\(\int x^3 \tan ^3(a+b x) \, dx\) [11]

Optimal result

Integrand size = 12, antiderivative size = 205 \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b} \]

[Out]

3/2*I*x^2/b^2+1/2*x^3/b-1/4*I*x^4-3*x*ln(1+exp(2*I*(b*x+a)))/b^3+x^3*ln(1+exp(2*I*(b*x+a)))/b+3/2*I*polylog(2,
-exp(2*I*(b*x+a)))/b^4-3/2*I*x^2*polylog(2,-exp(2*I*(b*x+a)))/b^2+3/2*x*polylog(3,-exp(2*I*(b*x+a)))/b^3+3/4*I
*polylog(4,-exp(2*I*(b*x+a)))/b^4-3/2*x^2*tan(b*x+a)/b^2+1/2*x^3*tan(b*x+a)^2/b

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3801, 3800, 2221, 2317, 2438, 30, 2611, 6744, 2320, 6724} \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4} \]

[In]

Int[x^3*Tan[a + b*x]^3,x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 3801

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(c + d*x)^m*((b*Tan[e
+ f*x])^(n - 1)/(f*(n - 1))), x] + (-Dist[b*d*(m/(f*(n - 1))), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[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]

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*} \text {integral}& = \frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {3 \int x^2 \tan ^2(a+b x) \, dx}{2 b}-\int x^3 \tan (a+b x) \, dx \\ & = -\frac {i x^4}{4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}} \, dx+\frac {3 \int x \tan (a+b x) \, dx}{b^2}+\frac {3 \int x^2 \, dx}{2 b} \\ & = \frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {(6 i) \int \frac {e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}} \, dx}{b^2}-\frac {3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b} \\ & = \frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {3 \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^3}+\frac {(3 i) \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right ) \, dx}{b^2} \\ & = \frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {(3 i) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 \int \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right ) \, dx}{2 b^3} \\ & = \frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {(3 i) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4} \\ & = \frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.54 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.80 \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {i e^{i a} \left (2 b^4 e^{-2 i a} x^4-4 i b^3 \left (1+e^{-2 i a}\right ) x^3 \log \left (1+e^{-2 i (a+b x)}\right )+6 b^2 \left (1+e^{-2 i a}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-6 i b \left (1+e^{-2 i a}\right ) x \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )-3 \left (1+e^{-2 i a}\right ) \operatorname {PolyLog}\left (4,-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{8 b^4}+\frac {x^3 \sec ^2(a+b x)}{2 b}-\frac {3 \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^4 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {3 x^2 \sec (a) \sec (a+b x) \sin (b x)}{2 b^2}-\frac {1}{4} x^4 \tan (a) \]

[In]

Integrate[x^3*Tan[a + b*x]^3,x]

[Out]

((I/8)*E^(I*a)*((2*b^4*x^4)/E^((2*I)*a) - (4*I)*b^3*(1 + E^((-2*I)*a))*x^3*Log[1 + E^((-2*I)*(a + b*x))] + 6*b
^2*(1 + E^((-2*I)*a))*x^2*PolyLog[2, -E^((-2*I)*(a + b*x))] - (6*I)*b*(1 + E^((-2*I)*a))*x*PolyLog[3, -E^((-2*
I)*(a + b*x))] - 3*(1 + E^((-2*I)*a))*PolyLog[4, -E^((-2*I)*(a + b*x))])*Sec[a])/b^4 + (x^3*Sec[a + b*x]^2)/(2
*b) - (3*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*
I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Co
t[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Se
c[a])/(2*b^4*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (3*x^2*Sec[a]*Sec[a + b*x]*Sin[b*x])/(2*b^2) - (x^4*Tan[a
])/4

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {3 i \operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}+\frac {x^{2} \left (2 b x \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i {\mathrm e}^{2 i \left (b x +a \right )}-3 i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {3 i x^{2}}{b^{2}}-\frac {i x^{4}}{4}+\frac {3 i \operatorname {Li}_{4}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {3 i a^{4}}{2 b^{4}}-\frac {3 i x^{2} \operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {6 i a x}{b^{3}}-\frac {2 i a^{3} x}{b^{3}}-\frac {6 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 i a^{2}}{b^{4}}+\frac {x^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {3 x \,\operatorname {Li}_{3}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {3 x \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}\)	\(251\)

[In]

int(x^3*tan(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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. 344 vs. \(2 (163) = 326\).

Time = 0.27 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.68 \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {4 \, b^{3} x^{3} \tan \left (b x + a\right )^{2} + 4 \, b^{3} x^{3} - 12 \, b^{2} x^{2} \tan \left (b x + a\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (-i \, b^{2} x^{2} + i\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 \, {\left (i \, b^{2} x^{2} - i\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^3 \tan ^3(a+b x) \, dx=\int x^{3} \tan ^{3}{\left (a + b x \right )}\, dx \]

[In]

integrate(x**3*tan(b*x+a)**3,x)

[Out]

Integral(x**3*tan(a + b*x)**3, x)

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. 1205 vs. \(2 (163) = 326\).

Time = 0.47 (sec) , antiderivative size = 1205, normalized size of antiderivative = 5.88 \[ \int x^3 \tan ^3(a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(a^3*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) - 2*(3*(b*x + a)^4 - 12*(b*x + a)^3*a + 18*(b*x +
a)^2*a^2 + 36*a^2 - 4*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x + a) + (4*(b*x + a)^3 - 9*(b*x + a)^
2*a + 9*(a^2 - 1)*(b*x + a) + 9*a)*cos(4*b*x + 4*a) + 2*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x +
a) + 9*a)*cos(2*b*x + 2*a) - (-4*I*(b*x + a)^3 + 9*I*(b*x + a)^2*a + 9*(-I*a^2 + I)*(b*x + a) - 9*I*a)*sin(4*b
*x + 4*a) - 2*(-4*I*(b*x + a)^3 + 9*I*(b*x + a)^2*a + 9*(-I*a^2 + I)*(b*x + a) - 9*I*a)*sin(2*b*x + 2*a) + 9*a
)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 3*((b*x + a)^4 - 4*(b*x + a)^3*a + 6*(a^2 - 2)*(b*x + a)^2
+ 24*(b*x + a)*a)*cos(4*b*x + 4*a) + 6*((b*x + a)^4 - 4*(b*x + a)^3*(a - I) + 6*(a^2 - 2*I*a - 1)*(b*x + a)^2
+ 12*(I*a^2 + a)*(b*x + a) + 6*a^2)*cos(2*b*x + 2*a) + 6*(4*(b*x + a)^2 - 6*(b*x + a)*a + 3*a^2 + (4*(b*x + a
)^2 - 6*(b*x + a)*a + 3*a^2 - 3)*cos(4*b*x + 4*a) + 2*(4*(b*x + a)^2 - 6*(b*x + a)*a + 3*a^2 - 3)*cos(2*b*x +
2*a) + (4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2 - 3*I)*sin(4*b*x + 4*a) + 2*(4*I*(b*x + a)^2 - 6*I*(b*x +
a)*a + 3*I*a^2 - 3*I)*sin(2*b*x + 2*a) - 3)*dilog(-e^(2*I*b*x + 2*I*a)) + 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2
*a + 9*(I*a^2 - I)*(b*x + a) + (4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*(I*a^2 - I)*(b*x + a) + 9*I*a)*cos(4*b
*x + 4*a) + 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*(I*a^2 - I)*(b*x + a) + 9*I*a)*cos(2*b*x + 2*a) - (4*(b
*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x + a) + 9*a)*sin(4*b*x + 4*a) - 2*(4*(b*x + a)^3 - 9*(b*x + a)^2
*a + 9*(a^2 - 1)*(b*x + a) + 9*a)*sin(2*b*x + 2*a) + 9*I*a)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*co
s(2*b*x + 2*a) + 1) - 12*(cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) + I*sin(4*b*x + 4*a) + 2*I*sin(2*b*x + 2*a) +
1)*polylog(4, -e^(2*I*b*x + 2*I*a)) + 6*(4*I*b*x + (4*I*b*x + I*a)*cos(4*b*x + 4*a) + 2*(4*I*b*x + I*a)*cos(2*
b*x + 2*a) - (4*b*x + a)*sin(4*b*x + 4*a) - 2*(4*b*x + a)*sin(2*b*x + 2*a) + I*a)*polylog(3, -e^(2*I*b*x + 2*I
*a)) + 3*(I*(b*x + a)^4 - 4*I*(b*x + a)^3*a + 6*(I*a^2 - 2*I)*(b*x + a)^2 + 24*I*(b*x + a)*a)*sin(4*b*x + 4*a)
+ 6*(I*(b*x + a)^4 + 4*(b*x + a)^3*(-I*a - 1) + 6*(I*a^2 + 2*a - I)*(b*x + a)^2 - 12*(a^2 - I*a)*(b*x + a) +
6*I*a^2)*sin(2*b*x + 2*a))/(-12*I*cos(4*b*x + 4*a) - 24*I*cos(2*b*x + 2*a) + 12*sin(4*b*x + 4*a) + 24*sin(2*b*
x + 2*a) - 12*I))/b^4

Giac [F]

\[ \int x^3 \tan ^3(a+b x) \, dx=\int { x^{3} \tan \left (b x + a\right )^{3} \,d x } \]

[In]

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

[Out]

integrate(x^3*tan(b*x + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \tan ^3(a+b x) \, dx=\int x^3\,{\mathrm {tan}\left (a+b\,x\right )}^3 \,d x \]

[In]

int(x^3*tan(a + b*x)^3,x)

[Out]

int(x^3*tan(a + b*x)^3, x)