Integrand size = 10, antiderivative size = 63 \[ \int x^2 \coth (a+b x) \, dx=-\frac {x^3}{3}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {x \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac {\operatorname {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3} \] Output:
-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*po lylog(3,exp(2*b*x+2*a))/b^3
Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.05 \[ \int x^2 \coth (a+b x) \, dx=-\frac {x^3}{3}+\frac {x^2 \log \left (1-e^{2 a+2 b x}\right )}{b}+\frac {x \operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{b^2}-\frac {\operatorname {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3} \] Input:
Integrate[x^2*Coth[a + b*x],x]
Output:
-1/3*x^3 + (x^2*Log[1 - E^(2*a + 2*b*x)])/b + (x*PolyLog[2, E^(2*a + 2*b*x )])/b^2 - PolyLog[3, E^(2*a + 2*b*x)]/(2*b^3)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.68, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 26, 4201, 2620, 3011, 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.
\(\displaystyle \int x^2 \coth (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i x^2 \tan \left (i a+i b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int x^2 \tan \left (\frac {1}{2} (2 i a+\pi )+i b x\right )dx\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle -i \left (2 i \int \frac {e^{2 a+2 b x-i \pi } x^2}{1+e^{2 a+2 b x-i \pi }}dx-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\int x \log \left (1+e^{2 a+2 b x-i \pi }\right )dx}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )\) |
Input:
Int[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))
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_))^(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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(165\) vs. \(2(59)=118\).
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.63
method | result | size |
risch | \(-\frac {x^{3}}{3}+\frac {a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{3}}+\frac {4 a^{3}}{3 b^{3}}+\frac {2 a^{2} x}{b^{2}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}\) | \(166\) |
Input:
int(x^2*coth(b*x+a),x,method=_RETURNVERBOSE)
Output:
-1/3*x^3+1/b^3*a^2*ln(exp(b*x+a)-1)-2/b^3*a^2*ln(exp(b*x+a))-1/b^3*ln(1-ex p(b*x+a))*a^2+4/3/b^3*a^3+2/b^2*a^2*x+1/b*ln(1+exp(b*x+a))*x^2+2/b^2*polyl og(2,-exp(b*x+a))*x+1/b*ln(1-exp(b*x+a))*x^2+2/b^2*polylog(2,exp(b*x+a))*x -2/b^3*polylog(3,-exp(b*x+a))-2/b^3*polylog(3,exp(b*x+a))
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (58) = 116\).
Time = 0.10 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.67 \[ \int x^2 \coth (a+b x) \, dx=-\frac {b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 6 \, b x {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, b x {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{3 \, b^{3}} \] Input:
integrate(x^2*coth(b*x+a),x, algorithm="fricas")
Output:
-1/3*(b^3*x^3 - 3*b^2*x^2*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 6*b*x*d ilog(cosh(b*x + a) + sinh(b*x + a)) - 6*b*x*dilog(-cosh(b*x + a) - sinh(b* x + a)) - 3*a^2*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 3*(b^2*x^2 - a^2) *log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*polylog(3, cosh(b*x + a) + si nh(b*x + a)) + 6*polylog(3, -cosh(b*x + a) - sinh(b*x + a)))/b^3
\[ \int x^2 \coth (a+b x) \, dx=\int x^{2} \coth {\left (a + b x \right )}\, dx \] Input:
integrate(x**2*coth(b*x+a),x)
Output:
Integral(x**2*coth(a + b*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (58) = 116\).
Time = 0.05 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.19 \[ \int x^2 \coth (a+b x) \, dx=\frac {1}{3} \, x^{3} \coth \left (b x + a\right ) - \frac {1}{3} \, {\left (\frac {2 \, x^{3}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} + \frac {2 \, x^{3}}{b} - \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} - \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}}\right )} b \] Input:
integrate(x^2*coth(b*x+a),x, algorithm="maxima")
Output:
1/3*x^3*coth(b*x + a) - 1/3*(2*x^3/(b*e^(2*b*x + 2*a) - b) + 2*x^3/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^4 - 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^4)*b
\[ \int x^2 \coth (a+b x) \, dx=\int { x^{2} \coth \left (b x + a\right ) \,d x } \] Input:
integrate(x^2*coth(b*x+a),x, algorithm="giac")
Output:
integrate(x^2*coth(b*x + a), x)
Timed out. \[ \int x^2 \coth (a+b x) \, dx=\int x^2\,\mathrm {coth}\left (a+b\,x\right ) \,d x \] Input:
int(x^2*coth(a + b*x),x)
Output:
int(x^2*coth(a + b*x), x)
\[ \int x^2 \coth (a+b x) \, dx=2 e^{2 a} \left (\int \frac {e^{2 b x} x^{2}}{e^{2 b x +2 a}-1}d x \right )-\frac {x^{3}}{3} \] Input:
int(x^2*coth(b*x+a),x)
Output:
(6*e**(2*a)*int((e**(2*b*x)*x**2)/(e**(2*a + 2*b*x) - 1),x) - x**3)/3