Integrand size = 18, antiderivative size = 130 \[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\frac {x^2}{4 b}+\frac {x^3}{3}-\frac {x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}+\frac {\operatorname {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}-\frac {x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac {\sinh ^2(a+b x)}{4 b^3}+\frac {x^2 \sinh ^2(a+b x)}{2 b} \]
1/4*x^2/b+1/3*x^3-x^2*ln(1+exp(2*b*x+2*a))/b-x*polylog(2,-exp(2*b*x+2*a))/ b^2+1/2*polylog(3,-exp(2*b*x+2*a))/b^3-1/2*x*cosh(b*x+a)*sinh(b*x+a)/b^2+1 /4*sinh(b*x+a)^2/b^3+1/2*x^2*sinh(b*x+a)^2/b
Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.94 \[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\frac {\cosh (a) (\cosh (a)+\sinh (a)) \left (-8 b^3 x^3+3 \cosh (2 (a+b x))+6 b^2 x^2 \cosh (2 (a+b x))-24 b^2 x^2 \log \left (1+e^{-2 (a+b x)}\right )+24 b x \operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )+12 \operatorname {PolyLog}\left (3,-e^{-2 (a+b x)}\right )-6 b x \sinh (2 (a+b x))\right )}{12 b^3 \left (1+e^{2 a}\right )} \]
(Cosh[a]*(Cosh[a] + Sinh[a])*(-8*b^3*x^3 + 3*Cosh[2*(a + b*x)] + 6*b^2*x^2 *Cosh[2*(a + b*x)] - 24*b^2*x^2*Log[1 + E^(-2*(a + b*x))] + 24*b*x*PolyLog [2, -E^(-2*(a + b*x))] + 12*PolyLog[3, -E^(-2*(a + b*x))] - 6*b*x*Sinh[2*( a + b*x)]))/(12*b^3*(1 + E^(2*a)))
Result contains complex when optimal does not.
Time = 0.89 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.18, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {5972, 3042, 26, 4201, 2620, 3011, 2720, 5895, 3042, 25, 3791, 15, 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.
| \(\displaystyle \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx\) |
| \(\Big \downarrow \) 5972 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-\int x^2 \tanh (a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-\int -i x^2 \tan (i a+i b x)dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+i \int x^2 \tan (i a+i b x)dx\) |
| \(\Big \downarrow \) 4201 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+i \left (2 i \int \frac {e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}}dx-\frac {i x^3}{3}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\int x \log \left (1+e^{2 (a+b x)}\right )dx}{b}\right )-\frac {i x^3}{3}\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
| \(\Big \downarrow \) 5895 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )-\frac {\int x \sinh ^2(a+b x)dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )-\frac {\int -x \sin (i a+i b x)^2dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\int x \sin (i a+i b x)^2dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3791 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\int xdx}{2}+\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x^2}{4}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\operatorname {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x^2}{4}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
I*((-1/3*I)*x^3 + (2*I)*((x^2*Log[1 + E^(2*(a + b*x))])/(2*b) - (-1/2*(x*P olyLog[2, -E^(2*(a + b*x))])/b + PolyLog[3, -E^(2*(a + b*x))]/(4*b^2))/b)) + (x^2*Sinh[a + b*x]^2)/(2*b) + (x^2/4 - (x*Cosh[a + b*x]*Sinh[a + b*x])/ (2*b) + Sinh[a + b*x]^2/(4*b^2))/b
3.4.78.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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[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[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(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*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.) ]^(p_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Simp[(m - n + 1)/(b*n*(p + 1)) Int[x^(m - n)*Sinh[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_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b* x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
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]
Time = 1.65 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.17
| method | result | size |
| risch | \(\frac {x^{3}}{3}+\frac {\left (2 x^{2} b^{2}-2 b x +1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{3}}+\frac {\left (2 x^{2} b^{2}+2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 a^{2} x}{b^{2}}-\frac {4 a^{3}}{3 b^{3}}-\frac {x^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}\) | \(152\) |
1/3*x^3+1/16*(2*b^2*x^2-2*b*x+1)/b^3*exp(2*b*x+2*a)+1/16*(2*b^2*x^2+2*b*x+ 1)/b^3*exp(-2*b*x-2*a)+2/b^3*a^2*ln(exp(b*x+a))-2/b^2*a^2*x-4/3/b^3*a^3-x^ 2*ln(1+exp(2*b*x+2*a))/b-x*polylog(2,-exp(2*b*x+2*a))/b^2+1/2*polylog(3,-e xp(2*b*x+2*a))/b^3
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 789, normalized size of antiderivative = 6.07 \[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\text {Too large to display} \]
1/48*(3*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^4 + 12*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + 3*(2*b^2*x^2 - 2*b*x + 1)*sinh(b*x + a) ^4 + 6*b^2*x^2 + 16*(b^3*x^3 + 2*a^3)*cosh(b*x + a)^2 + 2*(8*b^3*x^3 + 16* a^3 + 9*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 6*b*x - 96*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b* x + a)^2)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 96*(b*x*cosh(b*x + a) ^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2)*dilog(-I*cos h(b*x + a) - I*sinh(b*x + a)) - 48*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) + I) - 48*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2* sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 48*((b^2*x^2 - a ^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a) + (b^2 *x^2 - a^2)*sinh(b*x + a)^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 48*((b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh (b*x + a) + (b^2*x^2 - a^2)*sinh(b*x + a)^2)*log(-I*cosh(b*x + a) - I*sinh (b*x + a) + 1) + 96*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sin h(b*x + a)^2)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 96*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(3, -I*c osh(b*x + a) - I*sinh(b*x + a)) + 4*(3*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^3 + 8*(b^3*x^3 + 2*a^3)*cosh(b*x + a))*sinh(b*x + a) + 3)/(b^3*cosh(...
\[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\int x^{2} \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06 \[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\frac {2}{3} \, x^{3} - \frac {{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \, {\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 3 \, {\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, b^{3}} - \frac {2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} \]
2/3*x^3 - 1/48*(16*b^3*x^3*e^(2*a) - 3*(2*b^2*x^2*e^(4*a) - 2*b*x*e^(4*a) + e^(4*a))*e^(2*b*x) - 3*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x))*e^(-2*a)/b^3 - 1/2*(2*b^2*x^2*log(e^(2*b*x + 2*a) + 1) + 2*b*x*dilog(-e^(2*b*x + 2*a)) - polylog(3, -e^(2*b*x + 2*a)))/b^3
\[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\int { x^{2} \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx=\int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]