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

Optimal result

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

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*polyl 
og(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*p 
olylog(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
 

Mathematica [A] (warning: unable to verify)

Time = 6.48 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.80 \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {i e^{i a} \left (2 b^4 e^{-2 i a} x^4-4 i b^3 \left (1+e^{-2 i a}\right ) x^3 \log \left (1+e^{-2 i (a+b x)}\right )+6 b^2 \left (1+e^{-2 i a}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-6 i b \left (1+e^{-2 i a}\right ) x \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )-3 \left (1+e^{-2 i a}\right ) \operatorname {PolyLog}\left (4,-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{8 b^4}+\frac {x^3 \sec ^2(a+b x)}{2 b}-\frac {3 \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^4 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {3 x^2 \sec (a) \sec (a+b x) \sin (b x)}{2 b^2}-\frac {1}{4} x^4 \tan (a) \] Input:

Integrate[x^3*Tan[a + b*x]^3,x]
 

Output:

((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[C 
ot[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[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*P 
olyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[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
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.20, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.333, Rules used = {3042, 4203, 3042, 4202, 2620, 3011, 4203, 15, 3042, 4202, 2620, 2715, 2838, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^3 \tan ^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int x^3 \tan (a+b x)^3dx\)

\(\Big \downarrow \) 4203

\(\displaystyle -\int x^3 \tan (a+b x)dx-\frac {3 \int x^2 \tan ^2(a+b x)dx}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int x^3 \tan (a+b x)dx-\frac {3 \int x^2 \tan (a+b x)^2dx}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 4202

\(\displaystyle 2 i \int \frac {e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}}dx-\frac {3 \int x^2 \tan (a+b x)^2dx}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {3 \int x^2 \tan (a+b x)^2dx}{2 b}+2 i \left (\frac {3 i \int x^2 \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \int x^2 \tan (a+b x)^2dx}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 4203

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \int x \tan (a+b x)dx}{b}-\int x^2dx+\frac {x^2 \tan (a+b x)}{b}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 15

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \int x \tan (a+b x)dx}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \int x \tan (a+b x)dx}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 4202

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \int \frac {e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}}dx\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (\frac {i \int \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (\frac {\int e^{-2 i (a+b x)} \log \left (1+e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2838

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

\(\Big \downarrow \) 7143

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^2 \tan (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {i x^4}{4}\)

Input:

Int[x^3*Tan[a + b*x]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 15

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

rule 2620

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] - Si 
mp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 2838

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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[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] && IGt 
Q[m, 0]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7163

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.22

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

Input:

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

Output:

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

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 = 0.09 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.68 \[ \int x^3 \tan ^3(a+b x) \, dx=\frac {4 \, b^{3} x^{3} \tan \left (b x + a\right )^{2} + 4 \, b^{3} x^{3} - 12 \, b^{2} x^{2} \tan \left (b x + a\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (-i \, b^{2} x^{2} + i\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 \, {\left (i \, b^{2} x^{2} - i\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \] Input:

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

Output:

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*tan(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)*dilog(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)/(ta 
n(b*x + a)^2 + 1)))/b^4
 

Sympy [F]

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

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

Output:

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 = 0.34 (sec) , antiderivative size = 1205, normalized size of antiderivative = 5.88 \[ \int x^3 \tan ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

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 ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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