Integrand size = 8, antiderivative size = 62 \[ \int x \sin ^2(x) \tan (x) \, dx=\frac {x}{4}+\frac {i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {1}{4} \cos (x) \sin (x)-\frac {1}{2} x \sin ^2(x) \]
1/4*x+1/2*I*x^2-x*ln(1+exp(2*I*x))+1/2*I*polylog(2,-exp(2*I*x))-1/4*cos(x) *sin(x)-1/2*x*sin(x)^2
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int x \sin ^2(x) \tan (x) \, dx=\frac {i x^2}{2}+\frac {1}{4} x \cos (2 x)-x \log \left (1+e^{2 i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {1}{8} \sin (2 x) \]
(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
Time = 0.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {4907, 3042, 3924, 3042, 3115, 24, 4202, 2620, 2715, 2838}
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.
\(\displaystyle \int x \sin ^2(x) \tan (x) \, dx\) |
\(\Big \downarrow \) 4907 |
\(\displaystyle \int x \tan (x)dx-\int x \cos (x) \sin (x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x \tan (x)dx-\int x \cos (x) \sin (x)dx\) |
\(\Big \downarrow \) 3924 |
\(\displaystyle \frac {1}{2} \int \sin ^2(x)dx+\int x \tan (x)dx-\frac {1}{2} x \sin ^2(x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \sin (x)^2dx+\int x \tan (x)dx-\frac {1}{2} x \sin ^2(x)\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \int x \tan (x)dx+\frac {1}{2} \left (\frac {\int 1dx}{2}-\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{2} x \sin ^2(x)\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \int x \tan (x)dx-\frac {1}{2} x \sin ^2(x)+\frac {1}{2} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle -2 i \int \frac {e^{2 i x} x}{1+e^{2 i x}}dx+\frac {i x^2}{2}-\frac {1}{2} x \sin ^2(x)+\frac {1}{2} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -2 i \left (\frac {1}{2} i \int \log \left (1+e^{2 i x}\right )dx-\frac {1}{2} i x \log \left (1+e^{2 i x}\right )\right )+\frac {i x^2}{2}-\frac {1}{2} x \sin ^2(x)+\frac {1}{2} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -2 i \left (\frac {1}{4} \int e^{-2 i x} \log \left (1+e^{2 i x}\right )de^{2 i x}-\frac {1}{2} i x \log \left (1+e^{2 i x}\right )\right )+\frac {i x^2}{2}-\frac {1}{2} x \sin ^2(x)+\frac {1}{2} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 i \left (-\frac {1}{4} \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {1}{2} i x \log \left (1+e^{2 i x}\right )\right )+\frac {i x^2}{2}-\frac {1}{2} x \sin ^2(x)+\frac {1}{2} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )\) |
(I/2)*x^2 - (2*I)*((-1/2*I)*x*Log[1 + E^((2*I)*x)] - PolyLog[2, -E^((2*I)* x)]/4) - (x*Sin[x]^2)/2 + (x/2 - (Cos[x]*Sin[x])/2)/2
3.5.90.3.1 Defintions of rubi rules used
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]
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]
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]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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] - Simp[(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]
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]
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] /; Fr eeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {i x^{2}}{2}+\frac {\left (i+2 x \right ) {\mathrm e}^{2 i x}}{16}+\frac {\left (-i+2 x \right ) {\mathrm e}^{-2 i x}}{16}-x \ln \left ({\mathrm e}^{2 i x}+1\right )+\frac {i \operatorname {Li}_{2}\left (-{\mathrm e}^{2 i x}\right )}{2}\) | \(57\) |
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(exp(2*I*x )+1)+1/2*I*polylog(2,-exp(2*I*x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (41) = 82\).
Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.82 \[ \int x \sin ^2(x) \tan (x) \, dx=\frac {1}{2} \, x \cos \left (x\right )^{2} - \frac {1}{2} \, x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, \cos \left (x\right ) \sin \left (x\right ) - \frac {1}{4} \, x - \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) \]
1/2*x*cos(x)^2 - 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) - 1/2*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*dil og(-I*cos(x) - sin(x))
\[ \int x \sin ^2(x) \tan (x) \, dx=\int \frac {x \sin ^{3}{\left (x \right )}}{\cos {\left (x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int x \sin ^2(x) \tan (x) \, dx=\frac {1}{2} i \, x^{2} - i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{4} \, x \cos \left (2 \, x\right ) - \frac {1}{2} \, x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) - \frac {1}{8} \, \sin \left (2 \, x\right ) \]
1/2*I*x^2 - I*x*arctan2(sin(2*x), cos(2*x) + 1) + 1/4*x*cos(2*x) - 1/2*x*l og(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)
\[ \int x \sin ^2(x) \tan (x) \, dx=\int { \frac {x \sin \left (x\right )^{3}}{\cos \left (x\right )} \,d x } \]
Timed out. \[ \int x \sin ^2(x) \tan (x) \, dx=\int \frac {x\,{\sin \left (x\right )}^3}{\cos \left (x\right )} \,d x \]
\[ \int x \sin ^2(x) \tan (x) \, dx=\int \frac {\sin \left (x \right )^{3} x}{\cos \left (x \right )}d x \]