\(\int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx\) [11]

Optimal result

Integrand size = 20, antiderivative size = 152 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {i (c+d x)^3}{a f}+\frac {(c+d x)^4}{4 a d}-\frac {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {12 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac {12 d^3 \operatorname {PolyLog}\left (3,-e^{i (e+f x)}\right )}{a f^4}-\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4276, 3399, 4269, 3800, 2221, 2611, 2320, 6724} \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {12 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {i (c+d x)^3}{a f}+\frac {(c+d x)^4}{4 a d}-\frac {12 d^3 \operatorname {PolyLog}\left (3,-e^{i (e+f x)}\right )}{a f^4} \]

[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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3800

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 4269

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 4276

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 6724

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*} \text {integral}& = \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) \operatorname {PolyLog}\left (2,-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 \operatorname {PolyLog}\left (2,-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) \operatorname {PolyLog}\left (2,-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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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) \operatorname {PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac {12 d^3 \operatorname {PolyLog}\left (3,-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] (verified)

Time = 2.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.42 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \cos \left (\frac {1}{2} (e+f x)\right )+\frac {8 \cos \left (\frac {1}{2} (e+f x)\right ) \left (-\frac {i f^3 (c+d x)^3}{1+e^{i e}}-3 d f^2 (c+d x)^2 \log \left (1+e^{-i (e+f x)}\right )-6 i d^2 f (c+d x) \operatorname {PolyLog}\left (2,-e^{-i (e+f x)}\right )-6 d^3 \operatorname {PolyLog}\left (3,-e^{-i (e+f x)}\right )\right )}{f^4}-\frac {4 (c+d x)^3 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )}{f}\right )}{2 a (1+\sec (e+f x))} \]

[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]))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (138 ) = 276\).

Time = 0.55 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.74

method	result	size
risch	\(\frac {d^{3} x^{4}}{4 a}+\frac {d^{2} c \,x^{3}}{a}+\frac {3 d \,c^{2} x^{2}}{2 a}+\frac {c^{3} x}{a}+\frac {c^{4}}{4 a d}+\frac {12 i d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}-\frac {6 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{2}}{a \,f^{2}}-\frac {12 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \,f^{2}}+\frac {2 i d^{3} x^{3}}{a f}-\frac {12 d^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{a \,f^{2}}-\frac {4 i d^{3} e^{3}}{a \,f^{4}}+\frac {6 i d^{2} c \,x^{2}}{a f}-\frac {6 i d^{3} e^{2} x}{a \,f^{3}}+\frac {12 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{3}}+\frac {6 i d^{2} c \,e^{2}}{a \,f^{3}}+\frac {12 i d^{2} c e x}{a \,f^{2}}-\frac {12 d^{2} e c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}\)	\(417\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (135) = 270\).

Time = 0.30 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.39 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x + {\left (d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x\right )} \cos \left (f x + e\right ) - 24 \, {\left (i \, d^{3} f x + i \, c d^{2} f + {\left (i \, d^{3} f x + i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 24 \, {\left (-i \, d^{3} f x - i \, c d^{2} f + {\left (-i \, d^{3} f x - i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) - 24 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 24 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + c^{3} f^{3}\right )} \sin \left (f x + e\right )}{4 \, {\left (a f^{4} \cos \left (f x + e\right ) + a f^{4}\right )}} \]

[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 + I*c*d^2*f + (I*d^3*f*x + 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 - I*c*d^2*f + (-I*d^3*f*x - I*c*d^2*f)*cos(f*x + e))*dilog(-c
os(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)*polylog(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)

Sympy [F]

\[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {\int \frac {c^{3}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} x^{3}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} x^{2}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d x}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

[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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1285 vs. \(2 (135) = 270\).

Time = 0.46 (sec) , antiderivative size = 1285, normalized size of antiderivative = 8.45 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\text {Too large to display} \]

[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 - I*c*d^2*f)*(f*x + e)^3 + 24*((f*x + e
)^2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e) + ((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) - (-I*(f*x + e)^2*d^3 - I*d^3*e^2 + 2*(I*d^3*e - I*c*d^2*f)*(f*x + e))*sin(f*x + e))*arctan2(si
n(f*x + e), cos(f*x + e) + 1) + (I*(f*x + e)^4*d^3 - 4*(I*d^3*e - I*c*d^2*f + 2*d^3)*(f*x + e)^3 - 6*(-I*d^3*e
^2 - 4*d^3*e + 4*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
- d^3*e + c*d^2*f + ((f*x + e)*d^3 - d^3*e + c*d^2*f)*cos(f*x + e) + (I*(f*x + e)*d^3 - I*d^3*e + I*c*d^2*f)*s
in(f*x + e))*dilog(-e^(I*f*x + I*e)) - 12*(I*(f*x + e)^2*d^3 + I*d^3*e^2 + 2*(-I*d^3*e + I*c*d^2*f)*(f*x + e)
+ (I*(f*x + e)^2*d^3 + I*d^3*e^2 + 2*(-I*d^3*e + I*c*d^2*f)*(f*x + e))*cos(f*x + e) - ((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) - d^3*sin(f*x + e) + 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 - 4*I*d^3*e + 4*I*c*d^2*f)*(f*x + e)^2 - 4*(d^3*e^3 - 6*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

Giac [F]

\[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{a \sec \left (f x + e\right ) + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^3}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \]

[In]

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

[Out]

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