\(\int \frac {\operatorname {PolyLog}(2,a x^q)}{(d x)^{3/2}} \, dx\) [89]

Optimal result

Integrand size = 15, antiderivative size = 97 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^q\right )}{(d x)^{3/2}} \, dx=-\frac {8 a q^2 x^q \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {1}{q}\right ),\frac {1}{2} \left (4-\frac {1}{q}\right ),a x^q\right )}{d (1-2 q) \sqrt {d x}}+\frac {4 q \log \left (1-a x^q\right )}{d \sqrt {d x}}-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.49 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^q\right )}{(d x)^{3/2}} \, dx=-\frac {x G_{4,4}^{1,4}\left (-a x^q|\begin {array}{c} 1,1,1,1+\frac {1}{2 q} \\ 1,0,0,\frac {1}{2 q} \\\end {array}\right )}{q (d x)^{3/2}} \] Input:

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

Output:

-((x*MeijerG[{{1, 1, 1, 1 + 1/(2*q)}, {}}, {{1}, {0, 0, 1/(2*q)}}, -(a*x^q 
)])/(q*(d*x)^(3/2)))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 7145

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2905

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

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

-2*q*((4*a*q*x^q*Hypergeometric2F1[1, -((1/2 - q)/q), (4 - q^(-1))/2, a*x^ 
q])/(d*(1 - 2*q)*Sqrt[d*x]) - (2*Log[1 - a*x^q])/(d*Sqrt[d*x])) - (2*PolyL 
og[2, a*x^q])/(d*Sqrt[d*x])
 

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 = 0.69 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(d*x)*dilog(a*x^q)/(d^2*x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

8*q^3*integrate(1/((2*d^(3/2)*q + (2*a^2*d^(3/2)*q + a^2*d^(3/2))*x^(2*q) 
- 2*(2*a*d^(3/2)*q + a*d^(3/2))*x^q + d^(3/2))*x^(3/2)), x) + 2*(((2*a*sqr 
t(d)*q + a*sqrt(d))*x*x^q - (2*sqrt(d)*q + sqrt(d))*x)*dilog(a*x^q)/x^(3/2 
) - 2*((2*a*sqrt(d)*q^2 + a*sqrt(d)*q)*x*x^q - (2*sqrt(d)*q^2 + sqrt(d)*q) 
*x)*log(-a*x^q + 1)/x^(3/2) + 4*(2*sqrt(d)*q^3*x - (2*a*sqrt(d)*q^3 + a*sq 
rt(d)*q^2)*x*x^q)/x^(3/2))/(2*d^2*q + d^2 - (2*a*d^2*q + a*d^2)*x^q)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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