3.1.61 \(\int \frac {x^{3/2}}{a+b \text {csch}(c+d \sqrt {x})} \, dx\) [61]

3.1.61.1 Optimal result

Integrand size = 22, antiderivative size = 561 \[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{5/2}}{5 a}-\frac {2 b x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {24 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {24 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^4}+\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^4}+\frac {48 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^5}-\frac {48 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^5} \]

output

2/5*x^(5/2)/a-2*b*x^2*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^2*ln(1+a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d/(a^2 
+b^2)^(1/2)-8*b*x^(3/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2))) 
/a/d^2/(a^2+b^2)^(1/2)+8*b*x^(3/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b+(a^2+b 
^2)^(1/2)))/a/d^2/(a^2+b^2)^(1/2)+24*b*x*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)-24*b*x*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)+48*b*polylog(5,-a*exp(c+d*x^ 
(1/2))/(b-(a^2+b^2)^(1/2)))/a/d^5/(a^2+b^2)^(1/2)-48*b*polylog(5,-a*exp(c+ 
d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d^5/(a^2+b^2)^(1/2)-48*b*polylog(4,-a*ex 
p(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))*x^(1/2)/a/d^4/(a^2+b^2)^(1/2)+48*b*pol 
ylog(4,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))*x^(1/2)/a/d^4/(a^2+b^2)^(1 
/2)
 

3.1.61.2 Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 436, normalized size of antiderivative = 0.78 \[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {a^2+b^2} d^5 x^{5/2}-5 b d^4 x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )+5 b d^4 x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )-20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )+20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )+60 b d^2 x \operatorname {PolyLog}\left (3,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )-60 b d^2 x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )-120 b d \sqrt {x} \operatorname {PolyLog}\left (4,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )+120 b d \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )+120 b \operatorname {PolyLog}\left (5,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {a^2+b^2}}\right )-120 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )\right )}{5 a \sqrt {a^2+b^2} d^5} \]

input

Integrate[x^(3/2)/(a + b*Csch[c + d*Sqrt[x]]),x]
 

output

(2*(Sqrt[a^2 + b^2]*d^5*x^(5/2) - 5*b*d^4*x^2*Log[1 + (a*E^(c + d*Sqrt[x]) 
)/(b - Sqrt[a^2 + b^2])] + 5*b*d^4*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + 
Sqrt[a^2 + b^2])] - 20*b*d^3*x^(3/2)*PolyLog[2, (a*E^(c + d*Sqrt[x]))/(-b 
+ Sqrt[a^2 + b^2])] + 20*b*d^3*x^(3/2)*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/ 
(b + Sqrt[a^2 + b^2]))] + 60*b*d^2*x*PolyLog[3, (a*E^(c + d*Sqrt[x]))/(-b 
+ Sqrt[a^2 + b^2])] - 60*b*d^2*x*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + S 
qrt[a^2 + b^2]))] - 120*b*d*Sqrt[x]*PolyLog[4, (a*E^(c + d*Sqrt[x]))/(-b + 
 Sqrt[a^2 + b^2])] + 120*b*d*Sqrt[x]*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b 
 + Sqrt[a^2 + b^2]))] + 120*b*PolyLog[5, (a*E^(c + d*Sqrt[x]))/(-b + Sqrt[ 
a^2 + b^2])] - 120*b*PolyLog[5, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^ 
2]))]))/(5*a*Sqrt[a^2 + b^2]*d^5)
 

3.1.61.3 Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 562, 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.

\(\displaystyle \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx\)

\(\Big \downarrow \) 5960

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {24 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^5 \sqrt {a^2+b^2}}-\frac {24 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^5 \sqrt {a^2+b^2}}-\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^4 \sqrt {a^2+b^2}}+\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^4 \sqrt {a^2+b^2}}+\frac {12 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {12 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}-\frac {b x^2 \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {b x^2 \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {x^{5/2}}{5 a}\right )\)

input

Int[x^(3/2)/(a + b*Csch[c + d*Sqrt[x]]),x]
 

output

2*(x^(5/2)/(5*a) - (b*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^ 
2])])/(a*Sqrt[a^2 + b^2]*d) + (b*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sq 
rt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (4*b*x^(3/2)*PolyLog[2, -((a*E^(c 
 + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (4*b*x^( 
3/2)*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a 
^2 + b^2]*d^2) + (12*b*x*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 
+ b^2]))])/(a*Sqrt[a^2 + b^2]*d^3) - (12*b*x*PolyLog[3, -((a*E^(c + d*Sqrt 
[x]))/(b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^3) - (24*b*Sqrt[x]*Pol 
yLog[4, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2 
]*d^4) + (24*b*Sqrt[x]*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + 
b^2]))])/(a*Sqrt[a^2 + b^2]*d^4) + (24*b*PolyLog[5, -((a*E^(c + d*Sqrt[x]) 
)/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^5) - (24*b*PolyLog[5, -((a 
*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^5))
 

3.1.61.3.1 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]
 

3.1.61.4 Maple [F]

\[\int \frac {x^{\frac {3}{2}}}{a +b \,\operatorname {csch}\left (c +d \sqrt {x}\right )}d x\]

input

int(x^(3/2)/(a+b*csch(c+d*x^(1/2))),x)
 

output

int(x^(3/2)/(a+b*csch(c+d*x^(1/2))),x)
 

3.1.61.5 Fricas [F]

\[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{\frac {3}{2}}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

integrate(x^(3/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="fricas")
 

output

integral(x^(3/2)/(b*csch(d*sqrt(x) + c) + a), x)
 

3.1.61.6 Sympy [F]

\[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{\frac {3}{2}}}{a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}}\, dx \]

input

integrate(x**(3/2)/(a+b*csch(c+d*x**(1/2))),x)
 

output

Integral(x**(3/2)/(a + b*csch(c + d*sqrt(x))), x)
 

3.1.61.7 Maxima [F]

\[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{\frac {3}{2}}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

integrate(x^(3/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="maxima")
 

output

-2*b*integrate(x^(3/2)*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) + 2/5*x^(5/2)/a
 

3.1.61.8 Giac [F]

\[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{\frac {3}{2}}}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a} \,d x } \]

input

integrate(x^(3/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="giac")
 

output

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

3.1.61.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{3/2}}{a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}} \,d x \]

input

int(x^(3/2)/(a + b/sinh(c + d*x^(1/2))),x)
 

output

int(x^(3/2)/(a + b/sinh(c + d*x^(1/2))), x)