3.490 \(\int x \sin ^2(x) \tan (x) \, dx\)

Optimal. Leaf size=62 \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}+\frac{x}{4}-x \log \left (1+e^{2 i x}\right )-\frac{1}{2} x \sin ^2(x)-\frac{1}{4} \sin (x) \cos (x) \]

[Out]

x/4 + (I/2)*x^2 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] - (Cos[x]*Sin[x])/4 - (x*Sin[x]^2)/2

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Rubi [A]  time = 0.071817, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {4407, 3443, 2635, 8, 3719, 2190, 2279, 2391} \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}+\frac{x}{4}-x \log \left (1+e^{2 i x}\right )-\frac{1}{2} x \sin ^2(x)-\frac{1}{4} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[x]^2*Tan[x],x]

[Out]

x/4 + (I/2)*x^2 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] - (Cos[x]*Sin[x])/4 - (x*Sin[x]^2)/2

Rule 4407

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Int[
(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3443

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m - n
+ 1)*Sin[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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]

Rubi steps

\begin{align*} \int x \sin ^2(x) \tan (x) \, dx &=-\int x \cos (x) \sin (x) \, dx+\int x \tan (x) \, dx\\ &=\frac{i x^2}{2}-\frac{1}{2} x \sin ^2(x)-2 i \int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx+\frac{1}{2} \int \sin ^2(x) \, dx\\ &=\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )-\frac{1}{4} \cos (x) \sin (x)-\frac{1}{2} x \sin ^2(x)+\frac{\int 1 \, dx}{4}+\int \log \left (1+e^{2 i x}\right ) \, dx\\ &=\frac{x}{4}+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )-\frac{1}{4} \cos (x) \sin (x)-\frac{1}{2} x \sin ^2(x)-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\frac{x}{4}+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\frac{1}{2} i \text{Li}_2\left (-e^{2 i x}\right )-\frac{1}{4} \cos (x) \sin (x)-\frac{1}{2} x \sin ^2(x)\\ \end{align*}

Mathematica [A]  time = 0.0146834, size = 57, normalized size = 0.92 \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )-\frac{1}{8} \sin (2 x)+\frac{1}{4} x \cos (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[x]^2*Tan[x],x]

[Out]

(I/2)*x^2 + (x*Cos[2*x])/4 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] - Sin[2*x]/8

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Maple [A]  time = 0.06, size = 57, normalized size = 0.9 \begin{align*}{\frac{i}{2}}{x}^{2}+{\frac{ \left ( i+2\,x \right ){{\rm e}^{2\,ix}}}{16}}+{\frac{ \left ( -i+2\,x \right ){{\rm e}^{-2\,ix}}}{16}}-x\ln \left ( 1+{{\rm e}^{2\,ix}} \right ) +{\frac{i}{2}}{\it polylog} \left ( 2,-{{\rm e}^{2\,ix}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(x)^3/cos(x),x)

[Out]

1/2*I*x^2+1/16*(I+2*x)*exp(2*I*x)+1/16*(-I+2*x)*exp(-2*I*x)-x*ln(1+exp(2*I*x))+1/2*I*polylog(2,-exp(2*I*x))

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Maxima [A]  time = 1.44558, size = 89, normalized size = 1.44 \begin{align*} \frac{1}{2} i \, x^{2} - i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + \frac{1}{4} \, x \cos \left (2 \, x\right ) - \frac{1}{2} \, x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) - \frac{1}{8} \, \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(x)^3/cos(x),x, algorithm="maxima")

[Out]

1/2*I*x^2 - I*x*arctan2(sin(2*x), cos(2*x) + 1) + 1/4*x*cos(2*x) - 1/2*x*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2
*x) + 1) + 1/2*I*dilog(-e^(2*I*x)) - 1/8*sin(2*x)

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Fricas [B]  time = 2.0748, size = 432, normalized size = 6.97 \begin{align*} \frac{1}{2} \, x \cos \left (x\right )^{2} - \frac{1}{2} \, x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) - \frac{1}{2} \, x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) - \frac{1}{2} \, x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) - \frac{1}{2} \, x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) - \frac{1}{4} \, \cos \left (x\right ) \sin \left (x\right ) - \frac{1}{4} \, x - \frac{1}{2} i \,{\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + \frac{1}{2} i \,{\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - \frac{1}{2} i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(x)^3/cos(x),x, algorithm="fricas")

[Out]

1/2*x*cos(x)^2 - 1/2*x*log(I*cos(x) + sin(x) + 1) - 1/2*x*log(I*cos(x) - sin(x) + 1) - 1/2*x*log(-I*cos(x) + s
in(x) + 1) - 1/2*x*log(-I*cos(x) - sin(x) + 1) - 1/4*cos(x)*sin(x) - 1/4*x - 1/2*I*dilog(I*cos(x) + sin(x)) +
1/2*I*dilog(I*cos(x) - sin(x)) + 1/2*I*dilog(-I*cos(x) + sin(x)) - 1/2*I*dilog(-I*cos(x) - sin(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(x)**3/cos(x),x)

[Out]

Integral(x*sin(x)**3/cos(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(x)^3/cos(x),x, algorithm="giac")

[Out]

integrate(x*sin(x)^3/cos(x), x)