3.68 \(\int \frac{1}{\sqrt{x} (a+b \text{csch}(c+d \sqrt{x}))^2} \, dx\)

Optimal. Leaf size=118 \[ \frac{4 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a d \left (a^2+b^2\right ) \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

[Out]

(2*Sqrt[x])/a^2 + (4*b*(2*a^2 + b^2)*ArcTanh[(a - b*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2
)^(3/2)*d) - (2*b^2*Coth[c + d*Sqrt[x]])/(a*(a^2 + b^2)*d*(a + b*Csch[c + d*Sqrt[x]]))

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Rubi [A]  time = 0.212748, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {5437, 3785, 3919, 3831, 2660, 618, 204} \[ \frac{4 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a d \left (a^2+b^2\right ) \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])^2),x]

[Out]

(2*Sqrt[x])/a^2 + (4*b*(2*a^2 + b^2)*ArcTanh[(a - b*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2
)^(3/2)*d) - (2*b^2*Coth[c + d*Sqrt[x]])/(a*(a^2 + b^2)*d*(a + b*Csch[c + d*Sqrt[x]]))

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{1}{\sqrt{x} \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{(a+b \text{csch}(c+d x))^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}-\frac{2 \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,\sqrt{x}\right )}{a \left (a^2+b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}-\frac{\left (2 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,\sqrt{x}\right )}{a^2 \left (a^2+b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}-\frac{\left (2 \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \sinh (c+d x)}{b}} \, dx,x,\sqrt{x}\right )}{a^2 \left (a^2+b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}+\frac{\left (4 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 \sqrt{x}\right )\right )\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac{2 \sqrt{x}}{a^2}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}-\frac{\left (8 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 \sqrt{x}\right )\right )\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac{2 \sqrt{x}}{a^2}+\frac{4 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 \sqrt{x}\right )\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{3/2} d}-\frac{2 b^2 \coth \left (c+d \sqrt{x}\right )}{a \left (a^2+b^2\right ) d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.437044, size = 175, normalized size = 1.48 \[ \frac{2 \text{csch}\left (c+d \sqrt{x}\right ) \left (a \sinh \left (c+d \sqrt{x}\right )+b\right ) \left (-\frac{a b^2 \coth \left (c+d \sqrt{x}\right )}{a^2+b^2}+\frac{2 b \left (2 a^2+b^2\right ) \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right ) \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\left (c+d \sqrt{x}\right ) \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )\right )}{a^2 d \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])^2),x]

[Out]

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

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Maple [B]  time = 0.088, size = 257, normalized size = 2.2 \begin{align*} 4\,{\frac{b\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) }{d \left ( \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}b-2\,a\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -b \right ) \left ({a}^{2}+{b}^{2} \right ) }}+4\,{\frac{{b}^{2}}{ad \left ( \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}b-2\,a\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -b \right ) \left ({a}^{2}+{b}^{2} \right ) }}-8\,{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-4\,{\frac{{b}^{3}}{d{a}^{2} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{\ln \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -1 \right ) }{d{a}^{2}}}+2\,{\frac{\ln \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) +1 \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*csch(c+d*x^(1/2)))^2/x^(1/2),x)

[Out]

4/d*b/(tanh(1/2*c+1/2*d*x^(1/2))^2*b-2*a*tanh(1/2*c+1/2*d*x^(1/2))-b)/(a^2+b^2)*tanh(1/2*c+1/2*d*x^(1/2))+4/d/
a*b^2/(tanh(1/2*c+1/2*d*x^(1/2))^2*b-2*a*tanh(1/2*c+1/2*d*x^(1/2))-b)/(a^2+b^2)-8/d*b/(a^2+b^2)^(3/2)*arctanh(
1/2*(2*b*tanh(1/2*c+1/2*d*x^(1/2))-2*a)/(a^2+b^2)^(1/2))-4/d/a^2*b^3/(a^2+b^2)^(3/2)*arctanh(1/2*(2*b*tanh(1/2
*c+1/2*d*x^(1/2))-2*a)/(a^2+b^2)^(1/2))-2/d/a^2*ln(tanh(1/2*c+1/2*d*x^(1/2))-1)+2/d/a^2*ln(tanh(1/2*c+1/2*d*x^
(1/2))+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(c+d*x^(1/2)))^2/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.23051, size = 1611, normalized size = 13.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(c+d*x^(1/2)))^2/x^(1/2),x, algorithm="fricas")

[Out]

-2*(2*a^3*b^2 + 2*a*b^4 - (a^5 + 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt(x) + c)^2 - (a^5 + 2*a^3*b^2 + a*b^4
)*d*sqrt(x)*sinh(d*sqrt(x) + c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d*sqrt(x) - 2*(a^2*b^3 + b^5 + (a^4*b + 2*a^2*b^
3 + b^5)*d*sqrt(x))*cosh(d*sqrt(x) + c) - ((2*a^3*b + a*b^3)*sqrt(a^2 + b^2)*cosh(d*sqrt(x) + c)^2 + (2*a^3*b
+ a*b^3)*sqrt(a^2 + b^2)*sinh(d*sqrt(x) + c)^2 + 2*(2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)*cosh(d*sqrt(x) + c) + 2*(
(2*a^3*b + a*b^3)*sqrt(a^2 + b^2)*cosh(d*sqrt(x) + c) + (2*a^2*b^2 + b^4)*sqrt(a^2 + b^2))*sinh(d*sqrt(x) + c)
- (2*a^3*b + a*b^3)*sqrt(a^2 + b^2))*log((a*b + (a^2 + b^2 + sqrt(a^2 + b^2)*b)*cosh(d*sqrt(x) + c) - (b^2 +
sqrt(a^2 + b^2)*b)*sinh(d*sqrt(x) + c) + sqrt(a^2 + b^2)*a)/(a*sinh(d*sqrt(x) + c) + b)) - 2*(a^2*b^3 + b^5 +
(a^5 + 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt(x) + c) + (a^4*b + 2*a^2*b^3 + b^5)*d*sqrt(x))*sinh(d*sqrt(x)
+ c))/((a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c)^2 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d*sinh(d*sqrt(x) + c)
^2 + 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*d*cosh(d*sqrt(x) + 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*sqrt(x) + c) + (a^6*b + 2*a^4*b^3 + a^2*b^5)*d)*sinh(d*sqrt(x) + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x} \left (a + b \operatorname{csch}{\left (c + d \sqrt{x} \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(c+d*x**(1/2)))**2/x**(1/2),x)

[Out]

Integral(1/(sqrt(x)*(a + b*csch(c + d*sqrt(x)))**2), x)

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Giac [A]  time = 1.311, size = 240, normalized size = 2.03 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a e^{\left (d \sqrt{x} + c\right )} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d \sqrt{x} + c\right )} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} d + a^{2} b^{2} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{4 \,{\left (b^{3} e^{\left (d \sqrt{x} + c\right )} - a b^{2}\right )}}{{\left (a^{4} d + a^{2} b^{2} d\right )}{\left (a e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + 2 \, b e^{\left (d \sqrt{x} + c\right )} - a\right )}} + \frac{2 \,{\left (d \sqrt{x} + c\right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(c+d*x^(1/2)))^2/x^(1/2),x, algorithm="giac")

[Out]

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