### 3.2 $$\int x^2 \tan (a+b x) \, dx$$

Optimal. Leaf size=77 $\frac{i x \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^3}{3}$

[Out]

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

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Rubi [A]  time = 0.13408, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3719, 2190, 2531, 2282, 6589} $\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{\text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Tan[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0079411, size = 77, normalized size = 1. $\frac{i x \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Tan[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.035, size = 103, normalized size = 1.3 \begin{align*}{\frac{i}{3}}{x}^{3}-{\frac{2\,i{a}^{2}x}{{b}^{2}}}-{\frac{{\frac{4\,i}{3}}{a}^{3}}{{b}^{3}}}-{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{ix{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}+2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*tan(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.67243, size = 219, normalized size = 2.84 \begin{align*} -\frac{-2 i \,{\left (b x + a\right )}^{3} + 6 i \,{\left (b x + a\right )}^{2} a - 6 i \, b x{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, a^{2} \log \left (\sec \left (b x + a\right )\right ) +{\left (6 i \,{\left (b x + a\right )}^{2} - 12 i \,{\left (b x + a\right )} a\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\left ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\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 ) + 3 \,{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{6 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 1.62134, size = 562, normalized size = 7.3 \begin{align*} -\frac{2 \, b^{2} x^{2} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, b^{2} x^{2} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 i \, b x{\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 ) - 2 i \, b x{\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 ) +{\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 ) +{\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 )}{4 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \tan{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*tan(b*x+a),x)

[Out]

Integral(x**2*tan(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \tan \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*tan(b*x + a), x)