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

Optimal. Leaf size=66 \[ -\frac {a}{\left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}-\frac {b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3154, 3074, 206} \[ -\frac {a}{\left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}-\frac {b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2)) - a/((a^2 - b^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 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 3154

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 125, normalized size = 1.89 \[ -\frac {2 b^2 \sqrt {a+b} \sinh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+2 a b \sqrt {a+b} \cosh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+a \sqrt {a-b} (a+b)}{(a-b)^{3/2} (a+b)^2 (a \cosh (x)+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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*Cosh[x
] + b*Sinh[x])))

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fricas [B]  time = 0.45, size = 594, normalized size = 9.00 \[ \left [\frac {{\left ({\left (a b + b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a b + b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a b + b^{2}\right )} \sinh \relax (x)^{2} + a b - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - a + b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} + a - b}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cosh \relax (x) - 2 \, {\left (a^{3} - a b^{2}\right )} \sinh \relax (x)}{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}}, \frac {2 \, {\left ({\left ({\left (a b + b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a b + b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a b + b^{2}\right )} \sinh \relax (x)^{2} + a b - b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (\frac {\sqrt {a^{2} - b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right ) - {\left (a^{3} - a b^{2}\right )} \cosh \relax (x) - {\left (a^{3} - a b^{2}\right )} \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}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(((a*b + b^2)*cosh(x)^2 + 2*(a*b + b^2)*cosh(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 + a*b - b^2)*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*b^2)*cos
h(x) - 2*(a^3 - a*b^2)*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), 2*(((a*b + b^2)*cosh(x)^2 + 2*(a*b + b^2)*cos
h(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 + a*b - b^2)*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a
+ b)*sinh(x))) - (a^3 - a*b^2)*cosh(x) - (a^3 - a*b^2)*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 = 0.13, size = 72, normalized size = 1.09 \[ -\frac {2 \, b \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, a e^{x}}{{\left (a^{2} - b^{2}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 99, normalized size = 1.50 \[ \frac {-8 \tanh \left (\frac {x}{2}\right ) b -8 a}{\left (4 a^{2}-4 b^{2}\right ) \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}-\frac {8 b \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(-2*tanh(1/2*x)*b-2*a)/(4*a^2-4*b^2)/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)-8*b/(4*a^2-4*b^2)/(a^2-b^2)^(1/2)*a
rctan(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(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 = 1.65, size = 183, normalized size = 2.77 \[ -\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\left (b^2\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+a\,b\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}\right )}{a^4\,\sqrt {b^2}-2\,a^2\,{\left (b^2\right )}^{3/2}+b^4\,\sqrt {b^2}+a\,b\,{\left (b^2\right )}^{3/2}-a\,b^3\,\sqrt {b^2}}\right )\,\sqrt {b^2}}{\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}-\frac {2\,a\,{\mathrm {e}}^x}{\left (a+b\right )\,\left (a-b\right )\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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