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

Optimal. Leaf size=165 \[ \frac {a^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {2 a b \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a^2 b}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}-\frac {2 a b^2 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {a^3 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

[Out]

-a^3*arctan((b*cosh(x)+a*sinh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)-2*a*b^2*arctan((b*cosh(x)+a*sinh(x))/(a^2-b
^2)^(1/2))/(a^2-b^2)^(5/2)-2*a*b*cosh(x)/(a^2-b^2)^2+a^2*sinh(x)/(a^2-b^2)^2+b^2*sinh(x)/(a^2-b^2)^2-a^2*b/(a^
2-b^2)^2/(a*cosh(x)+b*sinh(x))

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Rubi [A]  time = 0.31, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3111, 3109, 2637, 2638, 3074, 206, 3099, 3154} \[ \frac {a^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {2 a b \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a^2 b}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}-\frac {a^3 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {2 a b^2 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2)) - (2*a*b^2*ArcTan[(b*Cosh[x] + a*Si
nh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (2*a*b*Cosh[x])/(a^2 - b^2)^2 + (a^2*Sinh[x])/(a^2 - b^2)^2 + (b^
2*Sinh[x])/(a^2 - b^2)^2 - (a^2*b)/((a^2 - b^2)^2*(a*Cosh[x] + b*Sinh[x]))

Rule 206

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

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

Rule 3099

Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(a*Sin[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a^2/(a^2 + b^2), Int[Sin[c + d*x]^(m - 2)/
(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Dist[b/(a^2 + b^2), Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3111

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1)*(a*Cos[c + d*
x] + b*Sin[c + d*x])^(p + 1), x], x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n*(a*Cos[c +
d*x] + b*Sin[c + d*x])^(p + 1), x], x] - Dist[(a*b)/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^(n - 1
)*(a*Cos[c + d*x] + b*Sin[c + d*x])^p, x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0] &&
IGtQ[n, 0] && ILtQ[p, 0]

Rule 3154

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> -Simp[(b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Co
s[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - c*C)/(a^2 - b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d +
e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 0]

Rubi steps

\begin {align*} \int \frac {\cosh (x) \sinh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx &=\frac {a \int \frac {\sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}-\frac {b \int \frac {\cosh (x) \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\sinh (x)}{(a \cosh (x)+b \sinh (x))^2} \, dx}{a^2-b^2}\\ &=\frac {a^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {a^2 b}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}-\frac {a^3 \int \frac {1}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}-2 \frac {(a b) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^2 \int \cosh (x) \, dx}{\left (a^2-b^2\right )^2}-2 \frac {\left (a b^2\right ) \int \frac {1}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {2 a b \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {a^2 b}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{\left (a^2-b^2\right )^2}-2 \frac {\left (i a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{\left (a^2-b^2\right )^2}\\ &=-\frac {a^3 \tan ^{-1}\left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {2 a b^2 \tan ^{-1}\left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {2 a b \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {a^2 b}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 222, normalized size = 1.35 \[ \frac {b \left (\sqrt {a-b} \sqrt {a+b} \left (a^2+b^2\right ) \sinh ^2(x)-2 a \left (a^2+2 b^2\right ) \sinh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+a^2 \left (-\sqrt {a-b}\right ) \sqrt {a+b}\right )+a \cosh (x) \left (\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right ) \sinh (x)-2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )\right )-2 a^2 b \sqrt {a-b} \sqrt {a+b} \cosh ^2(x)}{(a-b)^{5/2} (a+b)^{5/2} (a \cosh (x)+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sqrt[a - b]*b*Sqrt[a + b]*Cosh[x]^2 + a*Cosh[x]*(-2*a*(a^2 + 2*b^2)*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a -
b]*Sqrt[a + b])] + Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*Sinh[x]) + b*(-(a^2*Sqrt[a - b]*Sqrt[a + b]) - 2*a*(a^
2 + 2*b^2)*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])]*Sinh[x] + Sqrt[a - b]*Sqrt[a + b]*(a^2 + b^2)*S
inh[x]^2))/((a - b)^(5/2)*(a + b)^(5/2)*(a*Cosh[x] + b*Sinh[x]))

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fricas [B]  time = 0.55, size = 1819, normalized size = 11.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 - (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)
*cosh(x)^4 - 4*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*sinh(x)^3 - (a^5 - a^4*b - 2*a^3*b^
2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^4 + 2*(5*a^4*b - 4*a^2*b^3 - b^5)*cosh(x)^2 + 2*(5*a^4*b - 4*a^2*b^3 - b^
5 - 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 + 2*((a^4 + a^3*b + 2*a^2*b^2 +
2*a*b^3)*cosh(x)^3 + 3*(a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cosh(x)*sinh(x)^2 + (a^4 + a^3*b + 2*a^2*b^2 + 2*a
*b^3)*sinh(x)^3 + (a^4 - a^3*b + 2*a^2*b^2 - 2*a*b^3)*cosh(x) + (a^4 - a^3*b + 2*a^2*b^2 - 2*a*b^3 + 3*(a^4 +
a^3*b + 2*a^2*b^2 + 2*a*b^3)*cosh(x)^2)*sinh(x))*sqrt(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*s
inh(x) + (a + b)*sinh(x)^2 + 2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*co
sh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) - 4*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3
- (5*a^4*b - 4*a^2*b^3 - b^5)*cosh(x))*sinh(x))/((a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5
- a*b^6 - b^7)*cosh(x)^3 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh
(x)*sinh(x)^2 + (a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*sinh(x)^3 + (a^7 -
a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*cosh(x) + (a^7 - a^6*b - 3*a^5*b^2 + 3*a
^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5
- a*b^6 - b^7)*cosh(x)^2)*sinh(x)), -1/2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 - (a^5 - a^4*b - 2
*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 4*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*
sinh(x)^3 - (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^4 + 2*(5*a^4*b - 4*a^2*b^3 - b^5)*cosh
(x)^2 + 2*(5*a^4*b - 4*a^2*b^3 - b^5 - 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x
)^2 - 4*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cosh(x)^3 + 3*(a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cosh(x)*sinh(x)
^2 + (a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*sinh(x)^3 + (a^4 - a^3*b + 2*a^2*b^2 - 2*a*b^3)*cosh(x) + (a^4 - a^3*
b + 2*a^2*b^2 - 2*a*b^3 + 3*(a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*cosh(x)^2)*sinh(x))*sqrt(a^2 - b^2)*arctan(sqr
t(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) - 4*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cos
h(x)^3 - (5*a^4*b - 4*a^2*b^3 - b^5)*cosh(x))*sinh(x))/((a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a
^2*b^5 - a*b^6 - b^7)*cosh(x)^3 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7
)*cosh(x)*sinh(x)^2 + (a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*sinh(x)^3 +
(a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*cosh(x) + (a^7 - a^6*b - 3*a^5*b^2
+ 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^
2*b^5 - a*b^6 - b^7)*cosh(x)^2)*sinh(x))]

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giac [A]  time = 0.14, size = 179, normalized size = 1.08 \[ -\frac {2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {e^{x}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a^{3} e^{\left (2 \, x\right )} + 7 \, a^{2} b e^{\left (2 \, x\right )} + 3 \, a b^{2} e^{\left (2 \, x\right )} + b^{3} e^{\left (2 \, x\right )} + a^{3} + a^{2} b - a b^{2} - b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} + a e^{x} - b e^{x}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a^3 + 2*a*b^2)*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + 1/2*e^x
/(a^2 + 2*a*b + b^2) - 1/2*(a^3*e^(2*x) + 7*a^2*b*e^(2*x) + 3*a*b^2*e^(2*x) + b^3*e^(2*x) + a^3 + a^2*b - a*b^
2 - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(a*e^(3*x) + b*e^(3*x) + a*e^x - b*e^x))

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maple [A]  time = 0.27, size = 219, normalized size = 1.33 \[ -\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 a \tanh \left (\frac {x}{2}\right ) b^{2}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {2 a^{2} b}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {2 a^{3} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {4 b^{2} a \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.82, size = 397, normalized size = 2.41 \[ \frac {{\mathrm {e}}^x}{2\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^{-x}}{2\,{\left (a-b\right )}^2}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\left (a^3\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}+2\,a\,b^2\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}\right )}{a^5\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}-b^5\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}+2\,a^2\,b^3\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}-2\,a^3\,b^2\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}+a\,b^4\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}-a^4\,b\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}}\right )\,\sqrt {a^6+4\,a^4\,b^2+4\,a^2\,b^4}}{\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}-\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a+b\right )}^2\,{\left (a-b\right )}^2\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/(2*(a + b)^2) - exp(-x)/(2*(a - b)^2) - (2*atan((exp(x)*(a^3*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10
*a^6*b^4 - 5*a^8*b^2)^(1/2) + 2*a*b^2*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2)))/
(a^5*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) - b^5*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) + 2*a^2*b^3*(a^6 + 4*a^2*b^
4 + 4*a^4*b^2)^(1/2) - 2*a^3*b^2*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) + a*b^4*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/
2) - a^4*b*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2)))*(a^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2))/(a^10 - b^10 + 5*a^2*b^8
- 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - (2*a^2*b*exp(x))/((a + b)^2*(a - b)^2*(a - b + exp(2*x)*(a + b
)))

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

[Out]

Timed out

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