3.1.62 \(\int \frac {\sqrt {x}}{a+b \text {csch}(c+d \sqrt {x})} \, dx\) [62]

3.1.62.1 Optimal result

Integrand size = 22, antiderivative size = 337 \[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {4 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {4 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3} \]

output

2/3*x^(3/2)/a-2*b*x*ln(1+a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d/(a^2+ 
b^2)^(1/2)+2*b*x*ln(1+a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d/(a^2+b^2 
)^(1/2)+4*b*polylog(3,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d^3/(a^2+ 
b^2)^(1/2)-4*b*polylog(3,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d^3/(a 
^2+b^2)^(1/2)-4*b*polylog(2,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))*x^(1/ 
2)/a/d^2/(a^2+b^2)^(1/2)+4*b*polylog(2,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1 
/2)))*x^(1/2)/a/d^2/(a^2+b^2)^(1/2)
 

3.1.62.2 Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {a^2+b^2} d^3 x^{3/2}-3 b d^2 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )+3 b d^2 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )-6 b d \sqrt {x} \operatorname {PolyLog}\left (2,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )+6 b d \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )+6 b \operatorname {PolyLog}\left (3,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )-6 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )\right )}{3 a \sqrt {a^2+b^2} d^3} \]

input

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

output

(2*(Sqrt[a^2 + b^2]*d^3*x^(3/2) - 3*b*d^2*x*Log[1 + (a*E^(c + d*Sqrt[x]))/ 
(b - Sqrt[a^2 + b^2])] + 3*b*d^2*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt 
[a^2 + b^2])] - 6*b*d*Sqrt[x]*PolyLog[2, (a*E^(c + d*Sqrt[x]))/(-b + Sqrt[ 
a^2 + b^2])] + 6*b*d*Sqrt[x]*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[ 
a^2 + b^2]))] + 6*b*PolyLog[3, (a*E^(c + d*Sqrt[x]))/(-b + Sqrt[a^2 + b^2] 
)] - 6*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2]))]))/(3*a 
*Sqrt[a^2 + b^2]*d^3)
 

3.1.62.3 Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5960, 3042, 4679, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx\)

\(\Big \downarrow \) 5960

\(\displaystyle 2 \int \frac {x}{a+b \text {csch}\left (c+d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x}{a+i b \csc \left (i c+i d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (\frac {x}{a}-\frac {b x}{a \left (b+a \sinh \left (c+d \sqrt {x}\right )\right )}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {2 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {2 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}-\frac {b x \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {b x \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {x^{3/2}}{3 a}\right )\)

input

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

output

2*(x^(3/2)/(3*a) - (b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2] 
)])/(a*Sqrt[a^2 + b^2]*d) + (b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[a 
^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (2*b*Sqrt[x]*PolyLog[2, -((a*E^(c + d 
*Sqrt[x]))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (2*b*Sqrt[x] 
*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + 
 b^2]*d^2) + (2*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2]) 
)])/(a*Sqrt[a^2 + b^2]*d^3) - (2*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + 
 Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^3))
 

3.1.62.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 5960

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(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[Simplify[(m 
 + 1)/n], 0] && IntegerQ[p]
 

3.1.62.4 Maple [F]

\[\int \frac {\sqrt {x}}{a +b \,\operatorname {csch}\left (c +d \sqrt {x}\right )}d x\]

input

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

output

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

3.1.62.5 Fricas [F]

\[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {\sqrt {x}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

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

output

integral(sqrt(x)/(b*csch(d*sqrt(x) + c) + a), x)
 

3.1.62.6 Sympy [F]

\[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {\sqrt {x}}{a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}}\, dx \]

input

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

output

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

3.1.62.7 Maxima [F]

\[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {\sqrt {x}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

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

output

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

3.1.62.8 Giac [F]

\[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {\sqrt {x}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

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

output

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

3.1.62.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {\sqrt {x}}{a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}} \,d x \]

input

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

output

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