3.6.21 \(\int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx\) [521]

3.6.21.1 Optimal result
3.6.21.2 Mathematica [A] (verified)
3.6.21.3 Rubi [A] (verified)
3.6.21.4 Maple [C] (verified)
3.6.21.5 Fricas [C] (verification not implemented)
3.6.21.6 Sympy [F]
3.6.21.7 Maxima [F]
3.6.21.8 Giac [A] (verification not implemented)
3.6.21.9 Mupad [B] (verification not implemented)

3.6.21.1 Optimal result

Integrand size = 15, antiderivative size = 152 \[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sinh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\text {csch}^2(x)}{2 a}+\frac {\log (\sinh (x))}{a}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sinh (x)\right )}{3 a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sinh (x)+b^{2/3} \sinh ^2(x)\right )}{6 a^{5/3}}-\frac {\log \left (a+b \sinh ^3(x)\right )}{3 a} \]

output
-1/2*csch(x)^2/a+ln(sinh(x))/a-1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*sinh(x))/a^( 
5/3)+1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*sinh(x)+b^(2/3)*sinh(x)^2)/a^( 
5/3)-1/3*ln(a+b*sinh(x)^3)/a+1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*sin 
h(x))/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)
 
3.6.21.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89 \[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=-\frac {\text {csch}^2(x)}{2 a}+\frac {\log (\sinh (x))}{a}-\frac {\left (a^{2/3}+(-1)^{2/3} b^{2/3}\right ) \log \left (-(-1)^{2/3} \sqrt [3]{a}-\sqrt [3]{b} \sinh (x)\right )+\left (a^{2/3}+b^{2/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sinh (x)\right )+\left (a^{2/3}-\sqrt [3]{-1} b^{2/3}\right ) \log \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sinh (x)\right )}{3 a^{5/3}} \]

input
Integrate[Coth[x]^3/(a + b*Sinh[x]^3),x]
 
output
-1/2*Csch[x]^2/a + Log[Sinh[x]]/a - ((a^(2/3) + (-1)^(2/3)*b^(2/3))*Log[-( 
(-1)^(2/3)*a^(1/3)) - b^(1/3)*Sinh[x]] + (a^(2/3) + b^(2/3))*Log[a^(1/3) + 
 b^(1/3)*Sinh[x]] + (a^(2/3) - (-1)^(1/3)*b^(2/3))*Log[a^(1/3) + (-1)^(2/3 
)*b^(1/3)*Sinh[x]])/(3*a^(5/3))
 
3.6.21.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 3709, 2373, 2009}

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.

\(\displaystyle \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i}{\tan (i x)^3 \left (a+i b \sin (i x)^3\right )}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {1}{\left (i b \sin (i x)^3+a\right ) \tan (i x)^3}dx\)

\(\Big \downarrow \) 3709

\(\displaystyle \int \frac {\left (\sinh ^2(x)+1\right ) \text {csch}^3(x)}{a+b \sinh ^3(x)}d\sinh (x)\)

\(\Big \downarrow \) 2373

\(\displaystyle \int \left (\frac {-b \sinh ^2(x)-b}{a \left (a+b \sinh ^3(x)\right )}+\frac {\text {csch}^3(x)}{a}+\frac {\text {csch}(x)}{a}\right )d\sinh (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sinh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sinh (x)+b^{2/3} \sinh ^2(x)\right )}{6 a^{5/3}}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sinh (x)\right )}{3 a^{5/3}}-\frac {\log \left (a+b \sinh ^3(x)\right )}{3 a}-\frac {\text {csch}^2(x)}{2 a}+\frac {\log (\sinh (x))}{a}\)

input
Int[Coth[x]^3/(a + b*Sinh[x]^3),x]
 
output
(b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sinh[x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3] 
*a^(5/3)) - Csch[x]^2/(2*a) + Log[Sinh[x]]/a - (b^(2/3)*Log[a^(1/3) + b^(1 
/3)*Sinh[x]])/(3*a^(5/3)) + (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sinh[x] 
 + b^(2/3)*Sinh[x]^2])/(6*a^(5/3)) - Log[a + b*Sinh[x]^3]/(3*a)
 

3.6.21.3.1 Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2373
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[E 
xpandIntegrand[(c*x)^m*(Pq/(a + b*x^n)), x], x] /; FreeQ[{a, b, c, m}, x] & 
& PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3709
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si 
mp[ff^(m + 1)/f   Subst[Int[x^m*((a + b*(c*ff*x)^n)^p/(1 - ff^2*x^2)^((m + 
1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && 
ILtQ[(m - 1)/2, 0]
 
3.6.21.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 5.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.60

method result size
risch \(-\frac {2 \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2} a}+\frac {\ln \left ({\mathrm e}^{2 x}-1\right )}{a}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 a^{5} \textit {\_Z}^{3}+27 a^{4} \textit {\_Z}^{2}+9 a^{3} \textit {\_Z} +a^{2}+b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 x}+\left (-\frac {6 a^{2} \textit {\_R}}{b}-\frac {2 a}{b}\right ) {\mathrm e}^{x}-1\right )\right )\) \(91\)
default \(-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8 a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (-\textit {\_R}^{5} a -\textit {\_R}^{4} b +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b -\textit {\_R} a +b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}}{3 a}-\frac {1}{8 a \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a}\) \(132\)

input
int(coth(x)^3/(a+b*sinh(x)^3),x,method=_RETURNVERBOSE)
 
output
-2*exp(2*x)/(exp(2*x)-1)^2/a+1/a*ln(exp(2*x)-1)+sum(_R*ln(exp(2*x)+(-6*a^2 
/b*_R-2*a/b)*exp(x)-1),_R=RootOf(27*_Z^3*a^5+27*_Z^2*a^4+9*_Z*a^3+a^2+b^2) 
)
 
3.6.21.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.95 (sec) , antiderivative size = 1432, normalized size of antiderivative = 9.42 \[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\text {Too large to display} \]

input
integrate(coth(x)^3/(a+b*sinh(x)^3),x, algorithm="fricas")
 
output
-1/12*(2*(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 - 2*a*cosh(x)^ 
2 + 2*(3*a*cosh(x)^2 - a)*sinh(x)^2 + 4*(a*cosh(x)^3 - a*cosh(x))*sinh(x) 
+ a)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^2)/a^5)^(1/3 
) + 2/a)*log(b*cosh(x)^2 + b*sinh(x)^2 + (a^2*cosh(x) + a^2*sinh(x))*((1/2 
)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^2)/a^5)^(1/3) + 2/a) - 
 2*a*cosh(x) + 2*(b*cosh(x) - a)*sinh(x) - b) + 24*cosh(x)^2 + (6*cosh(x)^ 
4 + 24*cosh(x)*sinh(x)^3 + 6*sinh(x)^4 + 12*(3*cosh(x)^2 - 1)*sinh(x)^2 - 
(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 - 2*a*cosh(x)^2 + 2*(3* 
a*cosh(x)^2 - a)*sinh(x)^2 + 4*(a*cosh(x)^3 - a*cosh(x))*sinh(x) + a)*((1/ 
2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^2)/a^5)^(1/3) + 2/a) 
+ 3*sqrt(1/3)*(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 - 2*a*cos 
h(x)^2 + 2*(3*a*cosh(x)^2 - a)*sinh(x)^2 + 4*(a*cosh(x)^3 - a*cosh(x))*sin 
h(x) + a)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^ 
2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a 
^5 - (a^2 + b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2) - 12*cosh(x)^2 + 24*(cosh(x 
)^3 - cosh(x))*sinh(x) + 6)*log(b*cosh(x)^2 + b*sinh(x)^2 - 1/2*(a^2*cosh( 
x) + a^2*sinh(x))*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b 
^2)/a^5)^(1/3) + 2/a) + 3/2*sqrt(1/3)*(a^2*cosh(x) + a^2*sinh(x))*sqrt(-(( 
(1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^2)/a^5)^(1/3) + 2/ 
a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 + b^2...
 
3.6.21.6 Sympy [F]

\[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\int \frac {\coth ^{3}{\left (x \right )}}{a + b \sinh ^{3}{\left (x \right )}}\, dx \]

input
integrate(coth(x)**3/(a+b*sinh(x)**3),x)
 
output
Integral(coth(x)**3/(a + b*sinh(x)**3), x)
 
3.6.21.7 Maxima [F]

\[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\int { \frac {\coth \left (x\right )^{3}}{b \sinh \left (x\right )^{3} + a} \,d x } \]

input
integrate(coth(x)^3/(a+b*sinh(x)^3),x, algorithm="maxima")
 
output
2*b*(x/(a*b) - integrate((b*e^(5*x) - 3*b*e^(3*x) + 8*a*e^(2*x) + 3*b*e^x) 
*e^x/(b*e^(6*x) - 3*b*e^(4*x) + 8*a*e^(3*x) + 3*b*e^(2*x) - b), x)/(a*b)) 
- 6*b*integrate(e^(4*x)/(b*e^(6*x) - 3*b*e^(4*x) + 8*a*e^(3*x) + 3*b*e^(2* 
x) - b), x)/a - 2*(x*e^(4*x) - (2*x - 1)*e^(2*x) + x)/(a*e^(4*x) - 2*a*e^( 
2*x) + a) + log(e^x + 1)/a + log(e^x - 1)/a + 8*integrate(e^(3*x)/(b*e^(6* 
x) - 3*b*e^(4*x) + 8*a*e^(3*x) + 3*b*e^(2*x) - b), x)
 
3.6.21.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.38 \[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\frac {b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} - e^{\left (-x\right )} + e^{x} \right |}\right )}{3 \, a^{2}} - \frac {\log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 8 \, a \right |}\right )}{3 \, a} + \frac {\log \left ({\left | -e^{\left (-x\right )} + e^{x} \right |}\right )}{a} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} - e^{\left (-x\right )} + e^{x}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} {\left (e^{\left (-x\right )} - e^{x}\right )} + 4 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2}} - \frac {3 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4}{2 \, a {\left (e^{\left (-x\right )} - e^{x}\right )}^{2}} \]

input
integrate(coth(x)^3/(a+b*sinh(x)^3),x, algorithm="giac")
 
output
1/3*b*(-a/b)^(1/3)*log(abs(-2*(-a/b)^(1/3) - e^(-x) + e^x))/a^2 - 1/3*log( 
abs(-b*(e^(-x) - e^x)^3 + 8*a))/a + log(abs(-e^(-x) + e^x))/a - 1/3*sqrt(3 
)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) - e^(-x) + e^x)/(-a/b)^( 
1/3))/a^2 - 1/6*(-a*b^2)^(1/3)*log((e^(-x) - e^x)^2 - 2*(-a/b)^(1/3)*(e^(- 
x) - e^x) + 4*(-a/b)^(2/3))/a^2 - 1/2*(3*(e^(-x) - e^x)^2 + 4)/(a*(e^(-x) 
- e^x)^2)
 
3.6.21.9 Mupad [B] (verification not implemented)

Time = 1.00 (sec) , antiderivative size = 1129, normalized size of antiderivative = 7.43 \[ \int \frac {\coth ^3(x)}{a+b \sinh ^3(x)} \, dx=\text {Too large to display} \]

input
int(coth(x)^3/(a + b*sinh(x)^3),x)
 
output
2/(a - a*exp(2*x)) - 2/(a - 2*a*exp(2*x) + a*exp(4*x)) + symsum(log((50331 
648*a^6*exp(2*x) + 786432*b^6*exp(2*x) - 452984832*root(27*a^5*z^3 + 27*a^ 
4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^7 - 50331648*a^6 - 786432*b^6 - 13589 
54496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)^2*a^8 - 13 
58954496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)^3*a^9 - 
 50593792*a^2*b^4 - 102498304*a^4*b^2 + 1358954496*root(27*a^5*z^3 + 27*a^ 
4*z^2 + 9*a^3*z + b^2 + a^2, z, k)^2*a^8*exp(2*x) + 1358954496*root(27*a^5 
*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)^3*a^9*exp(2*x) + 50593792*a 
^2*b^4*exp(2*x) + 102498304*a^4*b^2*exp(2*x) - 7602176*root(27*a^5*z^3 + 2 
7*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^3*b^4 - 465305600*root(27*a^5*z^3 
 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^5*b^2 + 524288*a*b^5*exp(x) - 
 24379392*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)^2*a^4* 
b^4 - 1383333888*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k) 
^2*a^6*b^2 - 18874368*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, 
z, k)^3*a^5*b^4 - 1370750976*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 
+ a^2, z, k)^3*a^7*b^2 + 452984832*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z 
+ b^2 + a^2, z, k)*a^7*exp(2*x) + 5242880*a^3*b^3*exp(x) - 524288*root(27* 
a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^2*b^5*exp(x) + 8912896 
*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^4*b^3*exp(x) 
+ 7602176*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + b^2 + a^2, z, k)*a^3...