Integrand size = 7, antiderivative size = 108 \[ \int x^2 \arctan (\sinh (x)) \, dx=-\frac {2}{3} x^3 \arctan \left (e^x\right )+\frac {1}{3} x^3 \arctan (\sinh (x))+i x^2 \operatorname {PolyLog}\left (2,-i e^x\right )-i x^2 \operatorname {PolyLog}\left (2,i e^x\right )-2 i x \operatorname {PolyLog}\left (3,-i e^x\right )+2 i x \operatorname {PolyLog}\left (3,i e^x\right )+2 i \operatorname {PolyLog}\left (4,-i e^x\right )-2 i \operatorname {PolyLog}\left (4,i e^x\right ) \]
-2/3*x^3*arctan(exp(x))+1/3*x^3*arctan(sinh(x))+I*x^2*polylog(2,-I*exp(x)) -I*x^2*polylog(2,I*exp(x))-2*I*x*polylog(3,-I*exp(x))+2*I*x*polylog(3,I*ex p(x))+2*I*polylog(4,-I*exp(x))-2*I*polylog(4,I*exp(x))
Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int x^2 \arctan (\sinh (x)) \, dx=\frac {1}{3} x^3 \arctan (\sinh (x))-\frac {1}{3} i \left (x^3 \log \left (1-i e^x\right )-x^3 \log \left (1+i e^x\right )-3 x^2 \operatorname {PolyLog}\left (2,-i e^x\right )+3 x^2 \operatorname {PolyLog}\left (2,i e^x\right )+6 x \operatorname {PolyLog}\left (3,-i e^x\right )-6 x \operatorname {PolyLog}\left (3,i e^x\right )-6 \operatorname {PolyLog}\left (4,-i e^x\right )+6 \operatorname {PolyLog}\left (4,i e^x\right )\right ) \]
(x^3*ArcTan[Sinh[x]])/3 - (I/3)*(x^3*Log[1 - I*E^x] - x^3*Log[1 + I*E^x] - 3*x^2*PolyLog[2, (-I)*E^x] + 3*x^2*PolyLog[2, I*E^x] + 6*x*PolyLog[3, (-I )*E^x] - 6*x*PolyLog[3, I*E^x] - 6*PolyLog[4, (-I)*E^x] + 6*PolyLog[4, I*E ^x])
Time = 0.59 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5728, 3042, 4668, 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.
\(\displaystyle \int x^2 \arctan (\sinh (x)) \, dx\) |
\(\Big \downarrow \) 5728 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))-\frac {1}{3} \int x^3 \text {sech}(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))-\frac {1}{3} \int x^3 \csc \left (i x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 4668 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))+\frac {1}{3} \left (3 i \int x^2 \log \left (1-i e^x\right )dx-3 i \int x^2 \log \left (1+i e^x\right )dx-2 x^3 \arctan \left (e^x\right )\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))+\frac {1}{3} \left (-3 i \left (2 \int x \operatorname {PolyLog}\left (2,-i e^x\right )dx-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )+3 i \left (2 \int x \operatorname {PolyLog}\left (2,i e^x\right )dx-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-2 x^3 \arctan \left (e^x\right )\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))+\frac {1}{3} \left (-3 i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\int \operatorname {PolyLog}\left (3,-i e^x\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )+3 i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\int \operatorname {PolyLog}\left (3,i e^x\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-2 x^3 \arctan \left (e^x\right )\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))+\frac {1}{3} \left (-3 i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,-i e^x\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )+3 i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,i e^x\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-2 x^3 \arctan \left (e^x\right )\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{3} x^3 \arctan (\sinh (x))+\frac {1}{3} \left (-2 x^3 \arctan \left (e^x\right )-3 i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\operatorname {PolyLog}\left (4,-i e^x\right )\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )+3 i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\operatorname {PolyLog}\left (4,i e^x\right )\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )\right )\) |
(x^3*ArcTan[Sinh[x]])/3 + (-2*x^3*ArcTan[E^x] - (3*I)*(-(x^2*PolyLog[2, (- I)*E^x]) + 2*(x*PolyLog[3, (-I)*E^x] - PolyLog[4, (-I)*E^x])) + (3*I)*(-(x ^2*PolyLog[2, I*E^x]) + 2*(x*PolyLog[3, I*E^x] - PolyLog[4, I*E^x])))/3
3.1.75.3.1 Defintions of rubi rules used
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]]]
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]
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ ))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[ 1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c , d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcTan[u])/(d*(m + 1))), x] - Simp[b/(d*(m + 1) ) Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/(1 + u^2)), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] & & !FunctionOfQ[(c + d*x)^(m + 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.74 (sec) , antiderivative size = 658, normalized size of antiderivative = 6.09
1/3*I*x^3*ln(exp(x)+I)-1/3*I*x^3*ln(exp(x)-I)+1/3*I*x^3*ln(1+I*exp(x))+I*x ^2*polylog(2,-I*exp(x))-2*I*x*polylog(3,-I*exp(x))+2*I*polylog(4,-I*exp(x) )-1/12*Pi*(csgn(I*(exp(x)-I))^2*csgn(I*(exp(x)-I)^2)-2*csgn(I*(exp(x)-I))* csgn(I*(exp(x)-I)^2)^2+csgn(I*(exp(x)-I)^2)^3+csgn(I*(exp(x)-I)^2)*csgn(I* exp(-x))*csgn(I*exp(-x)*(exp(x)-I)^2)-csgn(I*(exp(x)-I)^2)*csgn(I*exp(-x)* (exp(x)-I)^2)^2-csgn(I*(exp(x)+I))^2*csgn(I*(exp(x)+I)^2)+2*csgn(I*(exp(x) +I))*csgn(I*(exp(x)+I)^2)^2-csgn(I*(exp(x)+I)^2)^3-csgn(I*(exp(x)+I)^2)*cs gn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)+I)^2)+csgn(I*(exp(x)+I)^2)*csgn(I*exp (-x)*(exp(x)+I)^2)^2-csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-I)^2)^2+csgn(I *exp(-x))*csgn(I*exp(-x)*(exp(x)+I)^2)^2-csgn(I*exp(-x)*(exp(x)+I)^2)*csgn (exp(-x)*(exp(x)+I)^2)+csgn(exp(-x)*(exp(x)+I)^2)^2+csgn(I*exp(-x)*(exp(x) -I)^2)*csgn(exp(-x)*(exp(x)-I)^2)+csgn(exp(-x)*(exp(x)-I)^2)^2+csgn(I*exp( -x)*(exp(x)-I)^2)^3-csgn(I*exp(-x)*(exp(x)-I)^2)*csgn(exp(-x)*(exp(x)-I)^2 )^2-csgn(I*exp(-x)*(exp(x)+I)^2)^3+csgn(I*exp(-x)*(exp(x)+I)^2)*csgn(exp(- x)*(exp(x)+I)^2)^2-csgn(exp(-x)*(exp(x)+I)^2)^3-csgn(exp(-x)*(exp(x)-I)^2) ^3-2)*x^3-1/3*I*x^3*ln(1-I*exp(x))-I*x^2*polylog(2,I*exp(x))+2*I*x*polylog (3,I*exp(x))-2*I*polylog(4,I*exp(x))
Time = 0.36 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.16 \[ \int x^2 \arctan (\sinh (x)) \, dx=\frac {1}{3} \, x^{3} \arctan \left (\sinh \left (x\right )\right ) + \frac {1}{3} i \, x^{3} \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - \frac {1}{3} i \, x^{3} \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - i \, x^{2} {\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + i \, x^{2} {\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) + 2 i \, x {\rm polylog}\left (3, i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) - 2 i \, x {\rm polylog}\left (3, -i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) - 2 i \, {\rm polylog}\left (4, i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + 2 i \, {\rm polylog}\left (4, -i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) \]
1/3*x^3*arctan(sinh(x)) + 1/3*I*x^3*log(I*cosh(x) + I*sinh(x) + 1) - 1/3*I *x^3*log(-I*cosh(x) - I*sinh(x) + 1) - I*x^2*dilog(I*cosh(x) + I*sinh(x)) + I*x^2*dilog(-I*cosh(x) - I*sinh(x)) + 2*I*x*polylog(3, I*cosh(x) + I*sin h(x)) - 2*I*x*polylog(3, -I*cosh(x) - I*sinh(x)) - 2*I*polylog(4, I*cosh(x ) + I*sinh(x)) + 2*I*polylog(4, -I*cosh(x) - I*sinh(x))
\[ \int x^2 \arctan (\sinh (x)) \, dx=\int x^{2} \operatorname {atan}{\left (\sinh {\left (x \right )} \right )}\, dx \]
\[ \int x^2 \arctan (\sinh (x)) \, dx=\int { x^{2} \arctan \left (\sinh \left (x\right )\right ) \,d x } \]
\[ \int x^2 \arctan (\sinh (x)) \, dx=\int { x^{2} \arctan \left (\sinh \left (x\right )\right ) \,d x } \]
Timed out. \[ \int x^2 \arctan (\sinh (x)) \, dx=\int x^2\,\mathrm {atan}\left (\mathrm {sinh}\left (x\right )\right ) \,d x \]