3.129 \(\int x^3 \tan ^6(x) \, dx\)

Optimal. Leaf size=153 \[ -\frac{23}{5} i x \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{23}{10} \text{PolyLog}\left (3,-e^{2 i x}\right )-\frac{x^4}{4}-\frac{23 i x^3}{15}+\frac{19 x^2}{20}+\frac{23}{5} x^2 \log \left (1+e^{2 i x}\right )+\frac{1}{5} x^3 \tan ^5(x)-\frac{3}{20} x^2 \tan ^4(x)-\frac{1}{3} x^3 \tan ^3(x)+\frac{4}{5} x^2 \tan ^2(x)+x^3 \tan (x)+\frac{1}{10} x \tan ^3(x)-\frac{\tan ^2(x)}{20}-\frac{19}{10} x \tan (x)-2 \log (\cos (x)) \]

[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

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Rubi [A]  time = 0.40749, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 34, number of rules used = 9, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.125, Rules used = {3720, 3473, 3475, 30, 3719, 2190, 2531, 2282, 6589} \[ -\frac{23}{5} i x \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{23}{10} \text{PolyLog}\left (3,-e^{2 i x}\right )-\frac{x^4}{4}-\frac{23 i x^3}{15}+\frac{19 x^2}{20}+\frac{23}{5} x^2 \log \left (1+e^{2 i x}\right )+\frac{1}{5} x^3 \tan ^5(x)-\frac{3}{20} x^2 \tan ^4(x)-\frac{1}{3} x^3 \tan ^3(x)+\frac{4}{5} x^2 \tan ^2(x)+x^3 \tan (x)+\frac{1}{10} x \tan ^3(x)-\frac{\tan ^2(x)}{20}-\frac{19}{10} x \tan (x)-2 \log (\cos (x)) \]

Antiderivative was successfully verified.

[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 3720

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 3473

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 3475

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

Rule 30

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

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 x^3 \tan ^6(x) \, dx &=\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 \text{Li}_2\left (-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 \text{Li}_2\left (-e^{2 i x}\right ) \, dx+i \int \text{Li}_2\left (-e^{2 i x}\right ) \, dx+3 i \int \text{Li}_2\left (-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 \text{Li}_2\left (-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} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i x}\right )+\frac{3}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_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 \text{Li}_2\left (-e^{2 i x}\right )+\frac{23}{10} \text{Li}_3\left (-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]  time = 0.310905, size = 133, normalized size = 0.87 \[ \frac{1}{60} \left (-276 i x \text{PolyLog}\left (2,-e^{2 i x}\right )+138 \text{PolyLog}\left (3,-e^{2 i x}\right )-15 x^4-92 i x^3+276 x^2 \log \left (1+e^{2 i x}\right )+92 x^3 \tan (x)-9 x^2 \sec ^4(x)+66 x^2 \sec ^2(x)+12 x^3 \tan (x) \sec ^4(x)-44 x^3 \tan (x) \sec ^2(x)-120 x \tan (x)-3 \sec ^2(x)-120 \log (\cos (x))+6 x \tan (x) \sec ^2(x)\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.069, size = 237, normalized size = 1.6 \begin{align*} -{\frac{{x}^{4}}{4}}+{\frac{{\frac{i}{15}} \left ( -66\,i{x}^{2}{{\rm e}^{2\,ix}}+90\,{x}^{3}{{\rm e}^{8\,ix}}-162\,i{x}^{2}{{\rm e}^{6\,ix}}-162\,i{x}^{2}{{\rm e}^{4\,ix}}+180\,{x}^{3}{{\rm e}^{6\,ix}}-66\,x{{\rm e}^{8\,ix}}+3\,i{{\rm e}^{2\,ix}}+9\,i{{\rm e}^{6\,ix}}+280\,{x}^{3}{{\rm e}^{4\,ix}}-246\,x{{\rm e}^{6\,ix}}+3\,i{{\rm e}^{8\,ix}}+9\,i{{\rm e}^{4\,ix}}+140\,{x}^{3}{{\rm e}^{2\,ix}}-354\,x{{\rm e}^{4\,ix}}-66\,i{x}^{2}{{\rm e}^{8\,ix}}+46\,{x}^{3}-234\,x{{\rm e}^{2\,ix}}-60\,x \right ) }{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{5}}}+4\,\ln \left ({{\rm e}^{ix}} \right ) -2\,\ln \left ( 1+{{\rm e}^{2\,ix}} \right ) -{\frac{46\,i}{15}}{x}^{3}+{\frac{23\,{x}^{2}\ln \left ( 1+{{\rm e}^{2\,ix}} \right ) }{5}}-{\frac{23\,i}{5}}x{\it polylog} \left ( 2,-{{\rm e}^{2\,ix}} \right ) +{\frac{23\,{\it polylog} \left ( 3,-{{\rm e}^{2\,ix}} \right ) }{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 3.09721, size = 1034, normalized size = 6.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*I*x^4 + (276*x^2 + 12*(23*x^2 - 10)*cos(10*x) + 60*(23*x^2 - 10)*cos(8*x) + 120*(23*x^2 - 10)*cos(6*x) + 1
20*(23*x^2 - 10)*cos(4*x) + 60*(23*x^2 - 10)*cos(2*x) + (276*I*x^2 - 120*I)*sin(10*x) + (1380*I*x^2 - 600*I)*s
in(8*x) + (2760*I*x^2 - 1200*I)*sin(6*x) + (2760*I*x^2 - 1200*I)*sin(4*x) + (1380*I*x^2 - 600*I)*sin(2*x) - 12
0)*arctan2(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) + (150*I*x^4 - 1120*x^3 - 648*I*x^2 + 1416*x + 36*I)*cos(6*x) + (150*I*x^4 - 720*x^3
- 648*I*x^2 + 984*x + 36*I)*cos(4*x) + (75*I*x^4 - 360*x^3 - 264*I*x^2 + 264*x + 12*I)*cos(2*x) - (276*x*cos(1
0*x) + 1380*x*cos(8*x) + 2760*x*cos(6*x) + 2760*x*cos(4*x) + 1380*x*cos(2*x) + 276*I*x*sin(10*x) + 1380*I*x*si
n(8*x) + 2760*I*x*sin(6*x) + 2760*I*x*sin(4*x) + 1380*I*x*sin(2*x) + 276*x)*dilog(-e^(2*I*x)) + (-138*I*x^2 +
(-138*I*x^2 + 60*I)*cos(10*x) + (-690*I*x^2 + 300*I)*cos(8*x) + (-1380*I*x^2 + 600*I)*cos(6*x) + (-1380*I*x^2
+ 600*I)*cos(4*x) + (-690*I*x^2 + 300*I)*cos(2*x) + 6*(23*x^2 - 10)*sin(10*x) + 30*(23*x^2 - 10)*sin(8*x) + 60
*(23*x^2 - 10)*sin(6*x) + 60*(23*x^2 - 10)*sin(4*x) + 30*(23*x^2 - 10)*sin(2*x) + 60*I)*log(cos(2*x)^2 + sin(2
*x)^2 + 2*cos(2*x) + 1) + (-138*I*cos(10*x) - 690*I*cos(8*x) - 1380*I*cos(6*x) - 1380*I*cos(4*x) - 690*I*cos(2
*x) + 138*sin(10*x) + 690*sin(8*x) + 1380*sin(6*x) + 1380*sin(4*x) + 690*sin(2*x) - 138*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) - (150
*x^4 + 1120*I*x^3 - 648*x^2 - 1416*I*x + 36)*sin(6*x) - (150*x^4 + 720*I*x^3 - 648*x^2 - 984*I*x + 36)*sin(4*x
) - (75*x^4 + 360*I*x^3 - 264*x^2 - 264*I*x + 12)*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)

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Fricas [C]  time = 1.84195, size = 683, normalized size = 4.46 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \tan ^{6}{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*tan(x)**6, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \tan \left (x\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*tan(x)^6, x)