### 3.1 $$\int x^3 \tan (a+b x) \, dx$$

Optimal. Leaf size=106 $\frac{3 i x^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4}$

[Out]

(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

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Rubi [A]  time = 0.155191, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {3719, 2190, 2531, 6609, 2282, 6589} $\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Tan[a + b*x],x]

[Out]

(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

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 6609

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]

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 (a+b x) \, dx &=\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\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{(3 i) \int x \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 \int \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0639529, size = 106, normalized size = 1. $\frac{3 i x^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Tan[a + b*x],x]

[Out]

(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

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Maple [A]  time = 0.087, size = 125, normalized size = 1.2 \begin{align*}{\frac{i}{4}}{x}^{4}+{\frac{2\,i{a}^{3}x}{{b}^{3}}}+{\frac{{\frac{3\,i}{2}}{a}^{4}}{{b}^{4}}}-{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{\frac{3\,i}{2}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{{\frac{3\,i}{4}}{\it polylog} \left ( 4,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.69412, size = 325, normalized size = 3.07 \begin{align*} -\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 ) +{\left (16 i \,{\left (b x + a\right )}^{3} - 36 i \,{\left (b x + a\right )}^{2} a + 36 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 ) +{\left (-24 i \,{\left (b x + a\right )}^{2} + 36 i \,{\left (b x + a\right )} a - 18 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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)) + (16*I*(b*x +
a)^3 - 36*I*(b*x + a)^2*a + 36*I*(b*x + a)*a^2)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (-24*I*(b*x
+ a)^2 + 36*I*(b*x + a)*a - 18*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

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Fricas [C]  time = 1.61102, size = 802, normalized size = 7.57 \begin{align*} -\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)/(ta
n(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)/(ta
n(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)) - 3*I*poly
log(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*ta
n(b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^4

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*tan(a + b*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*tan(b*x + a), x)