3.1.39 \(\int (c+d x)^3 (a+b \tan (e+f x)) \, dx\) [39]

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

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3803, 3800, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^4}{4 d}-\frac {3 i b d^3 \text {Li}_4\left (-e^{2 i (e+f x)}\right )}{4 f^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3803

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

Rule 6744

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(546\) vs. \(2(152)=304\).
time = 7.08, size = 546, normalized size = 3.59 \begin {gather*} \frac {b c d^2 e^{-i e} \left (2 i f^2 x^2 \left (2 e^{2 i e} f x+3 i \left (1+e^{2 i e}\right ) \log \left (1+e^{2 i (e+f x)}\right )\right )+6 i \left (1+e^{2 i e}\right ) f x \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )-3 \left (1+e^{2 i e}\right ) \text {PolyLog}\left (3,-e^{2 i (e+f x)}\right )\right ) \sec (e)}{4 f^3}-\frac {1}{4} i b d^3 e^{i e} \left (-x^4+\left (1+e^{-2 i e}\right ) x^4-\frac {e^{-2 i e} \left (1+e^{2 i e}\right ) \left (2 f^4 x^4+4 i f^3 x^3 \log \left (1+e^{2 i (e+f x)}\right )+6 f^2 x^2 \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )+6 i f x \text {PolyLog}\left (3,-e^{2 i (e+f x)}\right )-3 \text {PolyLog}\left (4,-e^{2 i (e+f x)}\right )\right )}{2 f^4}\right ) \sec (e)+\frac {1}{4} x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \sec (e) (a \cos (e)+b \sin (e))-\frac {b c^3 \sec (e) (\cos (e) \log (\cos (e) \cos (f x)-\sin (e) \sin (f x))+f x \sin (e))}{f \left (\cos ^2(e)+\sin ^2(e)\right )}-\frac {3 b c^2 d \csc (e) \left (e^{-i \text {ArcTan}(\cot (e))} f^2 x^2-\frac {\cot (e) \left (i f x (-\pi -2 \text {ArcTan}(\cot (e)))-\pi \log \left (1+e^{-2 i f x}\right )-2 (f x-\text {ArcTan}(\cot (e))) \log \left (1-e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )+\pi \log (\cos (f x))-2 \text {ArcTan}(\cot (e)) \log (\sin (f x-\text {ArcTan}(\cot (e))))+i \text {PolyLog}\left (2,e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )\right )}{\sqrt {1+\cot ^2(e)}}\right ) \sec (e)}{2 f^2 \sqrt {\csc ^2(e) \left (\cos ^2(e)+\sin ^2(e)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d^2*((2*I)*f^2*x^2*(2*E^((2*I)*e)*f*x + (3*I)*(1 + E^((2*I)*e))*Log[1 + E^((2*I)*(e + f*x))]) + (6*I)*(1
+ E^((2*I)*e))*f*x*PolyLog[2, -E^((2*I)*(e + f*x))] - 3*(1 + E^((2*I)*e))*PolyLog[3, -E^((2*I)*(e + f*x))])*Se
c[e])/(4*E^(I*e)*f^3) - (I/4)*b*d^3*E^(I*e)*(-x^4 + (1 + E^((-2*I)*e))*x^4 - ((1 + E^((2*I)*e))*(2*f^4*x^4 + (
4*I)*f^3*x^3*Log[1 + E^((2*I)*(e + f*x))] + 6*f^2*x^2*PolyLog[2, -E^((2*I)*(e + f*x))] + (6*I)*f*x*PolyLog[3,
-E^((2*I)*(e + f*x))] - 3*PolyLog[4, -E^((2*I)*(e + f*x))]))/(2*E^((2*I)*e)*f^4))*Sec[e] + (x*(4*c^3 + 6*c^2*d
*x + 4*c*d^2*x^2 + d^3*x^3)*Sec[e]*(a*Cos[e] + b*Sin[e]))/4 - (b*c^3*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x] - Sin[
e]*Sin[f*x]] + f*x*Sin[e]))/(f*(Cos[e]^2 + Sin[e]^2)) - (3*b*c^2*d*Csc[e]*((f^2*x^2)/E^(I*ArcTan[Cot[e]]) - (C
ot[e]*(I*f*x*(-Pi - 2*ArcTan[Cot[e]]) - Pi*Log[1 + E^((-2*I)*f*x)] - 2*(f*x - ArcTan[Cot[e]])*Log[1 - E^((2*I)
*(f*x - ArcTan[Cot[e]]))] + Pi*Log[Cos[f*x]] - 2*ArcTan[Cot[e]]*Log[Sin[f*x - ArcTan[Cot[e]]]] + I*PolyLog[2,
E^((2*I)*(f*x - ArcTan[Cot[e]]))]))/Sqrt[1 + Cot[e]^2])*Sec[e])/(2*f^2*Sqrt[Csc[e]^2*(Cos[e]^2 + Sin[e]^2)])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (131 ) = 262\).
time = 0.34, size = 500, normalized size = 3.29

method	result	size
risch	\(\frac {3 i b \,c^{2} d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}+\frac {2 i b \,d^{3} e^{3} x}{f^{3}}+\frac {3 i b \,c^{2} d \,e^{2}}{f^{2}}-\frac {4 i b c \,d^{2} e^{3}}{f^{3}}+\frac {6 b c \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {b \,c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {2 b \,c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+d^{2} a c \,x^{3}+\frac {3 d a \,c^{2} x^{2}}{2}+a \,c^{3} x +\frac {3 i b \,d^{3} e^{4}}{2 f^{4}}-\frac {2 b \,d^{3} e^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {b \,d^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{3}}{f}-\frac {3 b c \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}-\frac {3 b \,d^{3} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{2 f^{3}}+\frac {3 i d b \,c^{2} x^{2}}{2}+i d^{2} b c \,x^{3}-\frac {3 i b \,d^{3} \polylog \left (4, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f^{4}}+\frac {d^{3} a \,x^{4}}{4}+\frac {a \,c^{4}}{4 d}+\frac {3 i b c \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {6 i b c \,d^{2} e^{2} x}{f^{2}}+\frac {6 i b \,c^{2} d e x}{f}+\frac {3 i b \,d^{3} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x^{2}}{2 f^{2}}+\frac {i d^{3} b \,x^{4}}{4}-i b \,c^{3} x -\frac {i b \,c^{4}}{4 d}-\frac {3 b \,c^{2} d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f}-\frac {6 b \,c^{2} d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {3 b c \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}\)	\(500\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 710 vs. \(2 (131) = 262\).
time = 0.57, size = 710, normalized size = 4.67 \begin {gather*} \frac {12 \, {\left (f x + e\right )} a c^{3} + \frac {3 \, {\left (f x + e\right )}^{4} a d^{3}}{f^{3}} + \frac {12 \, {\left (f x + e\right )}^{3} a c d^{2}}{f^{2}} + \frac {18 \, {\left (f x + e\right )}^{2} a c^{2} d}{f} - \frac {12 \, {\left (f x + e\right )}^{3} a d^{3} e}{f^{3}} - \frac {36 \, {\left (f x + e\right )}^{2} a c d^{2} e}{f^{2}} - \frac {36 \, {\left (f x + e\right )} a c^{2} d e}{f} + 12 \, b c^{3} \log \left (\sec \left (f x + e\right )\right ) - \frac {36 \, b c^{2} d e \log \left (\sec \left (f x + e\right )\right )}{f} + \frac {18 \, {\left (f x + e\right )}^{2} a d^{3} e^{2}}{f^{3}} + \frac {36 \, {\left (f x + e\right )} a c d^{2} e^{2}}{f^{2}} + \frac {36 \, b c d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, {\left (f x + e\right )} a d^{3} e^{3}}{f^{3}} - \frac {12 \, b d^{3} e^{3} \log \left (\sec \left (f x + e\right )\right )}{f^{3}} - \frac {-3 i \, {\left (f x + e\right )}^{4} b d^{3} + 12 i \, b d^{3} {\rm Li}_{4}(-e^{\left (2 i \, f x + 2 i \, e\right )}) - 12 \, {\left (i \, b c d^{2} f - i \, b d^{3} e\right )} {\left (f x + e\right )}^{3} - 18 \, {\left (i \, b c^{2} d f^{2} - 2 i \, b c d^{2} f e + i \, b d^{3} e^{2}\right )} {\left (f x + e\right )}^{2} - 4 \, {\left (-4 i \, {\left (f x + e\right )}^{3} b d^{3} + 9 \, {\left (-i \, b c d^{2} f + i \, b d^{3} e\right )} {\left (f x + e\right )}^{2} + 9 \, {\left (-i \, b c^{2} d f^{2} + 2 i \, b c d^{2} f e - i \, b d^{3} e^{2}\right )} {\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 6 \, {\left (4 i \, {\left (f x + e\right )}^{2} b d^{3} + 3 i \, b c^{2} d f^{2} - 6 i \, b c d^{2} f e + 3 i \, b d^{3} e^{2} + 6 \, {\left (i \, b c d^{2} f - i \, b d^{3} e\right )} {\left (f x + e\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 2 \, {\left (4 \, {\left (f x + e\right )}^{3} b d^{3} + 9 \, {\left (b c d^{2} f - b d^{3} e\right )} {\left (f x + e\right )}^{2} + 9 \, {\left (b c^{2} d f^{2} - 2 \, b c d^{2} f e + b d^{3} e^{2}\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + 6 \, {\left (4 \, {\left (f x + e\right )} b d^{3} + 3 \, b c d^{2} f - 3 \, b d^{3} e\right )} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )})}{f^{3}}}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(f*x + e)*a*c^3 + 3*(f*x + e)^4*a*d^3/f^3 + 12*(f*x + e)^3*a*c*d^2/f^2 + 18*(f*x + e)^2*a*c^2*d/f - 1
2*(f*x + e)^3*a*d^3*e/f^3 - 36*(f*x + e)^2*a*c*d^2*e/f^2 - 36*(f*x + e)*a*c^2*d*e/f + 12*b*c^3*log(sec(f*x + e
)) - 36*b*c^2*d*e*log(sec(f*x + e))/f + 18*(f*x + e)^2*a*d^3*e^2/f^3 + 36*(f*x + e)*a*c*d^2*e^2/f^2 + 36*b*c*d
^2*e^2*log(sec(f*x + e))/f^2 - 12*(f*x + e)*a*d^3*e^3/f^3 - 12*b*d^3*e^3*log(sec(f*x + e))/f^3 - (-3*I*(f*x +
e)^4*b*d^3 + 12*I*b*d^3*polylog(4, -e^(2*I*f*x + 2*I*e)) - 12*(I*b*c*d^2*f - I*b*d^3*e)*(f*x + e)^3 - 18*(I*b*
c^2*d*f^2 - 2*I*b*c*d^2*f*e + I*b*d^3*e^2)*(f*x + e)^2 - 4*(-4*I*(f*x + e)^3*b*d^3 + 9*(-I*b*c*d^2*f + I*b*d^3
*e)*(f*x + e)^2 + 9*(-I*b*c^2*d*f^2 + 2*I*b*c*d^2*f*e - I*b*d^3*e^2)*(f*x + e))*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e) + 1) - 6*(4*I*(f*x + e)^2*b*d^3 + 3*I*b*c^2*d*f^2 - 6*I*b*c*d^2*f*e + 3*I*b*d^3*e^2 + 6*(I*b*c*d^
2*f - I*b*d^3*e)*(f*x + e))*dilog(-e^(2*I*f*x + 2*I*e)) + 2*(4*(f*x + e)^3*b*d^3 + 9*(b*c*d^2*f - b*d^3*e)*(f*
x + e)^2 + 9*(b*c^2*d*f^2 - 2*b*c*d^2*f*e + b*d^3*e^2)*(f*x + e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2
+ 2*cos(2*f*x + 2*e) + 1) + 6*(4*(f*x + e)*b*d^3 + 3*b*c*d^2*f - 3*b*d^3*e)*polylog(3, -e^(2*I*f*x + 2*I*e)))/
f^3)/f

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (131) = 262\).
time = 0.37, size = 520, normalized size = 3.42 \begin {gather*} \frac {2 \, a d^{3} f^{4} x^{4} + 8 \, a c d^{2} f^{4} x^{3} + 12 \, a c^{2} d f^{4} x^{2} + 8 \, a c^{3} f^{4} x + 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (i \, b d^{3} f^{2} x^{2} + 2 i \, b c d^{2} f^{2} x + i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (-i \, b d^{3} f^{2} x^{2} - 2 i \, b c d^{2} f^{2} x - i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right )}{8 \, f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*d^3*f^4*x^4 + 8*a*c*d^2*f^4*x^3 + 12*a*c^2*d*f^4*x^2 + 8*a*c^3*f^4*x + 3*I*b*d^3*polylog(4, (tan(f*x
+ e)^2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 3*I*b*d^3*polylog(4, (tan(f*x + e)^2 - 2*I*tan(f*x + e)
- 1)/(tan(f*x + e)^2 + 1)) - 6*(I*b*d^3*f^2*x^2 + 2*I*b*c*d^2*f^2*x + I*b*c^2*d*f^2)*dilog(2*(I*tan(f*x + e)
- 1)/(tan(f*x + e)^2 + 1) + 1) - 6*(-I*b*d^3*f^2*x^2 - 2*I*b*c*d^2*f^2*x - I*b*c^2*d*f^2)*dilog(2*(-I*tan(f*x
+ e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*c^3*f^3)*log(
-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*c^3
*f^3)*log(-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(b*d^3*f*x + b*c*d^2*f)*polylog(3, (tan(f*x + e)^
2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(b*d^3*f*x + b*c*d^2*f)*polylog(3, (tan(f*x + e)^2 - 2*I*t
an(f*x + e) - 1)/(tan(f*x + e)^2 + 1)))/f^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d x\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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