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

Optimal. Leaf size=227 \[ -\frac {6 a d^2 (c+d x) \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {6 a d^2 (c+d x) \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3}+\frac {3 i a d (c+d x)^2 \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i a d (c+d x)^2 \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a (c+d x)^4}{4 d}-\frac {6 i a d^3 \text {Li}_4\left (-i e^{i (e+f x)}\right )}{f^4}+\frac {6 i a d^3 \text {Li}_4\left (i e^{i (e+f x)}\right )}{f^4} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4190, 4181, 2531, 6609, 2282, 6589} \[ -\frac {6 a d^2 (c+d x) \text {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 a d^2 (c+d x) \text {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {3 i a d (c+d x)^2 \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i a d (c+d x)^2 \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 i a d^3 \text {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {6 i a d^3 \text {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4}-\frac {2 i a (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 218, normalized size = 0.96 \[ a \left (\frac {3 i d \left (f^2 (c+d x)^2 \text {Li}_2\left (-i e^{i (e+f x)}\right )+2 i d f (c+d x) \text {Li}_3\left (-i e^{i (e+f x)}\right )-2 d^2 \text {Li}_4\left (-i e^{i (e+f x)}\right )\right )}{f^4}+\frac {3 d \left (2 d \left (f (c+d x) \text {Li}_3\left (i e^{i (e+f x)}\right )+i d \text {Li}_4\left (i e^{i (e+f x)}\right )\right )-i f^2 (c+d x)^2 \text {Li}_2\left (i e^{i (e+f x)}\right )\right )}{f^4}-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {(c+d x)^4}{4 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*((c + d*x)^4/(4*d) - ((2*I)*(c + d*x)^3*ArcTan[E^(I*(e + f*x))])/f + ((3*I)*d*(f^2*(c + d*x)^2*PolyLog[2, (-
I)*E^(I*(e + f*x))] + (2*I)*d*f*(c + d*x)*PolyLog[3, (-I)*E^(I*(e + f*x))] - 2*d^2*PolyLog[4, (-I)*E^(I*(e + f
*x))]))/f^4 + (3*d*((-I)*f^2*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))] + 2*d*(f*(c + d*x)*PolyLog[3, I*E^(I*(e
+ f*x))] + I*d*PolyLog[4, I*E^(I*(e + f*x))])))/f^4)

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fricas [C]  time = 1.25, size = 1081, normalized size = 4.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^3*f^4*x^4 + 4*a*c*d^2*f^4*x^3 + 6*a*c^2*d*f^4*x^2 + 4*a*c^3*f^4*x + 12*I*a*d^3*polylog(4, I*cos(f*x +
e) + sin(f*x + e)) + 12*I*a*d^3*polylog(4, I*cos(f*x + e) - sin(f*x + e)) - 12*I*a*d^3*polylog(4, -I*cos(f*x
+ e) + sin(f*x + e)) - 12*I*a*d^3*polylog(4, -I*cos(f*x + e) - sin(f*x + e)) + (-6*I*a*d^3*f^2*x^2 - 12*I*a*c*
d^2*f^2*x - 6*I*a*c^2*d*f^2)*dilog(I*cos(f*x + e) + sin(f*x + e)) + (-6*I*a*d^3*f^2*x^2 - 12*I*a*c*d^2*f^2*x -
6*I*a*c^2*d*f^2)*dilog(I*cos(f*x + e) - sin(f*x + e)) + (6*I*a*d^3*f^2*x^2 + 12*I*a*c*d^2*f^2*x + 6*I*a*c^2*d
*f^2)*dilog(-I*cos(f*x + e) + sin(f*x + e)) + (6*I*a*d^3*f^2*x^2 + 12*I*a*c*d^2*f^2*x + 6*I*a*c^2*d*f^2)*dilog
(-I*cos(f*x + e) - sin(f*x + e)) - 2*(a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(cos(f*x +
e) + I*sin(f*x + e) + I) + 2*(a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(cos(f*x + e) - I
*sin(f*x + e) + I) + 2*(a*d^3*f^3*x^3 + 3*a*c*d^2*f^3*x^2 + 3*a*c^2*d*f^3*x + a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*
a*c^2*d*e*f^2)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(a*d^3*f^3*x^3 + 3*a*c*d^2*f^3*x^2 + 3*a*c^2*d*f^3*x
+ a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2)*log(I*cos(f*x + e) - sin(f*x + e) + 1) + 2*(a*d^3*f^3*x^3 +
3*a*c*d^2*f^3*x^2 + 3*a*c^2*d*f^3*x + a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2)*log(-I*cos(f*x + e) + sin
(f*x + e) + 1) - 2*(a*d^3*f^3*x^3 + 3*a*c*d^2*f^3*x^2 + 3*a*c^2*d*f^3*x + a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^
2*d*e*f^2)*log(-I*cos(f*x + e) - sin(f*x + e) + 1) - 2*(a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*
f^3)*log(-cos(f*x + e) + I*sin(f*x + e) + I) + 2*(a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*l
og(-cos(f*x + e) - I*sin(f*x + e) + I) - 12*(a*d^3*f*x + a*c*d^2*f)*polylog(3, I*cos(f*x + e) + sin(f*x + e))
+ 12*(a*d^3*f*x + a*c*d^2*f)*polylog(3, I*cos(f*x + e) - sin(f*x + e)) - 12*(a*d^3*f*x + a*c*d^2*f)*polylog(3,
-I*cos(f*x + e) + sin(f*x + e)) + 12*(a*d^3*f*x + a*c*d^2*f)*polylog(3, -I*cos(f*x + e) - sin(f*x + e)))/f^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.01, size = 747, normalized size = 3.29 \[ \frac {3 a \,c^{2} d \,x^{2}}{2}+a c \,d^{2} x^{3}+\frac {6 i a \,d^{2} c \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {6 i a \,d^{3} \polylog \left (4, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}+\frac {a \,d^{3} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{3}}{f}+a \,c^{3} x +\frac {a \,d^{3} x^{4}}{4}+\frac {6 i a \,c^{2} d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {6 i a c \,d^{2} e^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {6 i a \,d^{2} c \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {3 i a \,c^{2} d \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {3 i a \,d^{3} \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f^{2}}+\frac {3 a \,c^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {3 a \,c^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {6 i a \,d^{3} \polylog \left (4, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}+\frac {3 a \,d^{2} c \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}-\frac {3 a \,c^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}-\frac {3 a c \,d^{2} e^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {3 a \,c^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {3 a c \,d^{2} e^{2} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{3}}-\frac {3 a \,d^{2} c \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{2}}{f}+\frac {2 i a \,d^{3} e^{3} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {3 i a \,d^{3} \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f^{2}}-\frac {3 i a \,c^{2} d \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {6 a \,d^{2} c \polylog \left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {6 a \,d^{3} \polylog \left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{3}}+\frac {6 a \,d^{2} c \polylog \left (3, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 i a \,c^{3} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {a \,d^{3} e^{3} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{4}}-\frac {a \,d^{3} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{3}}{f}+\frac {6 a \,d^{3} \polylog \left (3, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{3}}+\frac {a \,d^{3} e^{3} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.39, size = 926, normalized size = 4.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(f*x + e)*a*c^3 + (f*x + e)^4*a*d^3/f^3 - 4*(f*x + e)^3*a*d^3*e/f^3 + 6*(f*x + e)^2*a*d^3*e^2/f^3 - 4*(
f*x + e)*a*d^3*e^3/f^3 + 4*(f*x + e)^3*a*c*d^2/f^2 - 12*(f*x + e)^2*a*c*d^2*e/f^2 + 12*(f*x + e)*a*c*d^2*e^2/f
^2 + 6*(f*x + e)^2*a*c^2*d/f - 12*(f*x + e)*a*c^2*d*e/f + 4*a*c^3*log(sec(f*x + e) + tan(f*x + e)) - 4*a*d^3*e
^3*log(sec(f*x + e) + tan(f*x + e))/f^3 + 12*a*c*d^2*e^2*log(sec(f*x + e) + tan(f*x + e))/f^2 - 12*a*c^2*d*e*l
og(sec(f*x + e) + tan(f*x + e))/f + 2*(12*I*a*d^3*polylog(4, I*e^(I*f*x + I*e)) - 12*I*a*d^3*polylog(4, -I*e^(
I*f*x + I*e)) + (-2*I*(f*x + e)^3*a*d^3 + (6*I*a*d^3*e - 6*I*a*c*d^2*f)*(f*x + e)^2 + (-6*I*a*d^3*e^2 + 12*I*a
*c*d^2*e*f - 6*I*a*c^2*d*f^2)*(f*x + e))*arctan2(cos(f*x + e), sin(f*x + e) + 1) + (-2*I*(f*x + e)^3*a*d^3 + (
6*I*a*d^3*e - 6*I*a*c*d^2*f)*(f*x + e)^2 + (-6*I*a*d^3*e^2 + 12*I*a*c*d^2*e*f - 6*I*a*c^2*d*f^2)*(f*x + e))*ar
ctan2(cos(f*x + e), -sin(f*x + e) + 1) + (-6*I*(f*x + e)^2*a*d^3 - 6*I*a*d^3*e^2 + 12*I*a*c*d^2*e*f - 6*I*a*c^
2*d*f^2 + (12*I*a*d^3*e - 12*I*a*c*d^2*f)*(f*x + e))*dilog(I*e^(I*f*x + I*e)) + (6*I*(f*x + e)^2*a*d^3 + 6*I*a
*d^3*e^2 - 12*I*a*c*d^2*e*f + 6*I*a*c^2*d*f^2 + (-12*I*a*d^3*e + 12*I*a*c*d^2*f)*(f*x + e))*dilog(-I*e^(I*f*x
+ I*e)) + ((f*x + e)^3*a*d^3 - 3*(a*d^3*e - a*c*d^2*f)*(f*x + e)^2 + 3*(a*d^3*e^2 - 2*a*c*d^2*e*f + a*c^2*d*f^
2)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - ((f*x + e)^3*a*d^3 - 3*(a*d^3*e - a*
c*d^2*f)*(f*x + e)^2 + 3*(a*d^3*e^2 - 2*a*c*d^2*e*f + a*c^2*d*f^2)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e
)^2 - 2*sin(f*x + e) + 1) + 12*((f*x + e)*a*d^3 - a*d^3*e + a*c*d^2*f)*polylog(3, I*e^(I*f*x + I*e)) - 12*((f*
x + e)*a*d^3 - a*d^3*e + a*c*d^2*f)*polylog(3, -I*e^(I*f*x + I*e)))/f^3)/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )\,{\left (c+d\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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