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[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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Rubi [A]  time = 0.0737309, antiderivative size = 60, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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 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])

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*}

Mathematica [A]  time = 0.101181, 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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Maple [A]  time = 0.02, size = 97, normalized size = 1.6 \begin{align*} 4\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,a}{ \left ( -4\,{a}^{2}-4\,{b}^{2} \right ) \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-8\,{\frac{b}{ \left ( -4\,{a}^{2}-4\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0855, size = 855, normalized size = 14.25 \begin{align*} -\frac{2 \, a^{3} b + 2 \, a b^{3} +{\left (b^{3} \cosh \left (x\right )^{2} + b^{3} \sinh \left (x\right )^{2} + 2 \, a b^{2} \cosh \left (x\right ) - b^{3} + 2 \,{\left (b^{3} \cosh \left (x\right ) + a b^{2}\right )} \sinh \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \,{\left (a^{4} + a^{2} b^{2}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{4} + a^{2} b^{2}\right )} \sinh \left (x\right )}{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 \left (x\right )^{2} -{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \left (x\right ) - 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 \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.39155, size = 134, normalized size = 2.23 \begin{align*} \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 )}} \end{align*}

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))