3.115 \(\int \frac {\tanh ^3(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=113 \[ -\frac {a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac {b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}-\frac {\tanh ^2(x) (a-b \text {csch}(x))}{2 \left (a^2+b^2\right )}+\frac {b^4 \log (a+b \text {csch}(x))}{a \left (a^2+b^2\right )^2}-\frac {b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\log (\sinh (x))}{a} \]

[Out]

-b^3*arctan(sinh(x))/(a^2+b^2)^2-1/2*b*arctan(sinh(x))/(a^2+b^2)+b^4*ln(a+b*csch(x))/a/(a^2+b^2)^2+ln(sinh(x))
/a-a*(a^2+2*b^2)*ln(tanh(x))/(a^2+b^2)^2-1/2*(a-b*csch(x))*tanh(x)^2/(a^2+b^2)

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Rubi [A]  time = 0.16, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3885, 894, 639, 203, 635, 260} \[ -\frac {a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac {b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac {b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac {b^4 \log (a+b \text {csch}(x))}{a \left (a^2+b^2\right )^2}-\frac {\tanh ^2(x) (a-b \text {csch}(x))}{2 \left (a^2+b^2\right )}+\frac {\log (\sinh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Csch[x]),x]

[Out]

-((b^3*ArcTan[Sinh[x]])/(a^2 + b^2)^2) - (b*ArcTan[Sinh[x]])/(2*(a^2 + b^2)) + (b^4*Log[a + b*Csch[x]])/(a*(a^
2 + b^2)^2) + Log[Sinh[x]]/a - (a*(a^2 + 2*b^2)*Log[Tanh[x]])/(a^2 + b^2)^2 - ((a - b*Csch[x])*Tanh[x]^2)/(2*(
a^2 + b^2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\tanh ^3(x)}{a+b \text {csch}(x)} \, dx &=-\left (b^4 \operatorname {Subst}\left (\int \frac {1}{x (a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \text {csch}(x)\right )\right )\\ &=-\left (b^4 \operatorname {Subst}\left (\int \left (\frac {1}{a b^4 x}-\frac {1}{a \left (a^2+b^2\right )^2 (a+x)}+\frac {-b^2-a x}{b^2 \left (a^2+b^2\right ) \left (b^2+x^2\right )^2}+\frac {-b^4-a \left (a^2+2 b^2\right ) x}{b^4 \left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \text {csch}(x)\right )\right )\\ &=\frac {b^4 \log (a+b \text {csch}(x))}{a \left (a^2+b^2\right )^2}+\frac {\log (\sinh (x))}{a}-\frac {\operatorname {Subst}\left (\int \frac {-b^4-a \left (a^2+2 b^2\right ) x}{b^2+x^2} \, dx,x,b \text {csch}(x)\right )}{\left (a^2+b^2\right )^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {-b^2-a x}{\left (b^2+x^2\right )^2} \, dx,x,b \text {csch}(x)\right )}{a^2+b^2}\\ &=\frac {b^4 \log (a+b \text {csch}(x))}{a \left (a^2+b^2\right )^2}+\frac {\log (\sinh (x))}{a}-\frac {(a-b \text {csch}(x)) \tanh ^2(x)}{2 \left (a^2+b^2\right )}+\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \text {csch}(x)\right )}{\left (a^2+b^2\right )^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \text {csch}(x)\right )}{2 \left (a^2+b^2\right )}+\frac {\left (a \left (a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \text {csch}(x)\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac {b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac {b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac {b^4 \log (a+b \text {csch}(x))}{a \left (a^2+b^2\right )^2}+\frac {\log (\sinh (x))}{a}-\frac {a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac {(a-b \text {csch}(x)) \tanh ^2(x)}{2 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 191, normalized size = 1.69 \[ \frac {a^4 \log (-\sinh (x)+i)+a^4 \log (\sinh (x)+i)+i a^3 b \log (-\sinh (x)+i)-i a^3 b \log (\sinh (x)+i)+a^2 \left (a^2+b^2\right ) \text {sech}^2(x)+2 a^2 b^2 \log (-\sinh (x)+i)+2 a^2 b^2 \log (\sinh (x)+i)+a b \left (a^2+b^2\right ) \tan ^{-1}(\sinh (x))+a b \left (a^2+b^2\right ) \tanh (x) \text {sech}(x)+2 b^4 \log (a \sinh (x)+b)+2 i a b^3 \log (-\sinh (x)+i)-2 i a b^3 \log (\sinh (x)+i)}{2 a \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(a + b*Csch[x]),x]

[Out]

(a*b*(a^2 + b^2)*ArcTan[Sinh[x]] + a^4*Log[I - Sinh[x]] + I*a^3*b*Log[I - Sinh[x]] + 2*a^2*b^2*Log[I - Sinh[x]
] + (2*I)*a*b^3*Log[I - Sinh[x]] + a^4*Log[I + Sinh[x]] - I*a^3*b*Log[I + Sinh[x]] + 2*a^2*b^2*Log[I + Sinh[x]
] - (2*I)*a*b^3*Log[I + Sinh[x]] + 2*b^4*Log[b + a*Sinh[x]] + a^2*(a^2 + b^2)*Sech[x]^2 + a*b*(a^2 + b^2)*Sech
[x]*Tanh[x])/(2*a*(a^2 + b^2)^2)

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fricas [B]  time = 0.87, size = 965, normalized size = 8.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*csch(x)),x, algorithm="fricas")

[Out]

-((a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^4 + (a^4 + 2*a^2*b^2 + b^4)*x*sinh(x)^4 - (a^3*b + a*b^3)*cosh(x)^3 - (a^3
*b + a*b^3 - 4*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x))*sinh(x)^3 - 2*(a^4 + a^2*b^2 - (a^4 + 2*a^2*b^2 + b^4)*x)*co
sh(x)^2 - (2*a^4 + 2*a^2*b^2 - 6*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^2 - 2*(a^4 + 2*a^2*b^2 + b^4)*x + 3*(a^3*b
+ a*b^3)*cosh(x))*sinh(x)^2 + (a^4 + 2*a^2*b^2 + b^4)*x + ((a^3*b + 3*a*b^3)*cosh(x)^4 + 4*(a^3*b + 3*a*b^3)*c
osh(x)*sinh(x)^3 + (a^3*b + 3*a*b^3)*sinh(x)^4 + a^3*b + 3*a*b^3 + 2*(a^3*b + 3*a*b^3)*cosh(x)^2 + 2*(a^3*b +
3*a*b^3 + 3*(a^3*b + 3*a*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3*b + 3*a*b^3)*cosh(x)^3 + (a^3*b + 3*a*b^3)*cosh(x
))*sinh(x))*arctan(cosh(x) + sinh(x)) + (a^3*b + a*b^3)*cosh(x) - (b^4*cosh(x)^4 + 4*b^4*cosh(x)*sinh(x)^3 + b
^4*sinh(x)^4 + 2*b^4*cosh(x)^2 + b^4 + 2*(3*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 4*(b^4*cosh(x)^3 + b^4*cosh(x))*s
inh(x))*log(2*(a*sinh(x) + b)/(cosh(x) - sinh(x))) - ((a^4 + 2*a^2*b^2)*cosh(x)^4 + 4*(a^4 + 2*a^2*b^2)*cosh(x
)*sinh(x)^3 + (a^4 + 2*a^2*b^2)*sinh(x)^4 + a^4 + 2*a^2*b^2 + 2*(a^4 + 2*a^2*b^2)*cosh(x)^2 + 2*(a^4 + 2*a^2*b
^2 + 3*(a^4 + 2*a^2*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + 2*a^2*b^2)*cosh(x)^3 + (a^4 + 2*a^2*b^2)*cosh(x))*si
nh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + (4*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^3 + a^3*b + a*b^3 - 3*(a^3*b
+ a*b^3)*cosh(x)^2 - 4*(a^4 + a^2*b^2 - (a^4 + 2*a^2*b^2 + b^4)*x)*cosh(x))*sinh(x))/(a^5 + 2*a^3*b^2 + a*b^4
+ (a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^4 + 4*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)*sinh(x)^3 + (a^5 + 2*a^3*b^2 + a*b
^4)*sinh(x)^4 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4 + 3*(a^5 + 2*a^3*b^2 + a*b^
4)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*cosh(x))*sinh(x))

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giac [B]  time = 0.13, size = 234, normalized size = 2.07 \[ \frac {b^{4} \log \left ({\left | -a {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (a^{2} b + 3 \, b^{3}\right )}}{4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 2 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 2 \, a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} + 4 \, a b^{2}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*csch(x)),x, algorithm="giac")

[Out]

b^4*log(abs(-a*(e^(-x) - e^x) + 2*b))/(a^5 + 2*a^3*b^2 + a*b^4) - 1/4*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x))
)*(a^2*b + 3*b^3)/(a^4 + 2*a^2*b^2 + b^4) + 1/2*(a^3 + 2*a*b^2)*log((e^(-x) - e^x)^2 + 4)/(a^4 + 2*a^2*b^2 + b
^4) - 1/2*(a^3*(e^(-x) - e^x)^2 + 2*a*b^2*(e^(-x) - e^x)^2 + 2*a^2*b*(e^(-x) - e^x) + 2*b^3*(e^(-x) - e^x) + 4
*a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*((e^(-x) - e^x)^2 + 4))

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maple [B]  time = 0.19, size = 324, normalized size = 2.87 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}+\frac {b^{4} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a \left (a^{2}+b^{2}\right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a^{2} b}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) a^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) a \,b^{2}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right ) a^{2} b}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right ) b^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a^{3}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a \,b^{2}}{\left (a^{2}+b^{2}\right )^{2}}-\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b^{3}}{\left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a^{2} b}{\left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a+b*csch(x)),x)

[Out]

-1/a*ln(tanh(1/2*x)-1)-1/a*ln(tanh(1/2*x)+1)+b^4/a/(a^2+b^2)^2*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)-1/(a^2+b^
2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a^2*b-1/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*b^3-2/(a^2+b^2)^2
/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a^3-2/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a*b^2+1/(a^2+b^2)^2/(ta
nh(1/2*x)^2+1)^2*tanh(1/2*x)*a^2*b+1/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*b^3+1/(a^2+b^2)^2*ln(tanh(1/2
*x)^2+1)*a^3+2/(a^2+b^2)^2*ln(tanh(1/2*x)^2+1)*a*b^2-3/(a^2+b^2)^2*arctan(tanh(1/2*x))*b^3-1/(a^2+b^2)^2*arcta
n(tanh(1/2*x))*a^2*b

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maxima [A]  time = 0.42, size = 172, normalized size = 1.52 \[ \frac {b^{4} \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (a^{2} b + 3 \, b^{3}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} + 2 \, a b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {b e^{\left (-x\right )} + 2 \, a e^{\left (-2 \, x\right )} - b e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} + \frac {x}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*csch(x)),x, algorithm="maxima")

[Out]

b^4*log(-2*b*e^(-x) + a*e^(-2*x) - a)/(a^5 + 2*a^3*b^2 + a*b^4) + (a^2*b + 3*b^3)*arctan(e^(-x))/(a^4 + 2*a^2*
b^2 + b^4) + (a^3 + 2*a*b^2)*log(e^(-2*x) + 1)/(a^4 + 2*a^2*b^2 + b^4) + (b*e^(-x) + 2*a*e^(-2*x) - b*e^(-3*x)
)/(a^2 + b^2 + 2*(a^2 + b^2)*e^(-2*x) + (a^2 + b^2)*e^(-4*x)) + x/a

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mupad [B]  time = 4.01, size = 335, normalized size = 2.96 \[ \frac {\frac {{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{{\left (a^2+b^2\right )}^2}+\frac {2\,\left (a^4+a^2\,b^2\right )}{a\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,a}{a^2+b^2}+\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}-\frac {x}{a}+\frac {b^4\,\ln \left (4\,a^9\,{\mathrm {e}}^{2\,x}-4\,a\,b^8-4\,a^9+7\,a^3\,b^6-14\,a^5\,b^4-17\,a^7\,b^2+8\,b^9\,{\mathrm {e}}^x-7\,a^3\,b^6\,{\mathrm {e}}^{2\,x}+14\,a^5\,b^4\,{\mathrm {e}}^{2\,x}+17\,a^7\,b^2\,{\mathrm {e}}^{2\,x}+8\,a^8\,b\,{\mathrm {e}}^x+4\,a\,b^8\,{\mathrm {e}}^{2\,x}-14\,a^2\,b^7\,{\mathrm {e}}^x+28\,a^4\,b^5\,{\mathrm {e}}^x+34\,a^6\,b^3\,{\mathrm {e}}^x\right )}{a^5+2\,a^3\,b^2+a\,b^4}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,\left (3\,b+a\,2{}\mathrm {i}\right )}{2\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )}+\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (2\,a+b\,3{}\mathrm {i}\right )}{2\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a + b/sinh(x)),x)

[Out]

((exp(x)*(a^2*b + b^3))/(a^2 + b^2)^2 + (2*(a^4 + a^2*b^2))/(a*(a^2 + b^2)^2))/(exp(2*x) + 1) - ((2*a)/(a^2 +
b^2) + (2*b*exp(x))/(a^2 + b^2))/(2*exp(2*x) + exp(4*x) + 1) - x/a + (b^4*log(4*a^9*exp(2*x) - 4*a*b^8 - 4*a^9
+ 7*a^3*b^6 - 14*a^5*b^4 - 17*a^7*b^2 + 8*b^9*exp(x) - 7*a^3*b^6*exp(2*x) + 14*a^5*b^4*exp(2*x) + 17*a^7*b^2*
exp(2*x) + 8*a^8*b*exp(x) + 4*a*b^8*exp(2*x) - 14*a^2*b^7*exp(x) + 28*a^4*b^5*exp(x) + 34*a^6*b^3*exp(x)))/(a*
b^4 + a^5 + 2*a^3*b^2) + (log(exp(x)*1i + 1)*(a*2i + 3*b))/(2*(2*a*b + a^2*1i - b^2*1i)) + (log(exp(x) + 1i)*(
2*a + b*3i))/(2*(a*b*2i + a^2 - b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(a+b*csch(x)),x)

[Out]

Integral(tanh(x)**3/(a + b*csch(x)), x)

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