3.20 \(\int \frac{x}{a+b \text{csch}(c+d x^2)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0984599, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5437, 3783, 2660, 618, 204} \[ \frac{b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2+b^2}}\right )}{a d \sqrt{a^2+b^2}}+\frac{x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Csch[c + d*x^2]),x]

[Out]

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

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.134403, size = 71, normalized size = 1.18 \[ \frac{-\frac{2 b \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{-a^2-b^2}}\right )}{d \sqrt{-a^2-b^2}}+\frac{c}{d}+x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Csch[c + d*x^2]),x]

[Out]

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

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Maple [A]  time = 0.04, size = 94, normalized size = 1.6 \begin{align*} -{\frac{b}{da}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,b\tanh \left ( 1/2\,d{x}^{2}+c/2 \right ) -2\,a \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{1}{2\,da}\ln \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{2\,da}\ln \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.56451, size = 513, normalized size = 8.55 \begin{align*} \frac{{\left (a^{2} + b^{2}\right )} d x^{2} + \sqrt{a^{2} + b^{2}} b \log \left (\frac{a^{2} \cosh \left (d x^{2} + c\right )^{2} + a^{2} \sinh \left (d x^{2} + c\right )^{2} + 2 \, a b \cosh \left (d x^{2} + c\right ) + a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (d x^{2} + c\right ) + a b\right )} \sinh \left (d x^{2} + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right ) + b\right )}}{a \cosh \left (d x^{2} + c\right )^{2} + a \sinh \left (d x^{2} + c\right )^{2} + 2 \, b \cosh \left (d x^{2} + c\right ) + 2 \,{\left (a \cosh \left (d x^{2} + c\right ) + b\right )} \sinh \left (d x^{2} + c\right ) - a}\right )}{2 \,{\left (a^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 + b^2)*d*x^2 + sqrt(a^2 + b^2)*b*log((a^2*cosh(d*x^2 + c)^2 + a^2*sinh(d*x^2 + c)^2 + 2*a*b*cosh(d*x
^2 + c) + a^2 + 2*b^2 + 2*(a^2*cosh(d*x^2 + c) + a*b)*sinh(d*x^2 + c) + 2*sqrt(a^2 + b^2)*(a*cosh(d*x^2 + c) +
a*sinh(d*x^2 + c) + b))/(a*cosh(d*x^2 + c)^2 + a*sinh(d*x^2 + c)^2 + 2*b*cosh(d*x^2 + c) + 2*(a*cosh(d*x^2 +
c) + b)*sinh(d*x^2 + 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{x}{a + b \operatorname{csch}{\left (c + d x^{2} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*csch(d*x**2+c)),x)

[Out]

Integral(x/(a + b*csch(c + d*x**2)), x)

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Giac [A]  time = 1.20755, size = 124, normalized size = 2.07 \begin{align*} -\frac{b \log \left (\frac{{\left | 2 \, a e^{\left (d x^{2} + c\right )} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x^{2} + c\right )} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{2 \, \sqrt{a^{2} + b^{2}} a d} + \frac{d x^{2} + c}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*log(abs(2*a*e^(d*x^2 + c) + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^(d*x^2 + c) + 2*b + 2*sqrt(a^2 + b^2)))/
(sqrt(a^2 + b^2)*a*d) + 1/2*(d*x^2 + c)/(a*d)