3.1.76 \(\int \frac {\operatorname {PolyLog}(2,a x^2)}{(d x)^{5/2}} \, dx\) [76]

3.1.76.1 Optimal result
3.1.76.2 Mathematica [C] (verified)
3.1.76.3 Rubi [A] (verified)
3.1.76.4 Maple [A] (verified)
3.1.76.5 Fricas [C] (verification not implemented)
3.1.76.6 Sympy [F(-1)]
3.1.76.7 Maxima [A] (verification not implemented)
3.1.76.8 Giac [F]
3.1.76.9 Mupad [F(-1)]

3.1.76.1 Optimal result

Integrand size = 15, antiderivative size = 111 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\frac {16 a^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 d^{5/2}}+\frac {16 a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 d^{5/2}}+\frac {8 \log \left (1-a x^2\right )}{9 d (d x)^{3/2}}-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}} \]

output
16/9*a^(3/4)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(5/2)+16/9*a^(3/4)*arct 
anh(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(5/2)+8/9*ln(-a*x^2+1)/d/(d*x)^(3/2)-2/ 
3*polylog(2,a*x^2)/d/(d*x)^(3/2)
 
3.1.76.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.56 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\frac {x \operatorname {Gamma}\left (\frac {1}{4}\right ) \left (16 a x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},a x^2\right )+4 \log \left (1-a x^2\right )-3 \operatorname {PolyLog}\left (2,a x^2\right )\right )}{18 (d x)^{5/2} \operatorname {Gamma}\left (\frac {5}{4}\right )} \]

input
Integrate[PolyLog[2, a*x^2]/(d*x)^(5/2),x]
 
output
(x*Gamma[1/4]*(16*a*x^2*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 4*Log[1 - 
a*x^2] - 3*PolyLog[2, a*x^2]))/(18*(d*x)^(5/2)*Gamma[5/4])
 
3.1.76.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {7145, 25, 2905, 8, 266, 756, 218, 221}

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.

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

\(\Big \downarrow \) 7145

\(\displaystyle \frac {4}{3} \int -\frac {\log \left (1-a x^2\right )}{(d x)^{5/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {4}{3} \int \frac {\log \left (1-a x^2\right )}{(d x)^{5/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 2905

\(\displaystyle -\frac {4}{3} \left (-\frac {4 a \int \frac {x}{(d x)^{3/2} \left (1-a x^2\right )}dx}{3 d}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 8

\(\displaystyle -\frac {4}{3} \left (-\frac {4 a \int \frac {1}{\sqrt {d x} \left (1-a x^2\right )}dx}{3 d^2}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {4}{3} \left (-\frac {8 a \int \frac {1}{1-a x^2}d\sqrt {d x}}{3 d^3}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {4}{3} \left (-\frac {8 a \left (\frac {1}{2} d \int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}+\frac {1}{2} d \int \frac {1}{\sqrt {a} x d+d}d\sqrt {d x}\right )}{3 d^3}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {4}{3} \left (-\frac {8 a \left (\frac {1}{2} d \int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}+\frac {\sqrt {d} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}\right )}{3 d^3}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {4}{3} \left (-\frac {8 a \left (\frac {\sqrt {d} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}\right )}{3 d^3}-\frac {2 \log \left (1-a x^2\right )}{3 d (d x)^{3/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}\)

input
Int[PolyLog[2, a*x^2]/(d*x)^(5/2),x]
 
output
(-4*((-8*a*((Sqrt[d]*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(2*a^(1/4)) + (S 
qrt[d]*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(2*a^(1/4))))/(3*d^3) - (2*Lo 
g[1 - a*x^2])/(3*d*(d*x)^(3/2))))/3 - (2*PolyLog[2, a*x^2])/(3*d*(d*x)^(3/ 
2))
 

3.1.76.3.1 Defintions of rubi rules used

rule 8
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m   Int[u*(a* 
x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 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]
 
3.1.76.4 Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00

method result size
meijerg \(-\frac {x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{4}} \left (-\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {1}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{9 \left (a \,x^{2}\right )^{\frac {1}{4}}}+\frac {16 \left (-a \right )^{\frac {1}{4}} \ln \left (-a \,x^{2}+1\right )}{9 x^{\frac {3}{2}} a}-\frac {4 \left (-a \right )^{\frac {1}{4}} \operatorname {polylog}\left (2, a \,x^{2}\right )}{3 x^{\frac {3}{2}} a}\right )}{2 \left (d x \right )^{\frac {5}{2}}}\) \(111\)
derivativedivides \(\frac {-\frac {2 \operatorname {polylog}\left (2, a \,x^{2}\right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{9 \left (d x \right )^{\frac {3}{2}}}+\frac {8 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{9 d^{2}}}{d}\) \(116\)
default \(\frac {-\frac {2 \operatorname {polylog}\left (2, a \,x^{2}\right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{9 \left (d x \right )^{\frac {3}{2}}}+\frac {8 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{9 d^{2}}}{d}\) \(116\)

input
int(polylog(2,a*x^2)/(d*x)^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/2/(d*x)^(5/2)*x^(5/2)*(-a)^(3/4)*(-16/9*x^(1/2)*(-a)^(1/4)/(a*x^2)^(1/4 
)*(ln(1-(a*x^2)^(1/4))-ln(1+(a*x^2)^(1/4))-2*arctan((a*x^2)^(1/4)))+16/9/x 
^(3/2)*(-a)^(1/4)*ln(-a*x^2+1)/a-4/3/x^(3/2)*(-a)^(1/4)/a*polylog(2,a*x^2) 
)
 
3.1.76.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.80 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (4 \, d^{3} x^{2} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} \log \left (8 \, d^{3} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} + 8 \, \sqrt {d x} a\right ) + 4 i \, d^{3} x^{2} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} \log \left (8 i \, d^{3} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} + 8 \, \sqrt {d x} a\right ) - 4 i \, d^{3} x^{2} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} \log \left (-8 i \, d^{3} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} + 8 \, \sqrt {d x} a\right ) - 4 \, d^{3} x^{2} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} \log \left (-8 \, d^{3} \left (\frac {a^{3}}{d^{10}}\right )^{\frac {1}{4}} + 8 \, \sqrt {d x} a\right ) - \sqrt {d x} {\left (3 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )}\right )}}{9 \, d^{3} x^{2}} \]

input
integrate(polylog(2,a*x^2)/(d*x)^(5/2),x, algorithm="fricas")
 
output
2/9*(4*d^3*x^2*(a^3/d^10)^(1/4)*log(8*d^3*(a^3/d^10)^(1/4) + 8*sqrt(d*x)*a 
) + 4*I*d^3*x^2*(a^3/d^10)^(1/4)*log(8*I*d^3*(a^3/d^10)^(1/4) + 8*sqrt(d*x 
)*a) - 4*I*d^3*x^2*(a^3/d^10)^(1/4)*log(-8*I*d^3*(a^3/d^10)^(1/4) + 8*sqrt 
(d*x)*a) - 4*d^3*x^2*(a^3/d^10)^(1/4)*log(-8*d^3*(a^3/d^10)^(1/4) + 8*sqrt 
(d*x)*a) - sqrt(d*x)*(3*dilog(a*x^2) - 4*log(-a*x^2 + 1)))/(d^3*x^2)
 
3.1.76.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\text {Timed out} \]

input
integrate(polylog(2,a*x**2)/(d*x)**(5/2),x)
 
output
Timed out
 
3.1.76.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {8 \, a \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} d} - \frac {4 \, a \log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} d} - \frac {3 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a d^{2} x^{2} + d^{2}\right ) + 8 \, \log \left (d\right )}{\left (d x\right )^{\frac {3}{2}}}\right )}}{9 \, d} \]

input
integrate(polylog(2,a*x^2)/(d*x)^(5/2),x, algorithm="maxima")
 
output
2/9*(8*a*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)*d)*d) - 4 
*a*log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqr 
t(a)*d)))/(sqrt(sqrt(a)*d)*d) - (3*dilog(a*x^2) - 4*log(-a*d^2*x^2 + d^2) 
+ 8*log(d))/(d*x)^(3/2))/d
 
3.1.76.8 Giac [F]

\[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\int { \frac {{\rm Li}_2\left (a x^{2}\right )}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]

input
integrate(polylog(2,a*x^2)/(d*x)^(5/2),x, algorithm="giac")
 
output
integrate(dilog(a*x^2)/(d*x)^(5/2), x)
 
3.1.76.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{5/2}} \, dx=\int \frac {\mathrm {polylog}\left (2,a\,x^2\right )}{{\left (d\,x\right )}^{5/2}} \,d x \]

input
int(polylog(2, a*x^2)/(d*x)^(5/2),x)
 
output
int(polylog(2, a*x^2)/(d*x)^(5/2), x)