Integrand size = 18, antiderivative size = 126 \[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\frac {x^2}{4 b}-\frac {x^3}{3}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {x \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac {\operatorname {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}-\frac {x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac {\sinh ^2(a+b x)}{4 b^3}+\frac {x^2 \sinh ^2(a+b x)}{2 b} \] Output:
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 = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.41 \[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\frac {\sinh (a) (\cosh (a)+\sinh (a)) \left (8 b^3 x^3+3 \cosh (2 (a+b x))+6 b^2 x^2 \cosh (2 (a+b x))+24 b^2 x^2 \log \left (1-e^{-a-b x}\right )+24 b^2 x^2 \log \left (1+e^{-a-b x}\right )-48 b x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )-48 b x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )-48 \operatorname {PolyLog}\left (3,-e^{-a-b x}\right )-48 \operatorname {PolyLog}\left (3,e^{-a-b x}\right )-6 b x \sinh (2 (a+b x))\right )}{12 b^3 \left (-1+e^{2 a}\right )} \] Input:
Integrate[x^2*Cosh[a + b*x]^2*Coth[a + b*x],x]
Output:
(Sinh[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^(-a - b*x)] + 24*b^2*x^2*Log[1 + E^(-a - b*x)] - 48*b*x*PolyLog[2, -E^(-a - b*x)] - 48*b*x*PolyLog[2, E^(-a - b*x)] - 48*PolyLog[3, -E^(-a - b*x)] - 48*PolyLog[3, E^(-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 = 0.98 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.37, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {5973, 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.
\(\displaystyle \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx\) |
\(\Big \downarrow \) 5973 |
\(\displaystyle \int x^2 \coth (a+b x)dx+\int x^2 \cosh (a+b x) \sinh (a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx+\int -i x^2 \tan \left (i a+i b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-i \int x^2 \tan \left (\frac {1}{2} (2 i a+\pi )+i b x\right )dx\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-i \left (2 i \int \frac {e^{2 a+2 b x-i \pi } x^2}{1+e^{2 a+2 b x-i \pi }}dx-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\int x \log \left (1+e^{2 a+2 b x-i \pi }\right )dx}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int x^2 \cosh (a+b x) \sinh (a+b x)dx-i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 5895 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )-\frac {\int x \sinh ^2(a+b x)dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )-\frac {\int -x \sin (i a+i b x)^2dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\int x \sin (i a+i b x)^2dx}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\int xdx}{2}+\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x^2}{4}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )+\frac {\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x^2}{4}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
Input:
Int[x^2*Cosh[a + b*x]^2*Coth[a + b*x],x]
Output:
(-I)*((-1/3*I)*x^3 + (2*I)*((x^2*Log[1 + E^(2*a - I*Pi + 2*b*x)])/(2*b) - (-1/2*(x*PolyLog[2, -E^(2*a - I*Pi + 2*b*x)])/b + PolyLog[3, -E^(2*a - I*P i + 2*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
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[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b* x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[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 = 1.31 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.76
method | result | size |
risch | \(-\frac {x^{3}}{3}+\frac {\left (2 b^{2} x^{2}-2 b x +1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{3}}+\frac {\left (2 b^{2} x^{2}+2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{3}}+\frac {2 a^{2} x}{b^{2}}+\frac {4 a^{3}}{3 b^{3}}-\frac {2 \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{2}}{b}+\frac {2 x \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {2 x \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}\) | \(222\) |
Input:
int(x^2*cosh(b*x+a)^2*coth(b*x+a),x,method=_RETURNVERBOSE)
Output:
-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^2*a^2*x+4/3/b^3*a^3-2*polylog(3,-exp(b*x+a))/b ^3-2*polylog(3,exp(b*x+a))/b^3+1/b*ln(1-exp(b*x+a))*x^2-2/b^3*a^2*ln(exp(b *x+a))-1/b^3*ln(1-exp(b*x+a))*a^2+1/b^3*a^2*ln(exp(b*x+a)-1)+1/b*ln(exp(b* x+a)+1)*x^2+2*x*polylog(2,-exp(b*x+a))/b^2+2*x*polylog(2,exp(b*x+a))/b^2
Leaf count of result is larger than twice the leaf count of optimal. 697 vs. \(2 (113) = 226\).
Time = 0.11 (sec) , antiderivative size = 697, normalized size of antiderivative = 5.53 \[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx =\text {Too large to display} \] Input:
integrate(x^2*cosh(b*x+a)^2*coth(b*x+a),x, algorithm="fricas")
Output:
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(cosh(b*x + a) + 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(-cosh(b*x + a) - sinh(b*x + a)) + 48*(b^2*x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 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) - 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(-cosh(b*x + a) - sinh(b*x + a) + 1) - 96*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*po lylog(3, cosh(b*x + a) + 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, -cosh(b*x + a) - 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(b*x + a)^2 + 2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*x + a)^2)
\[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\int x^{2} \cosh ^{2}{\left (a + b x \right )} \coth {\left (a + b x \right )}\, dx \] Input:
integrate(x**2*cosh(b*x+a)**2*coth(b*x+a),x)
Output:
Integral(x**2*cosh(a + b*x)**2*coth(a + b*x), x)
Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.36 \[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=-\frac {2}{3} \, x^{3} + \frac {{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} + 3 \, {\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 3 \, {\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, b^{3}} + \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}} \] Input:
integrate(x^2*cosh(b*x+a)^2*coth(b*x+a),x, algorithm="maxima")
Output:
-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 + (b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylog(3 , -e^(b*x + a)))/b^3 + (b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b^3
\[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\int { x^{2} \cosh \left (b x + a\right )^{2} \coth \left (b x + a\right ) \,d x } \] Input:
integrate(x^2*cosh(b*x+a)^2*coth(b*x+a),x, algorithm="giac")
Output:
integrate(x^2*cosh(b*x + a)^2*coth(b*x + a), x)
Timed out. \[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\int x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {coth}\left (a+b\,x\right ) \,d x \] Input:
int(x^2*cosh(a + b*x)^2*coth(a + b*x),x)
Output:
int(x^2*cosh(a + b*x)^2*coth(a + b*x), x)
\[ \int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx=\frac {6 e^{4 b x +4 a} b^{2} x^{2}-6 e^{4 b x +4 a} b x +3 e^{4 b x +4 a}+96 e^{2 b x +2 a} \left (\int \frac {x^{2}}{e^{4 b x +4 a}-e^{2 b x +2 a}}d x \right ) b^{3}+16 e^{2 b x +2 a} b^{3} x^{3}-42 b^{2} x^{2}-42 b x -21}{48 e^{2 b x +2 a} b^{3}} \] Input:
int(x^2*cosh(b*x+a)^2*coth(b*x+a),x)
Output:
(6*e**(4*a + 4*b*x)*b**2*x**2 - 6*e**(4*a + 4*b*x)*b*x + 3*e**(4*a + 4*b*x ) + 96*e**(2*a + 2*b*x)*int(x**2/(e**(4*a + 4*b*x) - e**(2*a + 2*b*x)),x)* b**3 + 16*e**(2*a + 2*b*x)*b**3*x**3 - 42*b**2*x**2 - 42*b*x - 21)/(48*e** (2*a + 2*b*x)*b**3)