\(\int x^3 \tan ^6(x) \, dx\) [129]

Optimal result

Integrand size = 8, antiderivative size = 153 \[ \int x^3 \tan ^6(x) \, dx=\frac {19 x^2}{20}-\frac {23 i x^3}{15}-\frac {x^4}{4}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-2 \log (\cos (x))-\frac {23}{5} i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )+\frac {23}{10} \operatorname {PolyLog}\left (3,-e^{2 i x}\right )-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x) \]

[Out]

19/20*x^2-23/15*I*x^3-1/4*x^4+23/5*x^2*ln(1+exp(2*I*x))-2*ln(cos(x))-23/5*I*x*polylog(2,-exp(2*I*x))+23/10*pol
ylog(3,-exp(2*I*x))-19/10*x*tan(x)+x^3*tan(x)-1/20*tan(x)^2+4/5*x^2*tan(x)^2+1/10*x*tan(x)^3-1/3*x^3*tan(x)^3-
3/20*x^2*tan(x)^4+1/5*x^3*tan(x)^5

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 34, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {3801, 3554, 3556, 30, 3800, 2221, 2611, 2320, 6724} \[ \int x^3 \tan ^6(x) \, dx=-\frac {23}{5} i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )+\frac {23}{10} \operatorname {PolyLog}\left (3,-e^{2 i x}\right )-\frac {x^4}{4}-\frac {23 i x^3}{15}+\frac {1}{5} x^3 \tan ^5(x)-\frac {1}{3} x^3 \tan ^3(x)+x^3 \tan (x)+\frac {19 x^2}{20}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-\frac {3}{20} x^2 \tan ^4(x)+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {\tan ^2(x)}{20}-\frac {19}{10} x \tan (x)-2 \log (\cos (x)) \]

[In]

Int[x^3*Tan[x]^6,x]

[Out]

(19*x^2)/20 - ((23*I)/15)*x^3 - x^4/4 + (23*x^2*Log[1 + E^((2*I)*x)])/5 - 2*Log[Cos[x]] - ((23*I)/5)*x*PolyLog
[2, -E^((2*I)*x)] + (23*PolyLog[3, -E^((2*I)*x)])/10 - (19*x*Tan[x])/10 + x^3*Tan[x] - Tan[x]^2/20 + (4*x^2*Ta
n[x]^2)/5 + (x*Tan[x]^3)/10 - (x^3*Tan[x]^3)/3 - (3*x^2*Tan[x]^4)/20 + (x^3*Tan[x]^5)/5

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 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 3554

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^3 \tan ^5(x)-\frac {3}{5} \int x^2 \tan ^5(x) \, dx-\int x^3 \tan ^4(x) \, dx \\ & = -\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)+\frac {3}{10} \int x \tan ^4(x) \, dx+\frac {3}{5} \int x^2 \tan ^3(x) \, dx+\int x^3 \tan ^2(x) \, dx+\int x^2 \tan ^3(x) \, dx \\ & = x^3 \tan (x)+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)-\frac {1}{10} \int \tan ^3(x) \, dx-\frac {3}{10} \int x \tan ^2(x) \, dx-\frac {3}{5} \int x^2 \tan (x) \, dx-\frac {3}{5} \int x \tan ^2(x) \, dx-3 \int x^2 \tan (x) \, dx-\int x^3 \, dx-\int x^2 \tan (x) \, dx-\int x \tan ^2(x) \, dx \\ & = -\frac {23 i x^3}{15}-\frac {x^4}{4}-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)+\frac {6}{5} i \int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx+2 i \int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx+6 i \int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx+\frac {1}{10} \int \tan (x) \, dx+\frac {3 \int x \, dx}{10}+\frac {3}{10} \int \tan (x) \, dx+\frac {3 \int x \, dx}{5}+\frac {3}{5} \int \tan (x) \, dx+\int x \, dx+\int \tan (x) \, dx \\ & = \frac {19 x^2}{20}-\frac {23 i x^3}{15}-\frac {x^4}{4}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-2 \log (\cos (x))-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)-\frac {6}{5} \int x \log \left (1+e^{2 i x}\right ) \, dx-2 \int x \log \left (1+e^{2 i x}\right ) \, dx-6 \int x \log \left (1+e^{2 i x}\right ) \, dx \\ & = \frac {19 x^2}{20}-\frac {23 i x^3}{15}-\frac {x^4}{4}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-2 \log (\cos (x))-\frac {23}{5} i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)+\frac {3}{5} i \int \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \, dx+i \int \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \, dx+3 i \int \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \, dx \\ & = \frac {19 x^2}{20}-\frac {23 i x^3}{15}-\frac {x^4}{4}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-2 \log (\cos (x))-\frac {23}{5} i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x)+\frac {3}{10} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i x}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i x}\right )+\frac {3}{2} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i x}\right ) \\ & = \frac {19 x^2}{20}-\frac {23 i x^3}{15}-\frac {x^4}{4}+\frac {23}{5} x^2 \log \left (1+e^{2 i x}\right )-2 \log (\cos (x))-\frac {23}{5} i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )+\frac {23}{10} \operatorname {PolyLog}\left (3,-e^{2 i x}\right )-\frac {19}{10} x \tan (x)+x^3 \tan (x)-\frac {\tan ^2(x)}{20}+\frac {4}{5} x^2 \tan ^2(x)+\frac {1}{10} x \tan ^3(x)-\frac {1}{3} x^3 \tan ^3(x)-\frac {3}{20} x^2 \tan ^4(x)+\frac {1}{5} x^3 \tan ^5(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.87 \[ \int x^3 \tan ^6(x) \, dx=\frac {1}{60} \left (-92 i x^3-15 x^4+276 x^2 \log \left (1+e^{2 i x}\right )-120 \log (\cos (x))-276 i x \operatorname {PolyLog}\left (2,-e^{2 i x}\right )+138 \operatorname {PolyLog}\left (3,-e^{2 i x}\right )-3 \sec ^2(x)+66 x^2 \sec ^2(x)-9 x^2 \sec ^4(x)-120 x \tan (x)+92 x^3 \tan (x)+6 x \sec ^2(x) \tan (x)-44 x^3 \sec ^2(x) \tan (x)+12 x^3 \sec ^4(x) \tan (x)\right ) \]

[In]

Integrate[x^3*Tan[x]^6,x]

[Out]

((-92*I)*x^3 - 15*x^4 + 276*x^2*Log[1 + E^((2*I)*x)] - 120*Log[Cos[x]] - (276*I)*x*PolyLog[2, -E^((2*I)*x)] +
138*PolyLog[3, -E^((2*I)*x)] - 3*Sec[x]^2 + 66*x^2*Sec[x]^2 - 9*x^2*Sec[x]^4 - 120*x*Tan[x] + 92*x^3*Tan[x] +
6*x*Sec[x]^2*Tan[x] - 44*x^3*Sec[x]^2*Tan[x] + 12*x^3*Sec[x]^4*Tan[x])/60

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.55

method	result	size
risch	\(-\frac {x^{4}}{4}+\frac {i \left (9 i {\mathrm e}^{6 i x}+90 x^{3} {\mathrm e}^{8 i x}-162 i x^{2} {\mathrm e}^{4 i x}-66 i x^{2} {\mathrm e}^{2 i x}+180 x^{3} {\mathrm e}^{6 i x}-66 x \,{\mathrm e}^{8 i x}+3 i {\mathrm e}^{8 i x}+9 i {\mathrm e}^{4 i x}+280 \,{\mathrm e}^{4 i x} x^{3}-246 x \,{\mathrm e}^{6 i x}-162 i x^{2} {\mathrm e}^{6 i x}+3 i {\mathrm e}^{2 i x}+140 \,{\mathrm e}^{2 i x} x^{3}-354 x \,{\mathrm e}^{4 i x}-66 i x^{2} {\mathrm e}^{8 i x}+46 x^{3}-234 x \,{\mathrm e}^{2 i x}-60 x \right )}{15 \left ({\mathrm e}^{2 i x}+1\right )^{5}}-2 \ln \left ({\mathrm e}^{2 i x}+1\right )+4 \ln \left ({\mathrm e}^{i x}\right )-\frac {46 i x^{3}}{15}+\frac {23 x^{2} \ln \left ({\mathrm e}^{2 i x}+1\right )}{5}-\frac {23 i x \,\operatorname {Li}_{2}\left (-{\mathrm e}^{2 i x}\right )}{5}+\frac {23 \,\operatorname {Li}_{3}\left (-{\mathrm e}^{2 i x}\right )}{10}\)	\(237\)

[In]

int(x^3*tan(x)^6,x,method=_RETURNVERBOSE)

[Out]

-1/4*x^4+1/15*I*(9*I*exp(6*I*x)+90*x^3*exp(8*I*x)-162*I*x^2*exp(4*I*x)-66*I*x^2*exp(2*I*x)+180*x^3*exp(6*I*x)-
66*x*exp(8*I*x)+3*I*exp(8*I*x)+9*I*exp(4*I*x)+280*exp(4*I*x)*x^3-246*x*exp(6*I*x)-162*I*x^2*exp(6*I*x)+3*I*exp
(2*I*x)+140*exp(2*I*x)*x^3-354*x*exp(4*I*x)-66*I*x^2*exp(8*I*x)+46*x^3-234*x*exp(2*I*x)-60*x)/(exp(2*I*x)+1)^5
-2*ln(exp(2*I*x)+1)+4*ln(exp(I*x))-46/15*I*x^3+23/5*x^2*ln(exp(2*I*x)+1)-23/5*I*x*polylog(2,-exp(2*I*x))+23/10
*polylog(3,-exp(2*I*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.39 \[ \int x^3 \tan ^6(x) \, dx=\frac {1}{5} \, x^{3} \tan \left (x\right )^{5} - \frac {3}{20} \, x^{2} \tan \left (x\right )^{4} - \frac {1}{4} \, x^{4} - \frac {1}{30} \, {\left (10 \, x^{3} - 3 \, x\right )} \tan \left (x\right )^{3} + \frac {1}{20} \, {\left (16 \, x^{2} - 1\right )} \tan \left (x\right )^{2} + \frac {19}{20} \, x^{2} + \frac {23}{10} i \, x {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) - \frac {23}{10} i \, x {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) + \frac {1}{10} \, {\left (23 \, x^{2} - 10\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac {1}{10} \, {\left (23 \, x^{2} - 10\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac {1}{10} \, {\left (10 \, x^{3} - 19 \, x\right )} \tan \left (x\right ) + \frac {23}{20} \, {\rm polylog}\left (3, \frac {\tan \left (x\right )^{2} + 2 i \, \tan \left (x\right ) - 1}{\tan \left (x\right )^{2} + 1}\right ) + \frac {23}{20} \, {\rm polylog}\left (3, \frac {\tan \left (x\right )^{2} - 2 i \, \tan \left (x\right ) - 1}{\tan \left (x\right )^{2} + 1}\right ) \]

[In]

integrate(x^3*tan(x)^6,x, algorithm="fricas")

[Out]

1/5*x^3*tan(x)^5 - 3/20*x^2*tan(x)^4 - 1/4*x^4 - 1/30*(10*x^3 - 3*x)*tan(x)^3 + 1/20*(16*x^2 - 1)*tan(x)^2 + 1
9/20*x^2 + 23/10*I*x*dilog(2*(I*tan(x) - 1)/(tan(x)^2 + 1) + 1) - 23/10*I*x*dilog(2*(-I*tan(x) - 1)/(tan(x)^2
+ 1) + 1) + 1/10*(23*x^2 - 10)*log(-2*(I*tan(x) - 1)/(tan(x)^2 + 1)) + 1/10*(23*x^2 - 10)*log(-2*(-I*tan(x) -
1)/(tan(x)^2 + 1)) + 1/10*(10*x^3 - 19*x)*tan(x) + 23/20*polylog(3, (tan(x)^2 + 2*I*tan(x) - 1)/(tan(x)^2 + 1)
) + 23/20*polylog(3, (tan(x)^2 - 2*I*tan(x) - 1)/(tan(x)^2 + 1))

Sympy [F]

\[ \int x^3 \tan ^6(x) \, dx=\int x^{3} \tan ^{6}{\left (x \right )}\, dx \]

[In]

integrate(x**3*tan(x)**6,x)

[Out]

Integral(x**3*tan(x)**6, 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. 777 vs. \(2 (113) = 226\).

Time = 0.57 (sec) , antiderivative size = 777, normalized size of antiderivative = 5.08 \[ \int x^3 \tan ^6(x) \, dx=\text {Too large to display} \]

[In]

integrate(x^3*tan(x)^6,x, algorithm="maxima")

[Out]

(15*I*x^4 + 12*(23*x^2 + (23*x^2 - 10)*cos(10*x) + 5*(23*x^2 - 10)*cos(8*x) + 10*(23*x^2 - 10)*cos(6*x) + 10*(
23*x^2 - 10)*cos(4*x) + 5*(23*x^2 - 10)*cos(2*x) - (-23*I*x^2 + 10*I)*sin(10*x) - 5*(-23*I*x^2 + 10*I)*sin(8*x
) - 10*(-23*I*x^2 + 10*I)*sin(6*x) - 10*(-23*I*x^2 + 10*I)*sin(4*x) - 5*(-23*I*x^2 + 10*I)*sin(2*x) - 10)*arct
an2(sin(2*x), cos(2*x) + 1) + (15*I*x^4 - 184*x^3 + 240*x)*cos(10*x) + (75*I*x^4 - 560*x^3 - 264*I*x^2 + 936*x
+ 12*I)*cos(8*x) - 2*(-75*I*x^4 + 560*x^3 + 324*I*x^2 - 708*x - 18*I)*cos(6*x) - 6*(-25*I*x^4 + 120*x^3 + 108
*I*x^2 - 164*x - 6*I)*cos(4*x) - 3*(-25*I*x^4 + 120*x^3 + 88*I*x^2 - 88*x - 4*I)*cos(2*x) - 276*(x*cos(10*x) +
5*x*cos(8*x) + 10*x*cos(6*x) + 10*x*cos(4*x) + 5*x*cos(2*x) + I*x*sin(10*x) + 5*I*x*sin(8*x) + 10*I*x*sin(6*x
) + 10*I*x*sin(4*x) + 5*I*x*sin(2*x) + x)*dilog(-e^(2*I*x)) - 6*(23*I*x^2 + (23*I*x^2 - 10*I)*cos(10*x) + 5*(2
3*I*x^2 - 10*I)*cos(8*x) + 10*(23*I*x^2 - 10*I)*cos(6*x) + 10*(23*I*x^2 - 10*I)*cos(4*x) + 5*(23*I*x^2 - 10*I)
*cos(2*x) - (23*x^2 - 10)*sin(10*x) - 5*(23*x^2 - 10)*sin(8*x) - 10*(23*x^2 - 10)*sin(6*x) - 10*(23*x^2 - 10)*
sin(4*x) - 5*(23*x^2 - 10)*sin(2*x) - 10*I)*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1) - 138*(I*cos(10*x) +
5*I*cos(8*x) + 10*I*cos(6*x) + 10*I*cos(4*x) + 5*I*cos(2*x) - sin(10*x) - 5*sin(8*x) - 10*sin(6*x) - 10*sin(4
*x) - 5*sin(2*x) + I)*polylog(3, -e^(2*I*x)) - (15*x^4 + 184*I*x^3 - 240*I*x)*sin(10*x) - (75*x^4 + 560*I*x^3
- 264*x^2 - 936*I*x + 12)*sin(8*x) - 2*(75*x^4 + 560*I*x^3 - 324*x^2 - 708*I*x + 18)*sin(6*x) - 6*(25*x^4 + 12
0*I*x^3 - 108*x^2 - 164*I*x + 6)*sin(4*x) - 3*(25*x^4 + 120*I*x^3 - 88*x^2 - 88*I*x + 4)*sin(2*x))/(-60*I*cos(
10*x) - 300*I*cos(8*x) - 600*I*cos(6*x) - 600*I*cos(4*x) - 300*I*cos(2*x) + 60*sin(10*x) + 300*sin(8*x) + 600*
sin(6*x) + 600*sin(4*x) + 300*sin(2*x) - 60*I)

Giac [F]

\[ \int x^3 \tan ^6(x) \, dx=\int { x^{3} \tan \left (x\right )^{6} \,d x } \]

[In]

integrate(x^3*tan(x)^6,x, algorithm="giac")

[Out]

integrate(x^3*tan(x)^6, x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \tan ^6(x) \, dx=\int x^3\,{\mathrm {tan}\left (x\right )}^6 \,d x \]

[In]

int(x^3*tan(x)^6,x)

[Out]

int(x^3*tan(x)^6, x)