3.701 \(\int \frac {\cosh ^3(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=133 \[ -\frac {3 a b^2 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {2 b^3 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac {x}{2}\right )+a+2 b \tanh \left (\frac {x}{2}\right )\right )}+\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

[Out]

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

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Rubi [A]  time = 0.78, antiderivative size = 193, normalized size of antiderivative = 1.45, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 638, 618, 204} \[ -\frac {2 b^3 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac {x}{2}\right )+a+2 b \tanh \left (\frac {x}{2}\right )\right )}-\frac {2 b^4 \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac {2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a*Cosh[x] + b*Sinh[x])^2,x]

[Out]

(-2*b^4*ArcTan[(b + a*Tanh[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) - (2*b^2*(3*a^2 - b^2)*ArcTan[(b + a*
Tanh[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) + 1/((a + b)^2*(1 - Tanh[x/2])) - 1/((a - b)^2*(1 + Tanh[x/
2])) - (2*b^3*(a + b*Tanh[x/2]))/(a*(a^2 - b^2)^2*(a + 2*b*Tanh[x/2] + a*Tanh[x/2]^2))

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx &=2 \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{\left (1-x^2\right )^2 \left (a+2 b x+a x^2\right )^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {1}{2 (a+b)^2 (-1+x)^2}+\frac {1}{2 (a-b)^2 (1+x)^2}-\frac {2 b^3 x}{a \left (-a^2+b^2\right ) \left (a+2 b x+a x^2\right )^2}+\frac {-3 a^2 b^2+b^4}{a \left (a^2-b^2\right )^2 \left (a+2 b x+a x^2\right )}\right ) \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\\ &=\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (1+\tanh \left (\frac {x}{2}\right )\right )}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a+2 b x+a x^2\right )^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )}-\frac {\left (2 b^2 \left (3 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {2 b^3 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac {x}{2}\right )+a \tanh ^2\left (\frac {x}{2}\right )\right )}-\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}+\frac {\left (4 b^2 \left (3 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=-\frac {2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {2 b^3 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac {x}{2}\right )+a \tanh ^2\left (\frac {x}{2}\right )\right )}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=-\frac {2 b^4 \tan ^{-1}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac {2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac {1}{(a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{(a-b)^2 \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {2 b^3 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac {x}{2}\right )+a \tanh ^2\left (\frac {x}{2}\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 204, normalized size = 1.53 \[ \frac {-2 a^2 b \sqrt {a-b} (a+b) \cosh ^2(x)+b \sqrt {a-b} \left (a^3+a^2 b+a b^2+b^3\right ) \sinh ^2(x)-6 a b^3 \sqrt {a+b} \sinh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+b^3 \left (-\sqrt {a-b}\right ) (a+b)+a \cosh (x) \left ((a-b)^{3/2} (a+b)^2 \sinh (x)-6 a b^2 \sqrt {a+b} \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )\right )}{(a-b)^{5/2} (a+b)^3 (a \cosh (x)+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(a*Cosh[x] + b*Sinh[x])^2,x]

[Out]

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

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fricas [B]  time = 0.48, size = 1645, normalized size = 12.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(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 + 6*(a^4*b - b^5)*cosh(x)^2 + 6*(a^4*b - b^5 - (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 + 6*((a^2*b^2 + a*b^3)*cosh(x)^3 + 3*(a^2*b^2 + a*b^3)*cosh(x
)*sinh(x)^2 + (a^2*b^2 + a*b^3)*sinh(x)^3 + (a^2*b^2 - a*b^3)*cosh(x) + (a^2*b^2 - a*b^3 + 3*(a^2*b^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)*sinh(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)*cosh(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 - 3*(a^4*b - 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 + 6*(a^4*b - b^5)*cosh(x)^2 + 6*(a^4*b - b^5 - (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 - 12*((a^2*b^2 + a*b^3)*cosh(x)^3 + 3*(a^2*b^2 + a*b^3)*cosh(x)*si
nh(x)^2 + (a^2*b^2 + a*b^3)*sinh(x)^3 + (a^2*b^2 - a*b^3)*cosh(x) + (a^2*b^2 - a*b^3 + 3*(a^2*b^2 + a*b^3)*cos
h(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 - 3*(a^4*b - 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)*cos
h(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 = 174, normalized size = 1.31 \[ -\frac {6 \, a b^{2} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {e^{x}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{\left (2 \, x\right )} + 3 \, a b^{2} e^{\left (2 \, x\right )} + 5 \, b^{3} e^{\left (2 \, x\right )} + a^{3} + a^{2} b - a b^{2} - b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} + a e^{x} - b e^{x}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a*cosh(x)+b*sinh(x))^2,x, algorithm="giac")

[Out]

-6*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) + 3*a^2*b*e^(2*x) + 3*a*b^2*e^(2*x) + 5*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 = 167, normalized size = 1.26 \[ -\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 b^{4} \tanh \left (\frac {x}{2}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {2 b^{3}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {6 b^{2} a \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a*cosh(x)+b*sinh(x))^2,x)

[Out]

-1/(a+b)^2/(tanh(1/2*x)-1)-2*b^4/(a-b)^2/(a+b)^2/a*tanh(1/2*x)/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)-2*b^3/(a-b)
^2/(a+b)^2/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)-6*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)^3/(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 = 1.86, size = 255, normalized size = 1.92 \[ \frac {{\mathrm {e}}^x}{2\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^{-x}}{2\,{\left (a-b\right )}^2}-\frac {6\,\mathrm {atan}\left (\frac {a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}{a^5\,\sqrt {a^2\,b^4}-b^5\,\sqrt {a^2\,b^4}+2\,a^2\,b^3\,\sqrt {a^2\,b^4}-2\,a^3\,b^2\,\sqrt {a^2\,b^4}+a\,b^4\,\sqrt {a^2\,b^4}-a^4\,b\,\sqrt {a^2\,b^4}}\right )\,\sqrt {a^2\,b^4}}{\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}-\frac {2\,b^3\,{\mathrm {e}}^x}{{\left (a+b\right )}^2\,{\left (a-b\right )}^2\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a*cosh(x) + b*sinh(x))^2,x)

[Out]

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

[Out]

Timed out

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