\(\int \frac {x^3}{a+b \text {csch}(c+d x^2)} \, dx\) [21]

Optimal result

Integrand size = 18, antiderivative size = 225 \[ \int \frac {x^3}{a+b \text {csch}\left (c+d x^2\right )} \, dx=\frac {x^4}{4 a}-\frac {b x^2 \log \left (1+\frac {a e^{c+d x^2}}{b-\sqrt {a^2+b^2}}\right )}{2 a \sqrt {a^2+b^2} d}+\frac {b x^2 \log \left (1+\frac {a e^{c+d x^2}}{b+\sqrt {a^2+b^2}}\right )}{2 a \sqrt {a^2+b^2} d}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b-\sqrt {a^2+b^2}}\right )}{2 a \sqrt {a^2+b^2} d^2}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b+\sqrt {a^2+b^2}}\right )}{2 a \sqrt {a^2+b^2} d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{a+b \text {csch}\left (c+d x^2\right )} \, dx=\frac {d x^2 \left (\sqrt {a^2+b^2} d x^2-2 b \log \left (1+\frac {a e^{c+d x^2}}{b-\sqrt {a^2+b^2}}\right )+2 b \log \left (1+\frac {a e^{c+d x^2}}{b+\sqrt {a^2+b^2}}\right )\right )-2 b \operatorname {PolyLog}\left (2,\frac {a e^{c+d x^2}}{-b+\sqrt {a^2+b^2}}\right )+2 b \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b+\sqrt {a^2+b^2}}\right )}{4 a \sqrt {a^2+b^2} d^2} \] Input:

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

Output:

(d*x^2*(Sqrt[a^2 + b^2]*d*x^2 - 2*b*Log[1 + (a*E^(c + d*x^2))/(b - Sqrt[a^ 
2 + b^2])] + 2*b*Log[1 + (a*E^(c + d*x^2))/(b + Sqrt[a^2 + b^2])]) - 2*b*P 
olyLog[2, (a*E^(c + d*x^2))/(-b + Sqrt[a^2 + b^2])] + 2*b*PolyLog[2, -((a* 
E^(c + d*x^2))/(b + Sqrt[a^2 + b^2]))])/(4*a*Sqrt[a^2 + b^2]*d^2)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.

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

\(\Big \downarrow \) 5960

\(\displaystyle \frac {1}{2} \int \frac {x^2}{a+b \text {csch}\left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {x^2}{a+i b \csc \left (i d x^2+i c\right )}dx^2\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {1}{2} \int \left (\frac {x^2}{a}-\frac {b x^2}{a \left (b+a \sinh \left (d x^2+c\right )\right )}\right )dx^2\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (193) = 386\).

Time = 0.10 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.24 \[ \int \frac {x^3}{a+b \text {csch}\left (c+d x^2\right )} \, dx=\frac {{\left (a^{2} + b^{2}\right )} d^{2} x^{4} - 2 \, a b c \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} \log \left (2 \, a \cosh \left (d x^{2} + c\right ) + 2 \, a \sinh \left (d x^{2} + c\right ) + 2 \, a \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} + 2 \, b\right ) + 2 \, a b c \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} \log \left (2 \, a \cosh \left (d x^{2} + c\right ) + 2 \, a \sinh \left (d x^{2} + c\right ) - 2 \, a \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} + 2 \, b\right ) - 2 \, a b \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} {\rm Li}_2\left (\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) + {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a} + 1\right ) + 2 \, a b \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} {\rm Li}_2\left (\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) - {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a} + 1\right ) - 2 \, {\left (a b d x^{2} + a b c\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} \log \left (-\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) + {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a}\right ) + 2 \, {\left (a b d x^{2} + a b c\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} \log \left (-\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) - {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{a^{2}}} - a}{a}\right )}{4 \, {\left (a^{3} + a b^{2}\right )} d^{2}} \] Input:

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

Output:

1/4*((a^2 + b^2)*d^2*x^4 - 2*a*b*c*sqrt((a^2 + b^2)/a^2)*log(2*a*cosh(d*x^ 
2 + c) + 2*a*sinh(d*x^2 + c) + 2*a*sqrt((a^2 + b^2)/a^2) + 2*b) + 2*a*b*c* 
sqrt((a^2 + b^2)/a^2)*log(2*a*cosh(d*x^2 + c) + 2*a*sinh(d*x^2 + c) - 2*a* 
sqrt((a^2 + b^2)/a^2) + 2*b) - 2*a*b*sqrt((a^2 + b^2)/a^2)*dilog((b*cosh(d 
*x^2 + c) + b*sinh(d*x^2 + c) + (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sq 
rt((a^2 + b^2)/a^2) - a)/a + 1) + 2*a*b*sqrt((a^2 + b^2)/a^2)*dilog((b*cos 
h(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c)) 
*sqrt((a^2 + b^2)/a^2) - a)/a + 1) - 2*(a*b*d*x^2 + a*b*c)*sqrt((a^2 + b^2 
)/a^2)*log(-(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) + (a*cosh(d*x^2 + c) + 
a*sinh(d*x^2 + c))*sqrt((a^2 + b^2)/a^2) - a)/a) + 2*(a*b*d*x^2 + a*b*c)*s 
qrt((a^2 + b^2)/a^2)*log(-(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh 
(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt((a^2 + b^2)/a^2) - a)/a))/((a^3 + a* 
b^2)*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/4*x^4/a - 2*b*integrate(x^3*e^(d*x^2 + c)/(a^2*e^(2*d*x^2 + 2*c) + 2*a*b 
*e^(d*x^2 + c) - a^2), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{a+b \text {csch}\left (c+d x^2\right )} \, dx=e^{2 c} \left (\int \frac {e^{2 d \,x^{2}} x^{3}}{e^{2 d \,x^{2}+2 c} a +2 e^{d \,x^{2}+c} b -a}d x \right )-\left (\int \frac {x^{3}}{e^{2 d \,x^{2}+2 c} a +2 e^{d \,x^{2}+c} b -a}d x \right ) \] Input:

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

Output:

e**(2*c)*int((e**(2*d*x**2)*x**3)/(e**(2*c + 2*d*x**2)*a + 2*e**(c + d*x** 
2)*b - a),x) - int(x**3/(e**(2*c + 2*d*x**2)*a + 2*e**(c + d*x**2)*b - a), 
x)