3.102 \(\int \frac{1}{(a+b \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=79 \[ -\frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac{b \cosh (c+d x)}{d \left (a^2+b^2\right ) (a+b \sinh (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.0628787, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2664, 12, 2660, 618, 204} \[ -\frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac{b \cosh (c+d x)}{d \left (a^2+b^2\right ) (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^(-2),x]

[Out]

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

Rule 2664

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

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.25853, size = 85, normalized size = 1.08 \[ -\frac{\frac{2 a \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac{b \cosh (c+d x)}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x])^(-2),x]

[Out]

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

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Maple [A]  time = 0.034, size = 118, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a} \left ( -{\frac{{b}^{2}\tanh \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{b}{{a}^{2}+{b}^{2}}} \right ) }+2\,{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(d*x+c))^2,x)

[Out]

1/d*(-2*(-b^2/a/(a^2+b^2)*tanh(1/2*d*x+1/2*c)-b/(a^2+b^2))/(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)
+2*a/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-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(1/(a+b*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.10616, size = 1041, normalized size = 13.18 \begin{align*} -\frac{2 \, a^{2} b + 2 \, b^{3} -{\left (a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) - a b + 2 \,{\left (a b \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \,{\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - 2 \,{\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) - 2 \,{\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right ) -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d + 2 \,{\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

-(2*a^2*b + 2*b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) - a*b + 2*(a*b*cosh(d*x +
c) + a^2)*sinh(d*x + c))*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c)
+ 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x +
c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) -
b)) - 2*(a^3 + a*b^2)*cosh(d*x + c) - 2*(a^3 + a*b^2)*sinh(d*x + c))/((a^4*b + 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) - (a^4*b + 2*
a^2*b^3 + b^5)*d + 2*((a^4*b + 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c))

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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(1/(a+b*sinh(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.35334, size = 180, normalized size = 2.28 \begin{align*} -\frac{a \log \left (\frac{{\left | -2 \, b e^{\left (d x + c\right )} - 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | -2 \, b e^{\left (d x + c\right )} - 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} d + b^{2} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (a e^{\left (d x + c\right )} - b\right )}}{{\left (a^{2} d + b^{2} d\right )}{\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

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