*x + e) + sin(f*x + e))
+ (-12*I*a*b*d^3*f^2*x^2 - 12*I*a*b*c^2*d*f^2 - 12*I*b^2*c*d^2*f - 12*I*(2*a*b*c*d^2*f^2 + b^2*d^3*f)*x)*cos(f
*x + e)*dilog(I*cos(f*x + e) - sin(f*x + e)) + (12*I*a*b*d^3*f^2*x^2 + 12*I*a*b*c^2*d*f^2 - 12*I*b^2*c*d^2*f +
12*I*(2*a*b*c*d^2*f^2 - b^2*d^3*f)*x)*cos(f*x + e)*dilog(-I*cos(f*x + e) + sin(f*x + e)) + (12*I*a*b*d^3*f^2*
x^2 + 12*I*a*b*c^2*d*f^2 + 12*I*b^2*c*d^2*f + 12*I*(2*a*b*c*d^2*f^2 + b^2*d^3*f)*x)*cos(f*x + e)*dilog(-I*cos(
f*x + e) - sin(f*x + e)) - 2*(2*a*b*d^3*e^3 - 2*a*b*c^3*f^3 - 3*b^2*d^3*e^2 + 3*(2*a*b*c^2*d*e - b^2*c^2*d)*f^
2 - 6*(a*b*c*d^2*e^2 - b^2*c*d^2*e)*f)*cos(f*x + e)*log(cos(f*x + e) + I*sin(f*x + e) + I) + 2*(2*a*b*d^3*e^3
- 2*a*b*c^3*f^3 + 3*b^2*d^3*e^2 + 3*(2*a*b*c^2*d*e + b^2*c^2*d)*f^2 - 6*(a*b*c*d^2*e^2 + b^2*c*d^2*e)*f)*cos(f
*x + e)*log(cos(f*x + e) - I*sin(f*x + e) + I) + 2*(2*a*b*d^3*f^3*x^3 + 2*a*b*d^3*e^3 + 6*a*b*c^2*d*e*f^2 - 3*
b^2*d^3*e^2 + 3*(2*a*b*c*d^2*f^3 + b^2*d^3*f^2)*x^2 - 6*(a*b*c*d^2*e^2 - b^2*c*d^2*e)*f + 6*(a*b*c^2*d*f^3 + b
^2*c*d^2*f^2)*x)*cos(f*x + e)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(2*a*b*d^3*f^3*x^3 + 2*a*b*d^3*e^3 +
6*a*b*c^2*d*e*f^2 + 3*b^2*d^3*e^2 + 3*(2*a*b*c*d^2*f^3 - b^2*d^3*f^2)*x^2 - 6*(a*b*c*d^2*e^2 + b^2*c*d^2*e)*f
+ 6*(a*b*c^2*d*f^3 - b^2*c*d^2*f^2)*x)*cos(f*x + e)*log(I*cos(f*x + e) - sin(f*x + e) + 1) + 2*(2*a*b*d^3*f^3*
x^3 + 2*a*b*d^3*e^3 + 6*a*b*c^2*d*e*f^2 - 3*b^2*d^3*e^2 + 3*(2*a*b*c*d^2*f^3 + b^2*d^3*f^2)*x^2 - 6*(a*b*c*d^2
*e^2 - b^2*c*d^2*e)*f + 6*(a*b*c^2*d*f^3 + b^2*c*d^2*f^2)*x)*cos(f*x + e)*log(-I*cos(f*x + e) + sin(f*x + e) +
1) - 2*(2*a*b*d^3*f^3*x^3 + 2*a*b*d^3*e^3 + 6*a*b*c^2*d*e*f^2 + 3*b^2*d^3*e^2 + 3*(2*a*b*c*d^2*f^3 - b^2*d^3*
f^2)*x^2 - 6*(a*b*c*d^2*e^2 + b^2*c*d^2*e)*f + 6*(a*b*c^2*d*f^3 - b^2*c*d^2*f^2)*x)*cos(f*x + e)*log(-I*cos(f*
x + e) - sin(f*x + e) + 1) - 2*(2*a*b*d^3*e^3 - 2*a*b*c^3*f^3 - 3*b^2*d^3*e^2 + 3*(2*a*b*c^2*d*e - b^2*c^2*d)*
f^2 - 6*(a*b*c*d^2*e^2 - b^2*c*d^2*e)*f)*cos(f*x + e)*log(-cos(f*x + e) + I*sin(f*x + e) + I) + 2*(2*a*b*d^3*e
^3 - 2*a*b*c^3*f^3 + 3*b^2*d^3*e^2 + 3*(2*a*b*c^2*d*e + b^2*c^2*d)*f^2 - 6*(a*b*c*d^2*e^2 + b^2*c*d^2*e)*f)*co
s(f*x + e)*log(-cos(f*x + e) - I*sin(f*x + e) + I) - 12*(2*a*b*d^3*f*x + 2*a*b*c*d^2*f - b^2*d^3)*cos(f*x + e)
*polylog(3, I*cos(f*x + e) + sin(f*x + e)) + 12*(2*a*b*d^3*f*x + 2*a*b*c*d^2*f + b^2*d^3)*cos(f*x + e)*polylog
(3, I*cos(f*x + e) - sin(f*x + e)) - 12*(2*a*b*d^3*f*x + 2*a*b*c*d^2*f - b^2*d^3)*cos(f*x + e)*polylog(3, -I*c
os(f*x + e) + sin(f*x + e)) + 12*(2*a*b*d^3*f*x + 2*a*b*c*d^2*f + b^2*d^3)*cos(f*x + e)*polylog(3, -I*cos(f*x
+ e) - sin(f*x + e)) + (a^2*d^3*f^4*x^4 + 4*a^2*c*d^2*f^4*x^3 + 6*a^2*c^2*d*f^4*x^2 + 4*a^2*c^3*f^4*x)*cos(f*x
+ e) + 4*(b^2*d^3*f^3*x^3 + 3*b^2*c*d^2*f^3*x^2 + 3*b^2*c^2*d*f^3*x + b^2*c^3*f^3)*sin(f*x + e))/(f^4*cos(f*x
+ e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3}{\left (b \sec \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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