3.588 \(\int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx\)

Optimal. Leaf size=67 \[ \frac {2 \sinh (x)}{3 a \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}+\frac {a \sinh (x)+b \cosh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3076, 3075} \[ \frac {2 \sinh (x)}{3 a \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}+\frac {a \sinh (x)+b \cosh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-4),x]

[Out]

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

Rule 3075

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-2), x_Symbol] :> Simp[Sin[c + d*x]/(a*d*
(a*Cos[c + d*x] + b*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))^4} \, dx &=\frac {b \cosh (x)+a \sinh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}+\frac {2 \int \frac {1}{(a \cosh (x)+b \sinh (x))^2} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac {b \cosh (x)+a \sinh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}+\frac {2 \sinh (x)}{3 a \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 64, normalized size = 0.96 \[ \frac {\sinh (x) \left (\left (a^2+b^2\right ) \cosh (2 x)+2 a^2-b^2\right )+a b \cosh (3 x)}{3 a (a-b) (a+b) (a \cosh (x)+b \sinh (x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^(-4),x]

[Out]

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

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fricas [B]  time = 0.42, size = 527, normalized size = 7.87 \[ -\frac {8 \, {\left ({\left (2 \, a + b\right )} \cosh \relax (x) + {\left (a + 2 \, b\right )} \sinh \relax (x)\right )}}{3 \, {\left ({\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \relax (x)^{5} + 5 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \relax (x) \sinh \relax (x)^{4} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sinh \relax (x)^{5} + 3 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \relax (x)^{3} + {\left (3 \, a^{5} + 9 \, a^{4} b + 6 \, a^{3} b^{2} - 6 \, a^{2} b^{3} - 9 \, a b^{4} - 3 \, b^{5} + 10 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{3} + {\left (10 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \relax (x)^{3} + 9 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 2 \, {\left (2 \, a^{5} + a^{4} b - 4 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \cosh \relax (x) + {\left (2 \, a^{5} + 4 \, a^{4} b - 4 \, a^{3} b^{2} - 8 \, a^{2} b^{3} + 2 \, a b^{4} + 4 \, b^{5} + 5 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \relax (x)^{4} + 9 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="fricas")

[Out]

-8/3*((2*a + b)*cosh(x) + (a + 2*b)*sinh(x))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x
)^5 + 5*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)*sinh(x)^4 + (a^5 + 5*a^4*b + 10*a^3*
b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sinh(x)^5 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x
)^3 + (3*a^5 + 9*a^4*b + 6*a^3*b^2 - 6*a^2*b^3 - 9*a*b^4 - 3*b^5 + 10*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3
+ 5*a*b^4 + b^5)*cosh(x)^2)*sinh(x)^3 + (10*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)
^3 + 9*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x))*sinh(x)^2 + 2*(2*a^5 + a^4*b - 4*a^3*b
^2 - 2*a^2*b^3 + 2*a*b^4 + b^5)*cosh(x) + (2*a^5 + 4*a^4*b - 4*a^3*b^2 - 8*a^2*b^3 + 2*a*b^4 + 4*b^5 + 5*(a^5
+ 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^4 + 9*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*
a*b^4 - b^5)*cosh(x)^2)*sinh(x))

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giac [A]  time = 0.14, size = 53, normalized size = 0.79 \[ -\frac {4 \, {\left (3 \, a e^{\left (2 \, x\right )} + 3 \, b e^{\left (2 \, x\right )} + a - b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="giac")

[Out]

-4/3*(3*a*e^(2*x) + 3*b*e^(2*x) + a - b)/((a^2 + 2*a*b + b^2)*(a*e^(2*x) + b*e^(2*x) + a - b)^3)

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maple [A]  time = 0.28, size = 87, normalized size = 1.30 \[ -\frac {2 \left (-\frac {\tanh ^{5}\left (\frac {x}{2}\right )}{a}-\frac {2 b \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {2 \left (a^{2}+2 b^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{3 a^{3}}-\frac {2 b \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {\tanh \left (\frac {x}{2}\right )}{a}\right )}{\left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)+b*sinh(x))^4,x)

[Out]

-2*(-1/a*tanh(1/2*x)^5-2/a^2*b*tanh(1/2*x)^4-2/3/a^3*(a^2+2*b^2)*tanh(1/2*x)^3-2/a^2*b*tanh(1/2*x)^2-1/a*tanh(
1/2*x))/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^3

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maxima [B]  time = 0.62, size = 498, normalized size = 7.43 \[ \frac {4 \, {\left (a - b\right )} e^{\left (-2 \, x\right )}}{a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}} + \frac {4 \, a}{3 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}\right )}} + \frac {4 \, b}{3 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="maxima")

[Out]

4*(a - b)*e^(-2*x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*e^(-2*x) + 3*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*e^(-4*x) + (a^5 - 5*a^4*b
+ 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*e^(-6*x)) + 4/3*a/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b
^5 + 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*e^(-2*x) + 3*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3
- 3*a*b^4 + b^5)*e^(-4*x) + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*e^(-6*x)) + 4/3*b/(a^5
+ a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*e^(-2*x)
+ 3*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*e^(-4*x) + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b
^3 + 5*a*b^4 - b^5)*e^(-6*x))

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mupad [B]  time = 1.70, size = 47, normalized size = 0.70 \[ -\frac {a\,\left (4\,{\mathrm {e}}^{2\,x}+\frac {4}{3}\right )+b\,\left (4\,{\mathrm {e}}^{2\,x}-\frac {4}{3}\right )}{{\left (a+b\right )}^2\,{\left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x) + b*sinh(x))^4,x)

[Out]

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

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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))**4,x)

[Out]

Timed out

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