3.237 \(\int \frac{\tanh ^3(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=135 \[ \frac{a^3}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}-\frac{a^2 \left (a^2-3 b^2\right ) \log (a+b \sinh (x))}{\left (a^2+b^2\right )^3}+\frac{a b \left (3 a^2-b^2\right ) \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^3}+\frac{a^2 \left (a^2-3 b^2\right ) \log (\cosh (x))}{\left (a^2+b^2\right )^3}+\frac{\text{sech}^2(x) \left (a^2-2 a b \sinh (x)-b^2\right )}{2 \left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.355971, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2721, 1647, 1629, 635, 203, 260} \[ \frac{a^3}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}-\frac{a^2 \left (a^2-3 b^2\right ) \log (a+b \sinh (x))}{\left (a^2+b^2\right )^3}+\frac{a b \left (3 a^2-b^2\right ) \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^3}+\frac{a^2 \left (a^2-3 b^2\right ) \log (\cosh (x))}{\left (a^2+b^2\right )^3}+\frac{\text{sech}^2(x) \left (a^2-2 a b \sinh (x)-b^2\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Sinh[x])^2,x]

[Out]

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

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.661667, size = 150, normalized size = 1.11 \[ \frac{\frac{2 a^3 \left (a^2+b^2\right )}{a+b \sinh (x)}+\left (a^4-b^4\right ) \text{sech}^2(x)-2 a^2 \left (a^2-3 b^2\right ) \log (a+b \sinh (x))-2 a b \left (a^2+b^2\right ) \tan ^{-1}(\sinh (x))-2 a b \left (a^2+b^2\right ) \tanh (x) \text{sech}(x)+a^2 (a-i b) (a-3 i b) \log (-\sinh (x)+i)+a^2 (a+i b) (a+3 i b) \log (\sinh (x)+i)}{2 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(a + b*Sinh[x])^2,x]

[Out]

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

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Maple [B]  time = 0.062, size = 491, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a+b*sinh(x))^2,x)

[Out]

2/(a^2+b^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a^3*b+2/(a^2+b^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a*b^3-2/
(a^2+b^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a^4+2/(a^2+b^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*b^4-2/(a^2+b
^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*a^3*b-2/(a^2+b^2)^3/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*a*b^3+1/(a^2+b^2)^3*
ln(tanh(1/2*x)^2+1)*a^4-3/(a^2+b^2)^3*ln(tanh(1/2*x)^2+1)*a^2*b^2+6/(a^2+b^2)^3*arctan(tanh(1/2*x))*a^3*b-2/(a
^2+b^2)^3*a*arctan(tanh(1/2*x))*b^3+2*a^4/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/
2*x)*b-a)*b+2*a^2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b^3-a^4/(a^4+2
*a^2*b^2+b^4)/(a^2+b^2)*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)+3*a^2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*ln(a*tanh(1/
2*x)^2-2*tanh(1/2*x)*b-a)*b^2

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Maxima [B]  time = 1.59428, size = 506, normalized size = 3.75 \begin{align*} -\frac{2 \,{\left (3 \, a^{3} b - a b^{3}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (a^{4} - 3 \, a^{2} b^{2}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{4} - 3 \, a^{2} b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (4 \, a^{3} e^{\left (-3 \, x\right )} +{\left (a^{3} - a b^{2}\right )} e^{\left (-x\right )} -{\left (a^{2} b + b^{3}\right )} e^{\left (-2 \, x\right )} +{\left (a^{2} b + b^{3}\right )} e^{\left (-4 \, x\right )} +{\left (a^{3} - a b^{2}\right )} e^{\left (-5 \, x\right )}\right )}}{a^{4} b + 2 \, a^{2} b^{3} + b^{5} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-x\right )} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-2 \, x\right )} + 4 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-3 \, x\right )} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-4 \, x\right )} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-5 \, x\right )} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-6 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

-2*(3*a^3*b - a*b^3)*arctan(e^(-x))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (a^4 - 3*a^2*b^2)*log(-2*a*e^(-x) +
b*e^(-2*x) - b)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 - 3*a^2*b^2)*log(e^(-2*x) + 1)/(a^6 + 3*a^4*b^2 + 3
*a^2*b^4 + b^6) + 2*(4*a^3*e^(-3*x) + (a^3 - a*b^2)*e^(-x) - (a^2*b + b^3)*e^(-2*x) + (a^2*b + b^3)*e^(-4*x) +
(a^3 - a*b^2)*e^(-5*x))/(a^4*b + 2*a^2*b^3 + b^5 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*e^(-x) + (a^4*b + 2*a^2*b^3 +
b^5)*e^(-2*x) + 4*(a^5 + 2*a^3*b^2 + a*b^4)*e^(-3*x) - (a^4*b + 2*a^2*b^3 + b^5)*e^(-4*x) + 2*(a^5 + 2*a^3*b^2
+ a*b^4)*e^(-5*x) - (a^4*b + 2*a^2*b^3 + b^5)*e^(-6*x))

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Fricas [B]  time = 2.79176, size = 6589, normalized size = 48.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

-(2*(a^5 - a*b^4)*cosh(x)^5 + 2*(a^5 - a*b^4)*sinh(x)^5 - 2*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^4 - 2*(a^4*b + 2
*a^2*b^3 + b^5 - 5*(a^5 - a*b^4)*cosh(x))*sinh(x)^4 + 8*(a^5 + a^3*b^2)*cosh(x)^3 + 4*(2*a^5 + 2*a^3*b^2 + 5*(
a^5 - a*b^4)*cosh(x)^2 - 2*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x))*sinh(x)^3 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^
2 + 2*(a^4*b + 2*a^2*b^3 + b^5 + 10*(a^5 - a*b^4)*cosh(x)^3 - 6*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^2 + 12*(a^5
+ a^3*b^2)*cosh(x))*sinh(x)^2 + 2*((3*a^3*b^2 - a*b^4)*cosh(x)^6 + (3*a^3*b^2 - a*b^4)*sinh(x)^6 + 2*(3*a^4*b
- a^2*b^3)*cosh(x)^5 + 2*(3*a^4*b - a^2*b^3 + 3*(3*a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^5 - 3*a^3*b^2 + a*b^4 + (
3*a^3*b^2 - a*b^4)*cosh(x)^4 + (3*a^3*b^2 - a*b^4 + 15*(3*a^3*b^2 - a*b^4)*cosh(x)^2 + 10*(3*a^4*b - a^2*b^3)*
cosh(x))*sinh(x)^4 + 4*(3*a^4*b - a^2*b^3)*cosh(x)^3 + 4*(3*a^4*b - a^2*b^3 + 5*(3*a^3*b^2 - a*b^4)*cosh(x)^3
+ 5*(3*a^4*b - a^2*b^3)*cosh(x)^2 + (3*a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^3 - (3*a^3*b^2 - a*b^4)*cosh(x)^2 - (
3*a^3*b^2 - a*b^4 - 15*(3*a^3*b^2 - a*b^4)*cosh(x)^4 - 20*(3*a^4*b - a^2*b^3)*cosh(x)^3 - 6*(3*a^3*b^2 - a*b^4
)*cosh(x)^2 - 12*(3*a^4*b - a^2*b^3)*cosh(x))*sinh(x)^2 + 2*(3*a^4*b - a^2*b^3)*cosh(x) + 2*(3*(3*a^3*b^2 - a*
b^4)*cosh(x)^5 + 3*a^4*b - a^2*b^3 + 5*(3*a^4*b - a^2*b^3)*cosh(x)^4 + 2*(3*a^3*b^2 - a*b^4)*cosh(x)^3 + 6*(3*
a^4*b - a^2*b^3)*cosh(x)^2 - (3*a^3*b^2 - a*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + 2*(a^5 - a*b^4)
*cosh(x) - ((a^4*b - 3*a^2*b^3)*cosh(x)^6 + (a^4*b - 3*a^2*b^3)*sinh(x)^6 + 2*(a^5 - 3*a^3*b^2)*cosh(x)^5 + 2*
(a^5 - 3*a^3*b^2 + 3*(a^4*b - 3*a^2*b^3)*cosh(x))*sinh(x)^5 - a^4*b + 3*a^2*b^3 + (a^4*b - 3*a^2*b^3)*cosh(x)^
4 + (a^4*b - 3*a^2*b^3 + 15*(a^4*b - 3*a^2*b^3)*cosh(x)^2 + 10*(a^5 - 3*a^3*b^2)*cosh(x))*sinh(x)^4 + 4*(a^5 -
3*a^3*b^2)*cosh(x)^3 + 4*(a^5 - 3*a^3*b^2 + 5*(a^4*b - 3*a^2*b^3)*cosh(x)^3 + 5*(a^5 - 3*a^3*b^2)*cosh(x)^2 +
(a^4*b - 3*a^2*b^3)*cosh(x))*sinh(x)^3 - (a^4*b - 3*a^2*b^3)*cosh(x)^2 - (a^4*b - 3*a^2*b^3 - 15*(a^4*b - 3*a
^2*b^3)*cosh(x)^4 - 20*(a^5 - 3*a^3*b^2)*cosh(x)^3 - 6*(a^4*b - 3*a^2*b^3)*cosh(x)^2 - 12*(a^5 - 3*a^3*b^2)*co
sh(x))*sinh(x)^2 + 2*(a^5 - 3*a^3*b^2)*cosh(x) + 2*(3*(a^4*b - 3*a^2*b^3)*cosh(x)^5 + a^5 - 3*a^3*b^2 + 5*(a^5
- 3*a^3*b^2)*cosh(x)^4 + 2*(a^4*b - 3*a^2*b^3)*cosh(x)^3 + 6*(a^5 - 3*a^3*b^2)*cosh(x)^2 - (a^4*b - 3*a^2*b^3
)*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + ((a^4*b - 3*a^2*b^3)*cosh(x)^6 + (a^4*b - 3*a
^2*b^3)*sinh(x)^6 + 2*(a^5 - 3*a^3*b^2)*cosh(x)^5 + 2*(a^5 - 3*a^3*b^2 + 3*(a^4*b - 3*a^2*b^3)*cosh(x))*sinh(x
)^5 - a^4*b + 3*a^2*b^3 + (a^4*b - 3*a^2*b^3)*cosh(x)^4 + (a^4*b - 3*a^2*b^3 + 15*(a^4*b - 3*a^2*b^3)*cosh(x)^
2 + 10*(a^5 - 3*a^3*b^2)*cosh(x))*sinh(x)^4 + 4*(a^5 - 3*a^3*b^2)*cosh(x)^3 + 4*(a^5 - 3*a^3*b^2 + 5*(a^4*b -
3*a^2*b^3)*cosh(x)^3 + 5*(a^5 - 3*a^3*b^2)*cosh(x)^2 + (a^4*b - 3*a^2*b^3)*cosh(x))*sinh(x)^3 - (a^4*b - 3*a^2
*b^3)*cosh(x)^2 - (a^4*b - 3*a^2*b^3 - 15*(a^4*b - 3*a^2*b^3)*cosh(x)^4 - 20*(a^5 - 3*a^3*b^2)*cosh(x)^3 - 6*(
a^4*b - 3*a^2*b^3)*cosh(x)^2 - 12*(a^5 - 3*a^3*b^2)*cosh(x))*sinh(x)^2 + 2*(a^5 - 3*a^3*b^2)*cosh(x) + 2*(3*(a
^4*b - 3*a^2*b^3)*cosh(x)^5 + a^5 - 3*a^3*b^2 + 5*(a^5 - 3*a^3*b^2)*cosh(x)^4 + 2*(a^4*b - 3*a^2*b^3)*cosh(x)^
3 + 6*(a^5 - 3*a^3*b^2)*cosh(x)^2 - (a^4*b - 3*a^2*b^3)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) +
2*(a^5 - a*b^4 + 5*(a^5 - a*b^4)*cosh(x)^4 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^3 + 12*(a^5 + a^3*b^2)*cosh(
x)^2 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x))*sinh(x))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7 - (a^6*b + 3*a^4*b^3
+ 3*a^2*b^5 + b^7)*cosh(x)^6 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sinh(x)^6 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b
^4 + a*b^6)*cosh(x)^5 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 3*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(
x))*sinh(x)^5 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^4 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7 + 15*(a
^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^2 + 10*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)^4 -
4*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^3 - 4*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 5*(a^6*b + 3*a^4*
b^3 + 3*a^2*b^5 + b^7)*cosh(x)^3 + 5*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^2 + (a^6*b + 3*a^4*b^3 + 3*
a^2*b^5 + b^7)*cosh(x))*sinh(x)^3 + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^2 + (a^6*b + 3*a^4*b^3 + 3*a
^2*b^5 + b^7 - 15*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^4 - 20*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*c
osh(x)^3 - 6*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^2 - 12*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x
))*sinh(x)^2 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x) - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 3*(a
^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^5 + 5*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^4 + 2*(a^6*b +
3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^3 + 6*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^2 - (a^6*b + 3*a^4*b
^3 + 3*a^2*b^5 + b^7)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(a+b*sinh(x))**2,x)

[Out]

Integral(tanh(x)**3/(a + b*sinh(x))**2, x)

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Giac [B]  time = 1.17526, size = 414, normalized size = 3.07 \begin{align*} \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}{\left (3 \, a^{3} b - a b^{3}\right )}}{2 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac{{\left (a^{4} - 3 \, a^{2} b^{2}\right )} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac{{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{2 \,{\left (a^{3}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - a b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + a^{2} b{\left (e^{\left (-x\right )} - e^{x}\right )} + b^{3}{\left (e^{\left (-x\right )} - e^{x}\right )} + 6 \, a^{3} - 2 \, a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 2 \, a{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4 \, b{\left (e^{\left (-x\right )} - e^{x}\right )} - 8 \, a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

1/2*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*(3*a^3*b - a*b^3)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 1/2*(a^4
- 3*a^2*b^2)*log((e^(-x) - e^x)^2 + 4)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (a^4*b - 3*a^2*b^3)*log(abs(-b*(
e^(-x) - e^x) + 2*a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 2*(a^3*(e^(-x) - e^x)^2 - a*b^2*(e^(-x) - e^x)^2
+ a^2*b*(e^(-x) - e^x) + b^3*(e^(-x) - e^x) + 6*a^3 - 2*a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(b*(e^(-x) - e^x)^3 -
2*a*(e^(-x) - e^x)^2 + 4*b*(e^(-x) - e^x) - 8*a))