\(\int (d x)^{3/2} \operatorname {PolyLog}(2,a x^q) \, dx\) [86]

Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\frac {8 a d q^2 x^{2+q} \sqrt {d x} \operatorname {Hypergeometric2F1}\left (1,\frac {\frac {5}{2}+q}{q},\frac {1}{2} \left (4+\frac {5}{q}\right ),a x^q\right )}{25 (5+2 q)}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d} \] Output:

8*a*d*q^2*x^(2+q)*(d*x)^(1/2)*hypergeom([1, (5/2+q)/q],[2+5/2/q],a*x^q)/(1 
25+50*q)+4/25*q*(d*x)^(5/2)*ln(1-a*x^q)/d+2/5*(d*x)^(5/2)*polylog(2,a*x^q) 
/d
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\frac {2 x (d x)^{3/2} \left (4 a q^2 x^q \operatorname {Hypergeometric2F1}\left (1,\frac {\frac {5}{2}+q}{q},2+\frac {5}{2 q},a x^q\right )+(5+2 q) \left (2 q \log \left (1-a x^q\right )+5 \operatorname {PolyLog}\left (2,a x^q\right )\right )\right )}{25 (5+2 q)} \] Input:

Integrate[(d*x)^(3/2)*PolyLog[2, a*x^q],x]
 

Output:

(2*x*(d*x)^(3/2)*(4*a*q^2*x^q*Hypergeometric2F1[1, (5/2 + q)/q, 2 + 5/(2*q 
), a*x^q] + (5 + 2*q)*(2*q*Log[1 - a*x^q] + 5*PolyLog[2, a*x^q])))/(25*(5 
+ 2*q))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7145, 25, 2905, 30, 888}

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.

\(\displaystyle \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx\)

\(\Big \downarrow \) 7145

\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}-\frac {2}{5} q \int -(d x)^{3/2} \log \left (1-a x^q\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{5} q \int (d x)^{3/2} \log \left (1-a x^q\right )dx+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\)

\(\Big \downarrow \) 2905

\(\displaystyle \frac {2}{5} q \left (\frac {2 a q \int \frac {x^{q-1} (d x)^{5/2}}{1-a x^q}dx}{5 d}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\)

\(\Big \downarrow \) 30

\(\displaystyle \frac {2}{5} q \left (\frac {2 a d q \sqrt {d x} \int \frac {x^{q+\frac {3}{2}}}{1-a x^q}dx}{5 \sqrt {x}}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {2}{5} q \left (\frac {4 a d q \sqrt {d x} x^{q+2} \operatorname {Hypergeometric2F1}\left (1,\frac {q+\frac {5}{2}}{q},\frac {1}{2} \left (4+\frac {5}{q}\right ),a x^q\right )}{5 (2 q+5)}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\)

Input:

Int[(d*x)^(3/2)*PolyLog[2, a*x^q],x]
 

Output:

(2*q*((4*a*d*q*x^(2 + q)*Sqrt[d*x]*Hypergeometric2F1[1, (5/2 + q)/q, (4 + 
5/q)/2, a*x^q])/(5*(5 + 2*q)) + (2*(d*x)^(5/2)*Log[1 - a*x^q])/(5*d)))/5 + 
 (2*(d*x)^(5/2)*PolyLog[2, a*x^q])/(5*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 2905

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ 
(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m 
+ 1))), x] - Simp[b*e*n*(p/(f*(m + 1)))   Int[x^(n - 1)*((f*x)^(m + 1)/(d + 
 e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
 

rule 7145

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbo 
l] :> Simp[(d*x)^(m + 1)*(PolyLog[n, a*(b*x^p)^q]/(d*(m + 1))), x] - Simp[p 
*(q/(m + 1))   Int[(d*x)^m*PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, 
 b, d, m, p, q}, x] && NeQ[m, -1] && GtQ[n, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5.

Time = 0.67 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.20

method	result	size
meijerg	\(-\frac {\left (d x \right )^{\frac {3}{2}} \left (-a \right )^{-\frac {5}{2 q}} \left (-\frac {4 q^{2} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \ln \left (1-a \,x^{q}\right )}{25}-\frac {2 q \,x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \left (\frac {2 q}{5}+1\right ) \operatorname {polylog}\left (2, a \,x^{q}\right )}{2 q +5}-\frac {4 q^{2} x^{\frac {5}{2}+q} a \left (-a \right )^{\frac {5}{2 q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {2 q +5}{2 q}\right )}{25}\right )}{x^{\frac {3}{2}} q}\)	\(121\)

Input:

int((d*x)^(3/2)*polylog(2,a*x^q),x,method=_RETURNVERBOSE)
 

Output:

-(d*x)^(3/2)/x^(3/2)*(-a)^(-5/2/q)/q*(-4/25*q^2*x^(5/2)*(-a)^(5/2/q)*ln(1- 
a*x^q)-2*q/(2*q+5)*x^(5/2)*(-a)^(5/2/q)*(2/5*q+1)*polylog(2,a*x^q)-4/25*q^ 
2*x^(5/2+q)*a*(-a)^(5/2/q)*LerchPhi(a*x^q,1,1/2*(2*q+5)/q))
 

Fricas [F]

\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_2\left (a x^{q}\right ) \,d x } \] Input:

integrate((d*x)^(3/2)*polylog(2,a*x^q),x, algorithm="fricas")
 

Output:

integral(sqrt(d*x)*d*x*dilog(a*x^q), x)
 

Sympy [F(-1)]

Timed out. \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\text {Timed out} \] Input:

integrate((d*x)**(3/2)*polylog(2,a*x**q),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_2\left (a x^{q}\right ) \,d x } \] Input:

integrate((d*x)^(3/2)*polylog(2,a*x^q),x, algorithm="maxima")
 

Output:

8*d^(3/2)*q^3*integrate(1/25*x^(3/2)/((2*a^2*q - 5*a^2)*x^(2*q) - 2*(2*a*q 
 - 5*a)*x^q + 2*q - 5), x) + 2/125*(25*((2*a*d^(3/2)*q - 5*a*d^(3/2))*x*x^ 
q - (2*d^(3/2)*q - 5*d^(3/2))*x)*x^(3/2)*dilog(a*x^q) + 10*((2*a*d^(3/2)*q 
^2 - 5*a*d^(3/2)*q)*x*x^q - (2*d^(3/2)*q^2 - 5*d^(3/2)*q)*x)*x^(3/2)*log(- 
a*x^q + 1) + 4*(2*d^(3/2)*q^3*x - (2*a*d^(3/2)*q^3 - 5*a*d^(3/2)*q^2)*x*x^ 
q)*x^(3/2))/((2*a*q - 5*a)*x^q - 2*q + 5)
 

Giac [F]

\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_2\left (a x^{q}\right ) \,d x } \] Input:

integrate((d*x)^(3/2)*polylog(2,a*x^q),x, algorithm="giac")
 

Output:

integrate((d*x)^(3/2)*dilog(a*x^q), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int {\left (d\,x\right )}^{3/2}\,\mathrm {polylog}\left (2,a\,x^q\right ) \,d x \] Input:

int((d*x)^(3/2)*polylog(2, a*x^q),x)
 

Output:

int((d*x)^(3/2)*polylog(2, a*x^q), x)
 

Reduce [F]

\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\sqrt {d}\, \left (\int \sqrt {x}\, \mathit {polylog}\left (2, x^{q} a \right ) x d x \right ) d \] Input:

int((d*x)^(3/2)*polylog(2,a*x^q),x)
 

Output:

sqrt(d)*int(sqrt(x)*polylog(2,x**q*a)*x,x)*d