3.587 \(\int \frac {1}{(a \cosh (x)+b \sinh (x))^3} \, dx\)

Optimal. Leaf size=77 \[ \frac {a \sinh (x)+b \cosh (x)}{2 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^2}+\frac {\tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3076, 3074, 206} \[ \frac {a \sinh (x)+b \cosh (x)}{2 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^2}+\frac {\tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-3),x]

[Out]

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

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 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 3076

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

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 96, normalized size = 1.25 \[ \frac {1}{2} \left (\frac {2 \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac {b}{a (a-b) (a+b) (a \cosh (x)+b \sinh (x))}+\frac {\sinh (x)}{a (a \cosh (x)+b \sinh (x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 1495, normalized size = 19.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^3,x, algorithm="fricas")

[Out]

[1/2*(2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^3 + 6*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)*sinh(x)^2 + 2*(a^3 + a^2
*b - a*b^2 - b^3)*sinh(x)^3 + ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2
+ 2*a*b + b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2
+ a^2 - 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*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)) - 2*(a^3 - a^2*b - a*b^2 +
b^3)*cosh(x) - 2*(a^3 - a^2*b - a*b^2 + b^3 - 3*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2)*sinh(x))/(a^6 - 2*a^5*b
- a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 + (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 +
b^6)*cosh(x)^4 + 4*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)*sinh(x)^3 + (a^6 +
2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*sinh(x)^4 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos
h(x)^2 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 3*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b
^6)*cosh(x)^2)*sinh(x)^2 + 4*((a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)^3 + (a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)), ((a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^3 + 3*(a^3 + a^2*b - a
*b^2 - b^3)*cosh(x)*sinh(x)^2 + (a^3 + a^2*b - a*b^2 - b^3)*sinh(x)^3 - ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^
2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a
*b + b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 + (a^2 - b^2
)*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) - (a^3 - a^2*b
- a*b^2 + b^3)*cosh(x) - (a^3 - a^2*b - a*b^2 + b^3 - 3*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2)*sinh(x))/(a^6
- 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 + (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2
*a*b^5 + b^6)*cosh(x)^4 + 4*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)*sinh(x)^3
+ (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*sinh(x)^4 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 -
b^6)*cosh(x)^2 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 3*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*
a*b^5 + b^6)*cosh(x)^2)*sinh(x)^2 + 4*((a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)
^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x))]

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giac [A]  time = 0.14, size = 88, normalized size = 1.14 \[ \frac {\arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} - a e^{x} + b e^{x}}{{\left (a^{2} - b^{2}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.26, size = 167, normalized size = 2.17 \[ \frac {-\frac {\left (a^{2}-2 b^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) a}+\frac {b \left (a^{2}+2 b^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) a^{2}}+\frac {\left (a^{2}+2 b^{2}\right ) \tanh \left (\frac {x}{2}\right )}{\left (a^{2}-b^{2}\right ) a}+\frac {2 b}{2 a^{2}-2 b^{2}}}{\left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )^{2}}+\frac {\arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)+b*sinh(x))^3,x)

[Out]

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

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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(1/(a*cosh(x)+b*sinh(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*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 = 1.61, size = 157, normalized size = 2.04 \[ \frac {{\mathrm {e}}^x}{\left (a+b\right )\,\left (a-b\right )\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )}-\frac {2\,{\mathrm {e}}^x}{\left (a+b\right )\,\left ({\mathrm {e}}^{4\,x}\,{\left (a+b\right )}^2+{\left (a-b\right )}^2+2\,{\mathrm {e}}^{2\,x}\,\left (a+b\right )\,\left (a-b\right )\right )}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{-a^3+a^2\,b+a\,b^2-b^3}\right )}{\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x) + b*sinh(x))^3,x)

[Out]

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

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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(1/(a*cosh(x)+b*sinh(x))**3,x)

[Out]

Timed out

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