Integrand size = 10, antiderivative size = 87 \[ \int x^3 \coth (a+b x) \, dx=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \operatorname {PolyLog}\left (4,e^{2 (a+b x)}\right )}{4 b^4} \]
-1/4*x^4+x^3*ln(1-exp(2*b*x+2*a))/b+3/2*x^2*polylog(2,exp(2*b*x+2*a))/b^2- 3/2*x*polylog(3,exp(2*b*x+2*a))/b^3+3/4*polylog(4,exp(2*b*x+2*a))/b^4
Time = 0.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int x^3 \coth (a+b x) \, dx=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 a+2 b x}\right )}{b}+\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}+\frac {3 \operatorname {PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4} \]
-1/4*x^4 + (x^3*Log[1 - E^(2*a + 2*b*x)])/b + (3*x^2*PolyLog[2, E^(2*a + 2 *b*x)])/(2*b^2) - (3*x*PolyLog[3, E^(2*a + 2*b*x)])/(2*b^3) + (3*PolyLog[4 , E^(2*a + 2*b*x)])/(4*b^4)
Result contains complex when optimal does not.
Time = 0.65 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.63, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 26, 4201, 2620, 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.
\(\displaystyle \int x^3 \coth (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i x^3 \tan \left (i a+i b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int x^3 \tan \left (\frac {1}{2} (2 i a+\pi )+i b x\right )dx\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle -i \left (2 i \int \frac {e^{2 a+2 b x-i \pi } x^3}{1+e^{2 a+2 b x-i \pi }}dx-\frac {i x^4}{4}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -i \left (2 i \left (\frac {x^3 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {3 \int x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )dx}{2 b}\right )-\frac {i x^4}{4}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -i \left (2 i \left (\frac {x^3 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}\right )}{2 b}\right )-\frac {i x^4}{4}\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle -i \left (2 i \left (\frac {x^3 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}\right )}{2 b}\right )-\frac {i x^4}{4}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -i \left (2 i \left (\frac {x^3 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}\right )}{2 b}\right )-\frac {i x^4}{4}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -i \left (2 i \left (\frac {x^3 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,-e^{2 a+2 b x-i \pi }\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}\right )}{2 b}\right )-\frac {i x^4}{4}\right )\) |
(-I)*((-1/4*I)*x^4 + (2*I)*((x^3*Log[1 + E^(2*a - I*Pi + 2*b*x)])/(2*b) - (3*(-1/2*(x^2*PolyLog[2, -E^(2*a - I*Pi + 2*b*x)])/b + ((x*PolyLog[3, -E^( 2*a - I*Pi + 2*b*x)])/(2*b) - PolyLog[4, -E^(2*a - I*Pi + 2*b*x)]/(4*b^2)) /b))/(2*b)))
3.4.98.3.1 Defintions of rubi rules used
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(199\) vs. \(2(79)=158\).
Time = 0.32 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.30
method | result | size |
risch | \(-\frac {x^{4}}{4}+\frac {3 x^{2} \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{b}+\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}-\frac {3 a^{4}}{2 b^{4}}-\frac {2 a^{3} x}{b^{3}}+\frac {6 \operatorname {polylog}\left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 \operatorname {polylog}\left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}\) | \(200\) |
-1/4*x^4+3*x^2*polylog(2,exp(b*x+a))/b^2-6*x*polylog(3,exp(b*x+a))/b^3+1/b *ln(exp(b*x+a)+1)*x^3+3*x^2*polylog(2,-exp(b*x+a))/b^2-6*x*polylog(3,-exp( b*x+a))/b^3+1/b*ln(1-exp(b*x+a))*x^3-3/2/b^4*a^4-2/b^3*a^3*x+6*polylog(4,- exp(b*x+a))/b^4+6*polylog(4,exp(b*x+a))/b^4+1/b^4*ln(1-exp(b*x+a))*a^3+2/b ^4*a^3*ln(exp(b*x+a))-1/b^4*a^3*ln(exp(b*x+a)-1)
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (78) = 156\).
Time = 0.26 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.48 \[ \int x^3 \coth (a+b x) \, dx=-\frac {b^{4} x^{4} - 4 \, b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 12 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 12 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 24 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 24 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 24 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 24 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{4}} \]
-1/4*(b^4*x^4 - 4*b^3*x^3*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 12*b^2* x^2*dilog(cosh(b*x + a) + sinh(b*x + a)) - 12*b^2*x^2*dilog(-cosh(b*x + a) - sinh(b*x + a)) + 4*a^3*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 24*b*x* polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 24*b*x*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) - 4*(b^3*x^3 + a^3)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - 24*polylog(4, cosh(b*x + a) + sinh(b*x + a)) - 24*polylog(4, -cosh (b*x + a) - sinh(b*x + a)))/b^4
\[ \int x^3 \coth (a+b x) \, dx=\int x^{3} \cosh {\left (a + b x \right )} \operatorname {csch}{\left (a + b x \right )}\, dx \]
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.49 \[ \int x^3 \coth (a+b x) \, dx=-\frac {1}{4} \, x^{4} + \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} \]
-1/4*x^4 + (b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*b*x*polylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^4 + (b^3*x ^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*dilog(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)))/b^4
\[ \int x^3 \coth (a+b x) \, dx=\int { x^{3} \cosh \left (b x + a\right ) \operatorname {csch}\left (b x + a\right ) \,d x } \]
Timed out. \[ \int x^3 \coth (a+b x) \, dx=\int \frac {x^3\,\mathrm {cosh}\left (a+b\,x\right )}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]