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]

1/2*x^2/a^2+b*(2*a^2+b^2)*arctanh((a-b*tanh(1/2*d*x^2+1/2*c))/(a^2+b^2)^(1/2))/a^2/(a^2+b^2)^(3/2)/d-1/2*b^2*c
oth(d*x^2+c)/a/(a^2+b^2)/d/(a+b*csch(d*x^2+c))

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Rubi [A]  time = 0.23, antiderivative size = 113, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.48, 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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fricas [B]  time = 0.45, size = 711, normalized size = 6.29 \[ \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \cosh \left (d x^{2} + c\right )^{2} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \sinh \left (d x^{2} + c\right )^{2} - 2 \, a^{3} b^{2} - 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} - {\left (2 \, a^{3} b + a b^{3} - {\left (2 \, a^{3} b + a b^{3}\right )} \cosh \left (d x^{2} + c\right )^{2} - {\left (2 \, a^{3} b + a b^{3}\right )} \sinh \left (d x^{2} + c\right )^{2} - 2 \, {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x^{2} + c\right ) - 2 \, {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + a b^{3}\right )} \cosh \left (d x^{2} + c\right )\right )} \sinh \left (d x^{2} + c\right )\right )} \sqrt {a^{2} + b^{2}} \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^{2} b^{3} + b^{5} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2}\right )} \cosh \left (d x^{2} + c\right ) + 2 \, {\left (a^{2} b^{3} + b^{5} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \cosh \left (d x^{2} + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2}\right )} \sinh \left (d x^{2} + c\right )}{2 \, {\left ({\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cosh \left (d x^{2} + c\right )^{2} + {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sinh \left (d x^{2} + c\right )^{2} + 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cosh \left (d x^{2} + c\right ) - {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d + 2 \, {\left ({\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cosh \left (d x^{2} + c\right ) + {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )} \sinh \left (d x^{2} + c\right )\right )}} \]

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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giac [A]  time = 0.17, size = 177, normalized size = 1.57 \[ -\frac {{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a e^{\left (d x^{2} + c\right )} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x^{2} + c\right )} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{2 \, {\left (a^{4} d + a^{2} b^{2} d\right )} \sqrt {a^{2} + b^{2}}} + \frac {b^{3} e^{\left (d x^{2} + c\right )} - a b^{2}}{{\left (a^{4} d + a^{2} b^{2} d\right )} {\left (a e^{\left (2 \, d x^{2} + 2 \, c\right )} + 2 \, b e^{\left (d x^{2} + c\right )} - a\right )}} + \frac {d x^{2} + c}{2 \, a^{2} d} \]

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)

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maple [B]  time = 0.34, size = 255, normalized size = 2.26 \[ -\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}+\frac {b \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right ) b -2 a \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-b \right ) \left (a^{2}+b^{2}\right )}+\frac {b^{2}}{d a \left (\left (\tanh ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right ) b -2 a \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-b \right ) \left (a^{2}+b^{2}\right )}-\frac {2 b \arctanh \left (\frac {2 b \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {b^{3} \arctanh \left (\frac {2 b \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{2} \left (a^{2}+b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)+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))

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maxima [A]  time = 0.87, size = 200, normalized size = 1.77 \[ -\frac {{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac {a e^{\left (-d x^{2} - c\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-d x^{2} - c\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{2 \, {\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a^{2} + b^{2}} d} - \frac {b^{3} e^{\left (-d x^{2} - c\right )} + a b^{2}}{{\left (a^{5} + a^{3} b^{2} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} e^{\left (-d x^{2} - c\right )} - {\left (a^{5} + a^{3} b^{2}\right )} e^{\left (-2 \, d x^{2} - 2 \, c\right )}\right )} d} + \frac {d x^{2} + c}{2 \, a^{2} d} \]

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]

-1/2*(2*a^2*b + b^3)*log((a*e^(-d*x^2 - c) - b - sqrt(a^2 + b^2))/(a*e^(-d*x^2 - c) - b + sqrt(a^2 + b^2)))/((
a^4 + a^2*b^2)*sqrt(a^2 + b^2)*d) - (b^3*e^(-d*x^2 - c) + a*b^2)/((a^5 + a^3*b^2 + 2*(a^4*b + a^2*b^3)*e^(-d*x
^2 - c) - (a^5 + a^3*b^2)*e^(-2*d*x^2 - 2*c))*d) + 1/2*(d*x^2 + c)/(a^2*d)

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mupad [B]  time = 1.97, size = 290, normalized size = 2.57 \[ \frac {x^2}{2\,a^2}-\frac {\frac {b^2}{d\,\left (a^3+a\,b^2\right )}-\frac {b^3\,{\mathrm {e}}^{d\,x^2+c}}{a\,d\,\left (a^3+a\,b^2\right )}}{2\,b\,{\mathrm {e}}^{d\,x^2+c}-a+a\,{\mathrm {e}}^{2\,d\,x^2+2\,c}}-\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2+b^2\right )}{a^3\,\left (a^2+b^2\right )}-\frac {2\,b\,x\,\left (2\,a^2+b^2\right )\,\left (a-b\,{\mathrm {e}}^{d\,x^2+c}\right )}{a^3\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )}{2\,a^2\,d\,{\left (a^2+b^2\right )}^{3/2}}+\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2+b^2\right )}{a^3\,\left (a^2+b^2\right )}+\frac {2\,b\,x\,\left (2\,a^2+b^2\right )\,\left (a-b\,{\mathrm {e}}^{d\,x^2+c}\right )}{a^3\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )}{2\,a^2\,d\,{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/sinh(c + d*x^2))^2,x)

[Out]

x^2/(2*a^2) - (b^2/(d*(a*b^2 + a^3)) - (b^3*exp(c + d*x^2))/(a*d*(a*b^2 + a^3)))/(2*b*exp(c + d*x^2) - a + a*e
xp(2*c + 2*d*x^2)) - (b*log((2*b*x*exp(c + d*x^2)*(2*a^2 + b^2))/(a^3*(a^2 + b^2)) - (2*b*x*(2*a^2 + b^2)*(a -
b*exp(c + d*x^2)))/(a^3*(a^2 + b^2)^(3/2)))*(2*a^2 + b^2))/(2*a^2*d*(a^2 + b^2)^(3/2)) + (b*log((2*b*x*exp(c
+ d*x^2)*(2*a^2 + b^2))/(a^3*(a^2 + b^2)) + (2*b*x*(2*a^2 + b^2)*(a - b*exp(c + d*x^2)))/(a^3*(a^2 + b^2)^(3/2
)))*(2*a^2 + b^2))/(2*a^2*d*(a^2 + b^2)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {csch}{\left (c + d x^{2} \right )}\right )^{2}}\, dx \]

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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