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

Optimal. Leaf size=163 \[ -\frac {2 a b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^2 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a b^2}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}+\frac {2 a^2 b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b^3 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 163, 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, 3100, 2637, 3074, 206, 3109, 2638, 3155} \[ -\frac {2 a b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {b^2 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^2 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a b^2}{\left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))}+\frac {b^3 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {2 a^2 b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*b*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b^3*ArcTan[(b*Cosh[x] + a*Sinh[
x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a^2*Cosh[x])/(a^2 - b^2)^2 + (b^2*Cosh[x])/(a^2 - b^2)^2 - (2*a*b*S
inh[x])/(a^2 - b^2)^2 + (a*b^2)/((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 3100

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(b*Cos[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), 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 3155

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

Rubi steps

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

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Mathematica [A]  time = 0.90, size = 264, normalized size = 1.62 \[ \frac {1}{4} \left (\frac {4 \left (a^2+b^2\right ) \cosh (x)}{(a-b)^2 (a+b)^2}+\frac {6 b \left (3 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}+\frac {a \left (a^2+3 b^2\right )}{(a-b)^2 (a+b)^2 (a \cosh (x)+b \sinh (x))}-\frac {8 a b \sinh (x)}{(a-b)^2 (a+b)^2}\right )-\frac {2 b^2 \sqrt {a+b} \sinh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+2 a b \sqrt {a+b} \cosh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+a \sqrt {a-b} (a+b)}{4 (a-b)^{3/2} (a+b)^2 (a \cosh (x)+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a*Sqrt[a - b]*(a + b) + 2*a*b*Sqrt[a + b]*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])]*Cosh[x] +
2*b^2*Sqrt[a + b]*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])]*Sinh[x])/((a - b)^(3/2)*(a + b)^2*(a*Cos
h[x] + b*Sinh[x])) + ((6*b*(3*a^2 + b^2)*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])])/((a - b)^(5/2)*(
a + b)^(5/2)) + (4*(a^2 + b^2)*Cosh[x])/((a - b)^2*(a + b)^2) - (8*a*b*Sinh[x])/((a - b)^2*(a + b)^2) + (a*(a^
2 + 3*b^2))/((a - b)^2*(a + b)^2*(a*Cosh[x] + b*Sinh[x])))/4

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fricas [B]  time = 0.47, size = 1805, normalized size = 11.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2*sinh(x)/(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*(a^5 + 4*a^3*b^2 - 5*a*b^4)*cosh(x)^2 + 2*(a^5 + 4*a^3*b^2 - 5*a*b^4
+ 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*((2*a^3*b + 2*a^2*b^2 + a*b^
3 + b^4)*cosh(x)^3 + 3*(2*a^3*b + 2*a^2*b^2 + a*b^3 + b^4)*cosh(x)*sinh(x)^2 + (2*a^3*b + 2*a^2*b^2 + a*b^3 +
b^4)*sinh(x)^3 + (2*a^3*b - 2*a^2*b^2 + a*b^3 - b^4)*cosh(x) + (2*a^3*b - 2*a^2*b^2 + a*b^3 - b^4 + 3*(2*a^3*b
+ 2*a^2*b^2 + a*b^3 + b^4)*cosh(x)^2)*sinh(x))*sqrt(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*si
nh(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)*cos
h(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
+ (a^5 + 4*a^3*b^2 - 5*a*b^4)*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)*si
nh(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*(a^5 + 4*a^3*b^2 - 5*a*b^4)*cosh(x
)^2 + 2*(a^5 + 4*a^3*b^2 - 5*a*b^4 + 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*((2*a^3*b + 2*a^2*b^2 + a*b^3 + b^4)*cosh(x)^3 + 3*(2*a^3*b + 2*a^2*b^2 + a*b^3 + b^4)*cosh(x)*sinh(x)^2
+ (2*a^3*b + 2*a^2*b^2 + a*b^3 + b^4)*sinh(x)^3 + (2*a^3*b - 2*a^2*b^2 + a*b^3 - b^4)*cosh(x) + (2*a^3*b - 2*
a^2*b^2 + a*b^3 - b^4 + 3*(2*a^3*b + 2*a^2*b^2 + a*b^3 + b^4)*cosh(x)^2)*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(
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)*cosh(
x)^3 + (a^5 + 4*a^3*b^2 - 5*a*b^4)*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 = 0.16, size = 179, normalized size = 1.10 \[ \frac {2 \, {\left (2 \, a^{2} b + b^{3}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {e^{x}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{\left (2 \, x\right )} + 7 \, a b^{2} e^{\left (2 \, x\right )} + b^{3} e^{\left (2 \, x\right )} + a^{3} + a^{2} b - a b^{2} - b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} + a e^{x} - b e^{x}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*a^2*b + b^3)*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) + 3*a^2*b*e^(2*x) + 7*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 = 0.26, size = 217, normalized size = 1.33 \[ -\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 b^{3} \tanh \left (\frac {x}{2}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}+\frac {2 b^{2} a}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}+\frac {4 a^{2} b \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 b^{3} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a+b)^2/(tanh(1/2*x)-1)+2/(a-b)^2/(a+b)^2*b^3*tanh(1/2*x)/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)+2*b^2/(a-b)^2
/(a+b)^2/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)*a+4*a^2/(a-b)^2/(a+b)^2*b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/
2*x)+2*b)/(a^2-b^2)^(1/2))+2*b^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))+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)^2*sinh(x)/(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 = 2.44 \[ \frac {{\mathrm {e}}^{-x}}{2\,{\left (a-b\right )}^2}+\frac {{\mathrm {e}}^x}{2\,{\left (a+b\right )}^2}+\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\left (b^3\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}+2\,a^2\,b\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}\right )}{a^5\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}-b^5\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}+2\,a^2\,b^3\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}-2\,a^3\,b^2\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}+a\,b^4\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}-a^4\,b\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}}\right )\,\sqrt {4\,a^4\,b^2+4\,a^2\,b^4+b^6}}{\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}+\frac {2\,a\,b^2\,{\mathrm {e}}^x}{{\left (a+b\right )}^2\,{\left (a-b\right )}^2\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)^2*sinh(x))/(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)*(b^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^2*b*(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*(b^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) - b^5*(b^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) + 2*a^2*b^3*(b^6 + 4*a^2*b^
4 + 4*a^4*b^2)^(1/2) - 2*a^3*b^2*(b^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2) + a*b^4*(b^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/
2) - a^4*b*(b^6 + 4*a^2*b^4 + 4*a^4*b^2)^(1/2)))*(b^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*b^2*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)**2*sinh(x)/(a*cosh(x)+b*sinh(x))**2,x)

[Out]

Timed out

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