\(\int x^3 \tan (a+b x) \, dx\) [1]

Optimal result

Integrand size = 10, antiderivative size = 106 \[ \int x^3 \tan (a+b x) \, dx=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int x^3 \tan (a+b x) \, dx=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 4202, 2620, 3011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 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 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 = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {i x^{4}}{4}+\frac {3 i x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {3 i \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 i a^{4}}{2 b^{4}}+\frac {2 i a^{3} x}{b^{3}}-\frac {3 x \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {x^{3} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}\)	\(125\)

Input:

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

Output:

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

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. 286 vs. \(2 (83) = 166\).

Time = 0.09 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.70 \[ \int x^3 \tan (a+b x) \, dx=-\frac {4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**3*tan(a + b*x), 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. 243 vs. \(2 (83) = 166\).

Time = 0.15 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.29 \[ \int x^3 \tan (a+b x) \, dx=-\frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} + 12 \, a^{3} \log \left (\sec \left (b x + a\right )\right ) - 4 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 9 i \, {\left (b x + a\right )}^{2} a - 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 12 i \, {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{12 \, b^{4}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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