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

Optimal. Leaf size=44 \[ -\frac{2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \sqrt{a^2+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.0355395, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2660, 618, 204} \[ -\frac{2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^(-1),x]

[Out]

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

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)} \, dx &=-\frac{(2 i) \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 )}{d}\\ &=\frac{(4 i) \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 )}{d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2} d}\\ \end{align*}

Mathematica [A]  time = 0.0360528, size = 52, normalized size = 1.18 \[ \frac{2 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{d \sqrt{-a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.015, size = 43, normalized size = 1. \begin{align*} 2\,{\frac{1}{d\sqrt{{a}^{2}+{b}^{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/(a^2+b^2)^(1/2)*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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99491, size = 424, normalized size = 9.64 \begin{align*} \frac{\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 )}{\sqrt{a^{2} + b^{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))/(sqrt(a^2 + b^2)*d)

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Sympy [A]  time = 20.6097, size = 189, normalized size = 4.3 \begin{align*} \begin{cases} - \frac{2 i \sqrt{b^{2}}}{- b^{2} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b d \sqrt{b^{2}}} & \text{for}\: a = - \sqrt{- b^{2}} \\- \frac{2 i \sqrt{b^{2}}}{b^{2} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b d \sqrt{b^{2}}} & \text{for}\: a = \sqrt{- b^{2}} \\\frac{\log{\left (\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{b d} & \text{for}\: a = 0 \\\frac{x}{a + b \sinh{\left (c \right )}} & \text{for}\: d = 0 \\- \frac{\log{\left (\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{b}{a} - \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{d \sqrt{a^{2} + b^{2}}} + \frac{\log{\left (\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{b}{a} + \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{d \sqrt{a^{2} + b^{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*I*sqrt(b**2)/(-b**2*d*tanh(c/2 + d*x/2) + I*b*d*sqrt(b**2)), Eq(a, -sqrt(-b**2))), (-2*I*sqrt(b*
*2)/(b**2*d*tanh(c/2 + d*x/2) + I*b*d*sqrt(b**2)), Eq(a, sqrt(-b**2))), (log(tanh(c/2 + d*x/2))/(b*d), Eq(a, 0
)), (x/(a + b*sinh(c)), Eq(d, 0)), (-log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(d*sqrt(a**2 + b**2))
+ log(tanh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/(d*sqrt(a**2 + b**2)), True))

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Giac [A]  time = 1.19226, size = 90, normalized size = 2.05 \begin{align*} \frac{\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 )}{\sqrt{a^{2} + b^{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)))/(sqrt(a^2 +
b^2)*d)