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

Optimal. Leaf size=113 \[ \frac{b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a d \left (a^2+b^2\right ) \left (a+b \text{csch}\left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.230873, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5437, 3785, 3919, 3831, 2660, 618, 204} \[ \frac{b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a d \left (a^2+b^2\right ) \left (a+b \text{csch}\left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Csch[c + d*x^2])^2,x]

[Out]

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

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 3785

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, 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}{\left (a+b \text{csch}\left (c+d x^2\right )\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b \text{csch}(c+d x))^2} \, dx,x,x^2\right )\\ &=-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a^2-b^2+a b \text{csch}(c+d x)}{a+b \text{csch}(c+d x)} \, dx,x,x^2\right )}{2 a \left (a^2+b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}-\frac{\left (b \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\text{csch}(c+d x)}{a+b \text{csch}(c+d x)} \, dx,x,x^2\right )}{2 a^2 \left (a^2+b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}-\frac{\left (2 a^2+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \sinh (c+d x)}{b}} \, dx,x,x^2\right )}{2 a^2 \left (a^2+b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}+\frac{\left (i \left (2 a^2+b^2\right )\right ) \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^2 \left (a^2+b^2\right ) d}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}-\frac{\left (2 i \left (2 a^2+b^2\right )\right ) \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^2 \left (a^2+b^2\right ) d}\\ &=\frac{x^2}{2 a^2}+\frac{b \left (2 a^2+b^2\right ) \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^2 \left (a^2+b^2\right )^{3/2} d}-\frac{b^2 \coth \left (c+d x^2\right )}{2 a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d x^2\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.450743, size = 161, normalized size = 1.42 \[ \frac{\text{csch}\left (c+d x^2\right ) \left (a \sinh \left (c+d x^2\right )+b\right ) \left (-\frac{a b^2 \coth \left (c+d x^2\right )}{a^2+b^2}+\frac{2 b \left (2 a^2+b^2\right ) \left (a+b \text{csch}\left (c+d x^2\right )\right ) \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\left (c+d x^2\right ) \left (a+b \text{csch}\left (c+d x^2\right )\right )\right )}{2 a^2 d \left (a+b \text{csch}\left (c+d x^2\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.065, size = 255, normalized size = 2.3 \begin{align*}{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) }\tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( 1/2\,d{x}^{2}+c/2 \right ) -b \right ) ^{-1}}+{\frac{{b}^{2}}{ad \left ({a}^{2}+{b}^{2} \right ) } \left ( \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( 1/2\,d{x}^{2}+c/2 \right ) -b \right ) ^{-1}}-2\,{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( 1/2\,d{x}^{2}+c/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{{b}^{3}}{d{a}^{2}}{\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 ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{2\,d{a}^{2}}\ln \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{2\,d{a}^{2}}\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))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.96933, size = 1596, normalized size = 14.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^5 + 2*a^3*b^2 + a*b^4)*d*x^2*cosh(d*x^2 + c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d*x^2*sinh(d*x^2 + c)^2 - 2
*a^3*b^2 - 2*a*b^4 - (a^5 + 2*a^3*b^2 + a*b^4)*d*x^2 - (2*a^3*b + a*b^3 - (2*a^3*b + a*b^3)*cosh(d*x^2 + c)^2
- (2*a^3*b + a*b^3)*sinh(d*x^2 + c)^2 - 2*(2*a^2*b^2 + b^4)*cosh(d*x^2 + c) - 2*(2*a^2*b^2 + b^4 + (2*a^3*b +
a*b^3)*cosh(d*x^2 + c))*sinh(d*x^2 + c))*sqrt(a^2 + b^2)*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*cos
h(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)) + 2*(a^2*b^3 + b^5 + (a^4*b + 2*a^2*b^3 + b^5)*d*x^2)*cosh(d*x^2 +
c) + 2*(a^2*b^3 + b^5 + (a^5 + 2*a^3*b^2 + a*b^4)*d*x^2*cosh(d*x^2 + c) + (a^4*b + 2*a^2*b^3 + b^5)*d*x^2)*si
nh(d*x^2 + c))/((a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x^2 + c)^2 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d*sinh(d*x^2 + c
)^2 + 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*d*cosh(d*x^2 + c) - (a^7 + 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 + 2*a^5*b^2
+ a^3*b^4)*d*cosh(d*x^2 + c) + (a^6*b + 2*a^4*b^3 + a^2*b^5)*d)*sinh(d*x^2 + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a + b \operatorname{csch}{\left (c + d x^{2} \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21029, size = 239, normalized size = 2.12 \begin{align*} -\frac{{\left (2 \, a^{2} b + b^{3}\right )} \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 \,{\left (a^{4} d + a^{2} b^{2} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{b^{3} e^{\left (d x^{2} + c\right )} - a b^{2}}{{\left (a^{4} d + a^{2} b^{2} d\right )}{\left (a e^{\left (2 \, d x^{2} + 2 \, c\right )} + 2 \, b e^{\left (d x^{2} + c\right )} - a\right )}} + \frac{d x^{2} + c}{2 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*b + b^3)*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)))/((a^4*d + a^2*b^2*d)*sqrt(a^2 + b^2)) + (b^3*e^(d*x^2 + c) - a*b^2)/((a^4*d + a^2*b^2*d)*(a*e^(2
*d*x^2 + 2*c) + 2*b*e^(d*x^2 + c) - a)) + 1/2*(d*x^2 + c)/(a^2*d)