3.74 \(\int \frac{1}{a+b \text{csch}(c+d x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{2 b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \sqrt{a^2+b^2}}+\frac{x}{a} \]

[Out]

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

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Rubi [A]  time = 0.0594489, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3783, 2660, 618, 204} \[ \frac{2 b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \sqrt{a^2+b^2}}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x])^(-1),x]

[Out]

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

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 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 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 \text{csch}(c+d x)} \, dx &=\frac{x}{a}-\frac{\int \frac{1}{1+\frac{a \sinh (c+d x)}{b}} \, dx}{a}\\ &=\frac{x}{a}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{1-\frac{2 i a x}{b}+x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a d}\\ &=\frac{x}{a}-\frac{(4 i) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1+\frac{a^2}{b^2}\right )-x^2} \, dx,x,-\frac{2 i a}{b}+2 \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a d}\\ &=\frac{x}{a}+\frac{2 b \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}-\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2}}\right )}{a \sqrt{a^2+b^2} d}\\ \end{align*}

Mathematica [A]  time = 0.103844, size = 64, normalized size = 1.19 \[ \frac{-\frac{2 b \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{d \sqrt{-a^2-b^2}}+\frac{c}{d}+x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 87, normalized size = 1.6 \begin{align*}{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{b}{da\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( 1/2\,dx+c/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-2/d/a*b/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*tanh(1/2*d*x+1/2*c)-2*a)/(a^2+b^2)^(1
/2))-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)

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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*csch(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60734, size = 473, normalized size = 8.76 \begin{align*} \frac{{\left (a^{2} + b^{2}\right )} d x + \sqrt{a^{2} + b^{2}} b \log \left (\frac{a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b\right )}}{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) + 2 \,{\left (a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) - a}\right )}{{\left (a^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \operatorname{csch}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a + b*csch(c + d*x)), x)

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Giac [A]  time = 1.17666, size = 115, normalized size = 2.13 \begin{align*} -\frac{b \log \left (\frac{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a d} + \frac{d x + c}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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