3.218 \(\int \frac {x \sinh (x)}{(a+b \cosh (x))^2} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}-\frac {x}{b (a+b \cosh (x))} \]

[Out]

-x/b/(a+b*cosh(x))+2*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5465, 2659, 208} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}-\frac {x}{b (a+b \cosh (x))} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sinh[x])/(a + b*Cosh[x])^2,x]

[Out]

(2*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]) - x/(b*(a + b*Cosh[x]))

Rule 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 5465

Int[(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.)*((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)], x_Symbo
l] :> Simp[((e + f*x)^m*(a + b*Cosh[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*d*(n + 1)), Int[(e +
f*x)^(m - 1)*(a + b*Cosh[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n,
-1]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 59, normalized size = 0.98 \[ -\frac {2 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{b \sqrt {b^2-a^2}}-\frac {x}{b (a+b \cosh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(b*Sqrt[-a^2 + b^2]) - x/(b*(a + b*Cosh[x]))

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fricas [B]  time = 0.50, size = 480, normalized size = 8.00 \[ \left [-\frac {2 \, {\left (a^{2} - b^{2}\right )} x \cosh \relax (x) + 2 \, {\left (a^{2} - b^{2}\right )} x \sinh \relax (x) - {\left (b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right )}{a^{2} b^{2} - b^{4} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x)^{2} + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cosh \relax (x) + 2 \, {\left (a^{3} b - a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}, -\frac {2 \, {\left ({\left (a^{2} - b^{2}\right )} x \cosh \relax (x) + {\left (a^{2} - b^{2}\right )} x \sinh \relax (x) + {\left (b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right )\right )}}{a^{2} b^{2} - b^{4} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x)^{2} + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cosh \relax (x) + 2 \, {\left (a^{3} b - a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*(a^2 - b^2)*x*cosh(x) + 2*(a^2 - b^2)*x*sinh(x) - (b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x)
+ a)*sinh(x) + b)*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*c
osh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x
) + 2*(b*cosh(x) + a)*sinh(x) + b)))/(a^2*b^2 - b^4 + (a^2*b^2 - b^4)*cosh(x)^2 + (a^2*b^2 - b^4)*sinh(x)^2 +
2*(a^3*b - a*b^3)*cosh(x) + 2*(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cosh(x))*sinh(x)), -2*((a^2 - b^2)*x*cosh(x) +
(a^2 - b^2)*x*sinh(x) + (b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)*sqrt(-a^2 +
b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)))/(a^2*b^2 - b^4 + (a^2*b^2 - b^4)*cosh(
x)^2 + (a^2*b^2 - b^4)*sinh(x)^2 + 2*(a^3*b - a*b^3)*cosh(x) + 2*(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cosh(x))*sin
h(x))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \relax (x)}{{\left (b \cosh \relax (x) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sinh(x)/(b*cosh(x) + a)^2, x)

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maple [B]  time = 0.18, size = 138, normalized size = 2.30 \[ -\frac {2 x \,{\mathrm e}^{x}}{b \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+b \right )}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x/b*exp(x)/(b*exp(2*x)+2*a*exp(x)+b)+1/(a^2-b^2)^(1/2)/b*ln(exp(x)+(a*(a^2-b^2)^(1/2)-a^2+b^2)/(a^2-b^2)^(1
/2)/b)-1/(a^2-b^2)^(1/2)/b*ln(exp(x)+(a*(a^2-b^2)^(1/2)+a^2-b^2)/(a^2-b^2)^(1/2)/b)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.09, size = 110, normalized size = 1.83 \[ \frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\left (b^4-a^2\,b^2\right )+a\,b^3+a^2\,b^2\,{\mathrm {e}}^x}{b^2\,\sqrt {b^4-a^2\,b^2}}\right )}{\sqrt {b^4-a^2\,b^2}}-\frac {2\,{\mathrm {e}}^x\,\left (a^2\,x-b^2\,x\right )}{\left (a^2\,b-b^3\right )\,\left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sinh(x))/(a + b*cosh(x))^2,x)

[Out]

(2*atan((exp(x)*(b^4 - a^2*b^2) + a*b^3 + a^2*b^2*exp(x))/(b^2*(b^4 - a^2*b^2)^(1/2))))/(b^4 - a^2*b^2)^(1/2)
- (2*exp(x)*(a^2*x - b^2*x))/((a^2*b - b^3)*(b + 2*a*exp(x) + b*exp(2*x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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