3.219 \(\int \frac{x \sinh (x)}{(a+b \cosh (x))^3} \, dx\)
Optimal. Leaf size=87 \[ -\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}-\frac{x}{2 b (a+b \cosh (x))^2}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b (a-b)^{3/2} (a+b)^{3/2}} \]
[Out]
(a*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*b*(a + b)^(3/2)) - x/(2*b*(a + b*Cosh[x])^2) -
Sinh[x]/(2*(a^2 - b^2)*(a + b*Cosh[x]))
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Rubi [A] time = 0.0892023, antiderivative size = 87, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used =
{5465, 2664, 12, 2659, 208} \[ -\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}-\frac{x}{2 b (a+b \cosh (x))^2}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b (a-b)^{3/2} (a+b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(x*Sinh[x])/(a + b*Cosh[x])^3,x]
[Out]
(a*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*b*(a + b)^(3/2)) - x/(2*b*(a + b*Cosh[x])^2) -
Sinh[x]/(2*(a^2 - b^2)*(a + b*Cosh[x]))
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]
Rule 2664
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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]
Rubi steps
\begin{align*} \int \frac{x \sinh (x)}{(a+b \cosh (x))^3} \, dx &=-\frac{x}{2 b (a+b \cosh (x))^2}+\frac{\int \frac{1}{(a+b \cosh (x))^2} \, dx}{2 b}\\ &=-\frac{x}{2 b (a+b \cosh (x))^2}-\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}+\frac{\int \frac{a}{a+b \cosh (x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{x}{2 b (a+b \cosh (x))^2}-\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}+\frac{a \int \frac{1}{a+b \cosh (x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{x}{2 b (a+b \cosh (x))^2}-\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}+\frac{a \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{b \left (a^2-b^2\right )}\\ &=\frac{a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b (a+b)^{3/2}}-\frac{x}{2 b (a+b \cosh (x))^2}-\frac{\sinh (x)}{2 \left (a^2-b^2\right ) (a+b \cosh (x))}\\ \end{align*}
Mathematica [A] time = 0.243949, size = 87, normalized size = 1. \[ \frac{1}{2} \left (\frac{\frac{2 a \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac{x}{(a+b \cosh (x))^2}}{b}-\frac{\sinh (x)}{(a-b) (a+b) (a+b \cosh (x))}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(x*Sinh[x])/(a + b*Cosh[x])^3,x]
[Out]
(((2*a*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - x/(a + b*Cosh[x])^2)/b - Sinh[x]/((a
- b)*(a + b)*(a + b*Cosh[x])))/2
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Maple [B] time = 0.083, size = 231, normalized size = 2.7 \begin{align*} -{\frac{2\,{a}^{2}x{{\rm e}^{2\,x}}-ab{{\rm e}^{3\,x}}-2\,{b}^{2}x{{\rm e}^{2\,x}}-2\,{a}^{2}{{\rm e}^{2\,x}}-{b}^{2}{{\rm e}^{2\,x}}-3\,a{{\rm e}^{x}}b-{b}^{2}}{b \left ( b{{\rm e}^{2\,x}}+2\,a{{\rm e}^{x}}+b \right ) ^{2} \left ({a}^{2}-{b}^{2} \right ) }}+{\frac{a}{ \left ( 2\,a+2\,b \right ) \left ( a-b \right ) b}\ln \left ({{\rm e}^{x}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}-{b}^{2}}-{a}^{2}+{b}^{2} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}-{\frac{a}{ \left ( 2\,a+2\,b \right ) \left ( a-b \right ) b}\ln \left ({{\rm e}^{x}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}-{b}^{2}}+{a}^{2}-{b}^{2} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*sinh(x)/(a+b*cosh(x))^3,x)
[Out]
-1/b*(2*a^2*x*exp(2*x)-a*b*exp(3*x)-2*b^2*x*exp(2*x)-2*a^2*exp(2*x)-b^2*exp(2*x)-3*a*exp(x)*b-b^2)/(b*exp(2*x)
+2*a*exp(x)+b)^2/(a^2-b^2)+1/2/(a^2-b^2)^(1/2)*a/(a+b)/(a-b)/b*ln(exp(x)+(a*(a^2-b^2)^(1/2)-a^2+b^2)/(a^2-b^2)
^(1/2)/b)-1/2/(a^2-b^2)^(1/2)*a/(a+b)/(a-b)/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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sinh(x)/(a+b*cosh(x))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.05765, size = 3969, normalized size = 45.62 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sinh(x)/(a+b*cosh(x))^3,x, algorithm="fricas")
[Out]
[1/2*(2*a^2*b^2 - 2*b^4 + 2*(a^3*b - a*b^3)*cosh(x)^3 + 2*(a^3*b - a*b^3)*sinh(x)^3 + 2*(2*a^4 - a^2*b^2 - b^4
- 2*(a^4 - 2*a^2*b^2 + b^4)*x)*cosh(x)^2 + 2*(2*a^4 - a^2*b^2 - b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*x + 3*(a^3*b
- a*b^3)*cosh(x))*sinh(x)^2 - (a*b^2*cosh(x)^4 + a*b^2*sinh(x)^4 + 4*a^2*b*cosh(x)^3 + 4*a^2*b*cosh(x) + 4*(a*
b^2*cosh(x) + a^2*b)*sinh(x)^3 + a*b^2 + 2*(2*a^3 + a*b^2)*cosh(x)^2 + 2*(3*a*b^2*cosh(x)^2 + 6*a^2*b*cosh(x)
+ 2*a^3 + a*b^2)*sinh(x)^2 + 4*(a*b^2*cosh(x)^3 + 3*a^2*b*cosh(x)^2 + a^2*b + (2*a^3 + a*b^2)*cosh(x))*sinh(x)
)*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*cosh(x) + a*b)*sin
h(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)) + 6*(a^3*b - a*b^3)*cosh(x) + 2*(3*a^3*b - 3*a*b^3 + 3*(a^3*b - a*b^3)*cosh(x)^2 + 2*(2*a^4
- a^2*b^2 - b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*x)*cosh(x))*sinh(x))/(a^4*b^3 - 2*a^2*b^5 + b^7 + (a^4*b^3 - 2*a^
2*b^5 + b^7)*cosh(x)^4 + (a^4*b^3 - 2*a^2*b^5 + b^7)*sinh(x)^4 + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cosh(x)^3 + 4
*(a^5*b^2 - 2*a^3*b^4 + a*b^6 + (a^4*b^3 - 2*a^2*b^5 + b^7)*cosh(x))*sinh(x)^3 + 2*(2*a^6*b - 3*a^4*b^3 + b^7)
*cosh(x)^2 + 2*(2*a^6*b - 3*a^4*b^3 + b^7 + 3*(a^4*b^3 - 2*a^2*b^5 + b^7)*cosh(x)^2 + 6*(a^5*b^2 - 2*a^3*b^4 +
a*b^6)*cosh(x))*sinh(x)^2 + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cosh(x) + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6 + (a^4*b
^3 - 2*a^2*b^5 + b^7)*cosh(x)^3 + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cosh(x)^2 + (2*a^6*b - 3*a^4*b^3 + b^7)*cosh
(x))*sinh(x)), (a^2*b^2 - b^4 + (a^3*b - a*b^3)*cosh(x)^3 + (a^3*b - a*b^3)*sinh(x)^3 + (2*a^4 - a^2*b^2 - b^4
- 2*(a^4 - 2*a^2*b^2 + b^4)*x)*cosh(x)^2 + (2*a^4 - a^2*b^2 - b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*x + 3*(a^3*b -
a*b^3)*cosh(x))*sinh(x)^2 - (a*b^2*cosh(x)^4 + a*b^2*sinh(x)^4 + 4*a^2*b*cosh(x)^3 + 4*a^2*b*cosh(x) + 4*(a*b^
2*cosh(x) + a^2*b)*sinh(x)^3 + a*b^2 + 2*(2*a^3 + a*b^2)*cosh(x)^2 + 2*(3*a*b^2*cosh(x)^2 + 6*a^2*b*cosh(x) +
2*a^3 + a*b^2)*sinh(x)^2 + 4*(a*b^2*cosh(x)^3 + 3*a^2*b*cosh(x)^2 + a^2*b + (2*a^3 + a*b^2)*cosh(x))*sinh(x))*
sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 3*(a^3*b - a*b^3)*cosh(x)
+ (3*a^3*b - 3*a*b^3 + 3*(a^3*b - a*b^3)*cosh(x)^2 + 2*(2*a^4 - a^2*b^2 - b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*x)*
cosh(x))*sinh(x))/(a^4*b^3 - 2*a^2*b^5 + b^7 + (a^4*b^3 - 2*a^2*b^5 + b^7)*cosh(x)^4 + (a^4*b^3 - 2*a^2*b^5 +
b^7)*sinh(x)^4 + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cosh(x)^3 + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6 + (a^4*b^3 - 2*a^2
*b^5 + b^7)*cosh(x))*sinh(x)^3 + 2*(2*a^6*b - 3*a^4*b^3 + b^7)*cosh(x)^2 + 2*(2*a^6*b - 3*a^4*b^3 + b^7 + 3*(a
^4*b^3 - 2*a^2*b^5 + b^7)*cosh(x)^2 + 6*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)^2 + 4*(a^5*b^2 - 2*a^3*
b^4 + a*b^6)*cosh(x) + 4*(a^5*b^2 - 2*a^3*b^4 + a*b^6 + (a^4*b^3 - 2*a^2*b^5 + b^7)*cosh(x)^3 + 3*(a^5*b^2 - 2
*a^3*b^4 + a*b^6)*cosh(x)^2 + (2*a^6*b - 3*a^4*b^3 + b^7)*cosh(x))*sinh(x))]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sinh(x)/(a+b*cosh(x))**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sinh \left (x\right )}{{\left (b \cosh \left (x\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sinh(x)/(a+b*cosh(x))^3,x, algorithm="giac")
[Out]
integrate(x*sinh(x)/(b*cosh(x) + a)^3, x)