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

Optimal. Leaf size=152 \[ \frac{12 i d^2 (c+d x) \text{PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac{12 d^3 \text{PolyLog}\left (3,-e^{i (e+f x)}\right )}{a f^4}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d} \]

[Out]

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

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Rubi [A]  time = 0.328892, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4191, 3318, 4184, 3719, 2190, 2531, 2282, 6589} \[ \frac{12 i d^2 (c+d x) \text{PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac{12 d^3 \text{PolyLog}\left (3,-e^{i (e+f x)}\right )}{a f^4}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3318

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

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]

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 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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^3}{a+a \sec (e+f x)} \, dx &=\int \left (\frac{(c+d x)^3}{a}-\frac{(c+d x)^3}{a+a \cos (e+f x)}\right ) \, dx\\ &=\frac{(c+d x)^4}{4 a d}-\int \frac{(c+d x)^3}{a+a \cos (e+f x)} \, dx\\ &=\frac{(c+d x)^4}{4 a d}-\frac{\int (c+d x)^3 \csc ^2\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{2 a}\\ &=\frac{(c+d x)^4}{4 a d}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{(3 d) \int (c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{(6 i d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)^2}{1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{\left (12 d^2\right ) \int (c+d x) \log \left (1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac{12 i d^2 (c+d x) \text{Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{\left (12 i d^3\right ) \int \text{Li}_2\left (-e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^3}\\ &=\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac{12 i d^2 (c+d x) \text{Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{\left (12 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{a f^4}\\ &=\frac{i (c+d x)^3}{a f}+\frac{(c+d x)^4}{4 a d}-\frac{6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac{12 i d^2 (c+d x) \text{Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}-\frac{12 d^3 \text{Li}_3\left (-e^{i (e+f x)}\right )}{a f^4}-\frac{(c+d x)^3 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}\\ \end{align*}

Mathematica [A]  time = 1.95161, size = 216, normalized size = 1.42 \[ \frac{\cos \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left (\frac{8 \cos \left (\frac{1}{2} (e+f x)\right ) \left (-6 i d^2 f (c+d x) \text{PolyLog}\left (2,-e^{-i (e+f x)}\right )-6 d^3 \text{PolyLog}\left (3,-e^{-i (e+f x)}\right )-\frac{i f^3 (c+d x)^3}{1+e^{i e}}-3 d f^2 (c+d x)^2 \log \left (1+e^{-i (e+f x)}\right )\right )}{f^4}+x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) \cos \left (\frac{1}{2} (e+f x)\right )-\frac{4 \sec \left (\frac{e}{2}\right ) (c+d x)^3 \sin \left (\frac{f x}{2}\right )}{f}\right )}{2 a (\sec (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*Sec[e + f*x]*(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Cos[(e + f*x)/2] + (8*Cos[(e + f
*x)/2]*(((-I)*f^3*(c + d*x)^3)/(1 + E^(I*e)) - 3*d*f^2*(c + d*x)^2*Log[1 + E^((-I)*(e + f*x))] - (6*I)*d^2*f*(
c + d*x)*PolyLog[2, -E^((-I)*(e + f*x))] - 6*d^3*PolyLog[3, -E^((-I)*(e + f*x))]))/f^4 - (4*(c + d*x)^3*Sec[e/
2]*Sin[(f*x)/2])/f))/(2*a*(1 + Sec[e + f*x]))

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Maple [B]  time = 0.155, size = 406, normalized size = 2.7 \begin{align*}{\frac{{d}^{3}{x}^{4}}{4\,a}}+{\frac{c{d}^{2}{x}^{3}}{a}}+{\frac{3\,{c}^{2}d{x}^{2}}{2\,a}}+{\frac{{c}^{3}x}{a}}+{\frac{2\,i{d}^{3}{x}^{3}}{fa}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }{a{f}^{2}}}+6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{2}}}+6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}+{\frac{12\,i{d}^{2}cex}{a{f}^{2}}}+{\frac{12\,i{d}^{2}c{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}+{\frac{12\,i{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{3}}}-6\,{\frac{{d}^{3}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ){x}^{2}}{a{f}^{2}}}+{\frac{6\,i{d}^{2}c{x}^{2}}{fa}}-12\,{\frac{{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}-12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}-{\frac{6\,i{d}^{3}{e}^{2}x}{a{f}^{3}}}-{\frac{4\,i{d}^{3}{e}^{3}}{{f}^{4}a}}+{\frac{6\,i{d}^{2}c{e}^{2}}{a{f}^{3}}}-12\,{\frac{c{d}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) x}{a{f}^{2}}}-{\frac{2\,i \left ({d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3} \right ) }{fa \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a*d^3*x^4+1/a*c*d^2*x^3+3/2/a*c^2*d*x^2+1/a*c^3*x+2*I*d^3/f/a*x^3-6*d/f^2/a*c^2*ln(exp(I*(f*x+e))+1)+6*d/f
^2/a*c^2*ln(exp(I*(f*x+e)))+6*d^3/f^4/a*e^2*ln(exp(I*(f*x+e)))+12*I*d^2/f^2/a*c*e*x+12*I*d^2/f^3/a*c*polylog(2
,-exp(I*(f*x+e)))+12*I*d^3/f^3/a*polylog(2,-exp(I*(f*x+e)))*x-6*d^3/f^2/a*ln(exp(I*(f*x+e))+1)*x^2+6*I*d^2/f/a
*c*x^2-12*d^3*polylog(3,-exp(I*(f*x+e)))/a/f^4-12*d^2/f^3/a*c*e*ln(exp(I*(f*x+e)))-6*I*d^3/f^3/a*e^2*x-4*I*d^3
/f^4/a*e^3+6*I*d^2/f^3/a*c*e^2-12*d^2/f^2/a*c*ln(exp(I*(f*x+e))+1)*x-2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/f
/a/(exp(I*(f*x+e))+1)

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Maxima [B]  time = 2.44992, size = 1732, normalized size = 11.39 \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+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

1/2*(6*c*d^2*e^2*(2*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/(a*f^2) - sin(f*x + e)/(a*f^2*(cos(f*x + e) + 1)))
- 6*c^2*d*e*(2*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/(a*f) - sin(f*x + e)/(a*f*(cos(f*x + e) + 1))) - 6*((f
*x + e)^2*cos(f*x + e)^2 + (f*x + e)^2*sin(f*x + e)^2 + 2*(f*x + e)^2*cos(f*x + e) + (f*x + e)^2 - 2*(cos(f*x
+ e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 4*(f
*x + e)*sin(f*x + e))*c*d^2*e/(a*f^2*cos(f*x + e)^2 + a*f^2*sin(f*x + e)^2 + 2*a*f^2*cos(f*x + e) + a*f^2) + 2
*c^3*(2*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 3*((f*x + e)^2*cos(
f*x + e)^2 + (f*x + e)^2*sin(f*x + e)^2 + 2*(f*x + e)^2*cos(f*x + e) + (f*x + e)^2 - 2*(cos(f*x + e)^2 + sin(f
*x + e)^2 + 2*cos(f*x + e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 4*(f*x + e)*sin(f*
x + e))*c^2*d/(a*f*cos(f*x + e)^2 + a*f*sin(f*x + e)^2 + 2*a*f*cos(f*x + e) + a*f) - 2*(I*(f*x + e)^4*d^3 + 6*
I*(f*x + e)^2*d^3*e^2 - 4*I*(f*x + e)*d^3*e^3 - 8*d^3*e^3 + (-4*I*d^3*e + 4*I*c*d^2*f)*(f*x + e)^3 + (24*(f*x
+ e)^2*d^3 + 24*d^3*e^2 - 48*(d^3*e - c*d^2*f)*(f*x + e) + 24*((f*x + e)^2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)
*(f*x + e))*cos(f*x + e) + (24*I*(f*x + e)^2*d^3 + 24*I*d^3*e^2 + (-48*I*d^3*e + 48*I*c*d^2*f)*(f*x + e))*sin(
f*x + e))*arctan2(sin(f*x + e), cos(f*x + e) + 1) + (I*(f*x + e)^4*d^3 + (-4*I*d^3*e + 4*I*c*d^2*f - 8*d^3)*(f
*x + e)^3 + (6*I*d^3*e^2 + 24*d^3*e - 24*c*d^2*f)*(f*x + e)^2 - 4*(I*d^3*e^3 + 6*d^3*e^2)*(f*x + e))*cos(f*x +
e) - (48*(f*x + e)*d^3 - 48*d^3*e + 48*c*d^2*f + 48*((f*x + e)*d^3 - d^3*e + c*d^2*f)*cos(f*x + e) - (-48*I*(
f*x + e)*d^3 + 48*I*d^3*e - 48*I*c*d^2*f)*sin(f*x + e))*dilog(-e^(I*f*x + I*e)) + (-12*I*(f*x + e)^2*d^3 - 12*
I*d^3*e^2 + (24*I*d^3*e - 24*I*c*d^2*f)*(f*x + e) + (-12*I*(f*x + e)^2*d^3 - 12*I*d^3*e^2 + (24*I*d^3*e - 24*I
*c*d^2*f)*(f*x + e))*cos(f*x + e) + 12*((f*x + e)^2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e))*sin(f*x + e
))*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) + (-48*I*d^3*cos(f*x + e) + 48*d^3*sin(f*x + e) -
48*I*d^3)*polylog(3, -e^(I*f*x + I*e)) - ((f*x + e)^4*d^3 - 4*(d^3*e - c*d^2*f - 2*I*d^3)*(f*x + e)^3 + (6*d^
3*e^2 - 24*I*d^3*e + 24*I*c*d^2*f)*(f*x + e)^2 - (4*d^3*e^3 - 24*I*d^3*e^2)*(f*x + e))*sin(f*x + e))/(-4*I*a*f
^3*cos(f*x + e) + 4*a*f^3*sin(f*x + e) - 4*I*a*f^3))/f

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Fricas [C]  time = 1.78961, size = 1237, normalized size = 8.14 \begin{align*} \frac{d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x +{\left (d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x\right )} \cos \left (f x + e\right ) +{\left (-24 i \, d^{3} f x - 24 i \, c d^{2} f +{\left (-24 i \, d^{3} f x - 24 i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) +{\left (24 i \, d^{3} f x + 24 i \, c d^{2} f +{\left (24 i \, d^{3} f x + 24 i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 12 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} +{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 12 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} +{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) - 24 \,{\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )}{\rm polylog}\left (3, -\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 24 \,{\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )}{\rm polylog}\left (3, -\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 4 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + c^{3} f^{3}\right )} \sin \left (f x + e\right )}{4 \,{\left (a f^{4} \cos \left (f x + e\right ) + a f^{4}\right )}} \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{3}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{3} x^{3}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c d^{2} x^{2}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c^{2} d x}{\sec{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c**3/(sec(e + f*x) + 1), x) + Integral(d**3*x**3/(sec(e + f*x) + 1), x) + Integral(3*c*d**2*x**2/(se
c(e + f*x) + 1), x) + Integral(3*c**2*d*x/(sec(e + f*x) + 1), x))/a

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

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)