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((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/b/(a+b)^(3/2)-1/2*x/b/(a+b*cosh(x))^2-1/2*sinh(x)/(
a^2-b^2)/(a+b*cosh(x))
________________________________________________________________________________________
Rubi [A] time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.00,
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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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))^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 = 0.26, size = 87, normalized size = 1.00 \[ \frac {1}{2} \left (\frac {\frac {2 a \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac {x}{(a+b \cosh (x))^2}}{b}-\frac {\sinh (x)}{(a-b) (a+b) (a+b \cosh (x))}\right ) \]
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
________________________________________________________________________________________
fricas [B] time = 0.52, size = 1692, normalized size = 19.45 \[ \text {result too large to display} \]
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))]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \relax (x)}{{\left (b \cosh \relax (x) + a\right )}^{3}}\,{d x} \]
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)
________________________________________________________________________________________
maple [B] time = 0.23, size = 231, normalized size = 2.66 \[ -\frac {2 a^{2} x \,{\mathrm e}^{2 x}-a b \,{\mathrm e}^{3 x}-2 b^{2} x \,{\mathrm e}^{2 x}-2 a^{2} {\mathrm e}^{2 x}-b^{2} {\mathrm e}^{2 x}-3 b a \,{\mathrm e}^{x}-b^{2}}{b \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+b \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b} \]
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*b*a*exp(x)-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)
________________________________________________________________________________________
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))^3,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?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {sinh}\relax (x)}{{\left (a+b\,\mathrm {cosh}\relax (x)\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x*sinh(x))/(a + b*cosh(x))^3,x)
[Out]
int((x*sinh(x))/(a + b*cosh(x))^3, x)
________________________________________________________________________________________
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))**3,x)
[Out]
Timed out
________________________________________________________________________________________