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

Optimal. Leaf size=60 \[ \frac {a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {2 b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {2754, 12, 2660, 618, 206} \[ \frac {a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {2 b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(a + b*Sinh[x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2754

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 68, normalized size = 1.13 \[ \frac {a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {2 b \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 341, normalized size = 5.68 \[ -\frac {2 \, a^{3} b + 2 \, a b^{3} + {\left (b^{3} \cosh \relax (x)^{2} + b^{3} \sinh \relax (x)^{2} + 2 \, a b^{2} \cosh \relax (x) - b^{3} + 2 \, {\left (b^{3} \cosh \relax (x) + a b^{2}\right )} \sinh \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x) - 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \sinh \relax (x)}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)^{2} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \relax (x)^{2} - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \relax (x) - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^3*b + 2*a*b^3 + (b^3*cosh(x)^2 + b^3*sinh(x)^2 + 2*a*b^2*cosh(x) - b^3 + 2*(b^3*cosh(x) + a*b^2)*sinh(x)
)*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sin
h(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x)
+ a)*sinh(x) - b)) - 2*(a^4 + a^2*b^2)*cosh(x) - 2*(a^4 + a^2*b^2)*sinh(x))/(a^4*b^2 + 2*a^2*b^4 + b^6 - (a^4*
b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 - (a^4*b^2 + 2*a^2*b^4 + b^6)*sinh(x)^2 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(
x) - 2*(a^5*b + 2*a^3*b^3 + a*b^5 + (a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x))*sinh(x))

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giac [A]  time = 0.20, size = 99, normalized size = 1.65 \[ \frac {b \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a^{2} e^{x} - a b\right )}}{{\left (a^{2} b + b^{3}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 97, normalized size = 1.62 \[ \frac {8 \tanh \left (\frac {x}{2}\right ) b +8 a}{\left (-4 a^{2}-4 b^{2}\right ) \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}-\frac {8 b \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (-4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(2*tanh(1/2*x)*b+2*a)/(-4*a^2-4*b^2)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)-8*b/(-4*a^2-4*b^2)/(a^2+b^2)^(1/2)*
arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))

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maxima [B]  time = 0.42, size = 117, normalized size = 1.95 \[ \frac {b \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a^{2} e^{\left (-x\right )} + a b\right )}}{a^{2} b^{2} + b^{4} + 2 \, {\left (a^{3} b + a b^{3}\right )} e^{\left (-x\right )} - {\left (a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.70, size = 142, normalized size = 2.37 \[ \frac {\frac {2\,a\,b}{a^2\,b+b^3}-\frac {2\,a^2\,{\mathrm {e}}^x}{a^2\,b+b^3}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}-\frac {b\,\ln \left (-\frac {2\,{\mathrm {e}}^x}{a^2+b^2}-\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}}+\frac {b\,\ln \left (\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {2\,{\mathrm {e}}^x}{a^2+b^2}\right )}{{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a + b*sinh(x))^2,x)

[Out]

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

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

[Out]

Timed out

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