\(\int x (a+b \text {csch}(c+d \sqrt {x})) \, dx\) [40]

Optimal result

Integrand size = 16, antiderivative size = 164 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {6 b x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4} \] Output:

1/2*a*x^2-4*b*x^(3/2)*arctanh(exp(c+d*x^(1/2)))/d-6*b*x*polylog(2,-exp(c+d 
*x^(1/2)))/d^2+6*b*x*polylog(2,exp(c+d*x^(1/2)))/d^2+12*b*x^(1/2)*polylog( 
3,-exp(c+d*x^(1/2)))/d^3-12*b*x^(1/2)*polylog(3,exp(c+d*x^(1/2)))/d^3-12*b 
*polylog(4,-exp(c+d*x^(1/2)))/d^4+12*b*polylog(4,exp(c+d*x^(1/2)))/d^4
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.10 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {2 b \left (d^3 x^{3/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^3 x^{3/2} \log \left (1+e^{c+d \sqrt {x}}\right )-3 d^2 x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+3 d^2 x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+6 d \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-6 d \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-6 \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+6 \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )\right )}{d^4} \] Input:

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

Output:

(a*x^2)/2 + (2*b*(d^3*x^(3/2)*Log[1 - E^(c + d*Sqrt[x])] - d^3*x^(3/2)*Log 
[1 + E^(c + d*Sqrt[x])] - 3*d^2*x*PolyLog[2, -E^(c + d*Sqrt[x])] + 3*d^2*x 
*PolyLog[2, E^(c + d*Sqrt[x])] + 6*d*Sqrt[x]*PolyLog[3, -E^(c + d*Sqrt[x]) 
] - 6*d*Sqrt[x]*PolyLog[3, E^(c + d*Sqrt[x])] - 6*PolyLog[4, -E^(c + d*Sqr 
t[x])] + 6*PolyLog[4, E^(c + d*Sqrt[x])]))/d^4
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 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.

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {12 b \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {6 b x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}\)

Input:

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

Output:

(a*x^2)/2 - (4*b*x^(3/2)*ArcTanh[E^(c + d*Sqrt[x])])/d - (6*b*x*PolyLog[2, 
 -E^(c + d*Sqrt[x])])/d^2 + (6*b*x*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (1 
2*b*Sqrt[x]*PolyLog[3, -E^(c + d*Sqrt[x])])/d^3 - (12*b*Sqrt[x]*PolyLog[3, 
 E^(c + d*Sqrt[x])])/d^3 - (12*b*PolyLog[4, -E^(c + d*Sqrt[x])])/d^4 + (12 
*b*PolyLog[4, E^(c + d*Sqrt[x])])/d^4
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {1}{2} \, a x^{2} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} \] Input:

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

Output:

1/2*a*x^2 - 2*(log(e^(d*sqrt(x) + c) + 1)*log(e^(d*sqrt(x)))^3 + 3*dilog(- 
e^(d*sqrt(x) + c))*log(e^(d*sqrt(x)))^2 - 6*log(e^(d*sqrt(x)))*polylog(3, 
-e^(d*sqrt(x) + c)) + 6*polylog(4, -e^(d*sqrt(x) + c)))*b/d^4 + 2*(log(-e^ 
(d*sqrt(x) + c) + 1)*log(e^(d*sqrt(x)))^3 + 3*dilog(e^(d*sqrt(x) + c))*log 
(e^(d*sqrt(x)))^2 - 6*log(e^(d*sqrt(x)))*polylog(3, e^(d*sqrt(x) + c)) + 6 
*polylog(4, e^(d*sqrt(x) + c)))*b/d^4
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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