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

Optimal. Leaf size=68 \[ \frac {x \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a}{\left (a^2-b^2\right ) (a \coth (x)+b)}-\frac {2 a b \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2} \]

[Out]

(a^2+b^2)*x/(a^2-b^2)^2-a/(a^2-b^2)/(b+a*coth(x))-2*a*b*ln(a*cosh(x)+b*sinh(x))/(a^2-b^2)^2

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Rubi [A]  time = 0.14, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3085, 3483, 3531, 3530} \[ \frac {x \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a}{\left (a^2-b^2\right ) (a \coth (x)+b)}-\frac {2 a b \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3085

Int[sin[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(b + a*Cot[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3483

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 61, normalized size = 0.90 \[ \frac {x \left (a^2+b^2\right )-\frac {a (a-b) (a+b) \sinh (x)}{a \cosh (x)+b \sinh (x)}-2 a b \log (a \cosh (x)+b \sinh (x))}{(a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 348, normalized size = 5.12 \[ \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \cosh \relax (x)^{2} + 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \cosh \relax (x) \sinh \relax (x) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \sinh \relax (x)^{2} + 2 \, a^{3} - 2 \, a^{2} b + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} x - 2 \, {\left (a^{2} b - a b^{2} + {\left (a^{2} b + a b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} b + a b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{2} b + a b^{2}\right )} \sinh \relax (x)^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*x*cosh(x)^2 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*x*cosh(x)*sinh(x) + (a^3 + 3*
a^2*b + 3*a*b^2 + b^3)*x*sinh(x)^2 + 2*a^3 - 2*a^2*b + (a^3 + a^2*b - a*b^2 - b^3)*x - 2*(a^2*b - a*b^2 + (a^2
*b + a*b^2)*cosh(x)^2 + 2*(a^2*b + a*b^2)*cosh(x)*sinh(x) + (a^2*b + a*b^2)*sinh(x)^2)*log(2*(a*cosh(x) + b*si
nh(x))/(cosh(x) - sinh(x))))/(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)^2 + 2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*cosh(x)*sinh(x) + (a
^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*sinh(x)^2)

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giac [A]  time = 0.14, size = 113, normalized size = 1.66 \[ -\frac {2 \, a b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {x}{a^{2} - 2 \, a b + b^{2}} + \frac {2 \, {\left (a b e^{\left (2 \, x\right )} + a^{2} - a b\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.26, size = 146, normalized size = 2.15 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{\left (a +b \right )^{2}}-\frac {2 a^{3} \tanh \left (\frac {x}{2}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}+\frac {2 a \tanh \left (\frac {x}{2}\right ) b^{2}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {2 a b \ln \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{\left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 104, normalized size = 1.53 \[ -\frac {2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, a^{2}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, x\right )}} + \frac {x}{a^{2} + 2 \, a b + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.47, size = 108, normalized size = 1.59 \[ \frac {\frac {a^2\,\mathrm {cosh}\relax (x)}{b\,\left (a^2-b^2\right )}+\frac {a\,x\,\mathrm {cosh}\relax (x)\,\left (a^2+b^2\right )}{{\left (a^2-b^2\right )}^2}+\frac {b\,x\,\mathrm {sinh}\relax (x)\,\left (a^2+b^2\right )}{{\left (a^2-b^2\right )}^2}}{a\,\mathrm {cosh}\relax (x)+b\,\mathrm {sinh}\relax (x)}+\ln \left (a\,\mathrm {cosh}\relax (x)+b\,\mathrm {sinh}\relax (x)\right )\,\left (\frac {1}{2\,{\left (a+b\right )}^2}-\frac {1}{2\,{\left (a-b\right )}^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.42, size = 983, normalized size = 14.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (2*x*sinh(x)**2/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*
cosh(x)**2) - 4*x*sinh(x)*cosh(x)/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + 2*x*cosh
(x)**2/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + 3*sinh(x)**2/(8*b**2*sinh(x)**2 - 1
6*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) - cosh(x)**2/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2
*cosh(x)**2), Eq(a, -b)), (2*x*sinh(x)**2/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) +
4*x*sinh(x)*cosh(x)/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + 2*x*cosh(x)**2/(8*b**2
*sinh(x)**2 + 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) - 3*sinh(x)**2/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x)
*cosh(x) + 8*b**2*cosh(x)**2) + cosh(x)**2/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2),
Eq(a, b)), ((x - sinh(x)/cosh(x))/a**2, Eq(b, 0)), (x/b**2, Eq(a, 0)), (a**4*cosh(x)/(a**5*b*cosh(x) + a**4*b*
*2*sinh(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x)) + a**3*b*x*cosh(x)/(a*
*5*b*cosh(x) + a**4*b**2*sinh(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x))
+ a**2*b**2*x*sinh(x)/(a**5*b*cosh(x) + a**4*b**2*sinh(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5
*cosh(x) + b**6*sinh(x)) - 2*a**2*b**2*log(cosh(x) + b*sinh(x)/a)*cosh(x)/(a**5*b*cosh(x) + a**4*b**2*sinh(x)
- 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x)) - a**2*b**2*cosh(x)/(a**5*b*cosh(
x) + a**4*b**2*sinh(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x)) + a*b**3*x
*cosh(x)/(a**5*b*cosh(x) + a**4*b**2*sinh(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b*
*6*sinh(x)) - 2*a*b**3*log(cosh(x) + b*sinh(x)/a)*sinh(x)/(a**5*b*cosh(x) + a**4*b**2*sinh(x) - 2*a**3*b**3*co
sh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x)) + b**4*x*sinh(x)/(a**5*b*cosh(x) + a**4*b**2*sinh
(x) - 2*a**3*b**3*cosh(x) - 2*a**2*b**4*sinh(x) + a*b**5*cosh(x) + b**6*sinh(x)), True))

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