\(\int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx\) [64]

Optimal result

Integrand size = 13, antiderivative size = 106 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=-\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}+\frac {8 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \operatorname {PolyLog}(2,a x)}{5 d (d x)^{5/2}} \] Output:

-8/75*a/d^2/(d*x)^(3/2)-8/25*a^2/d^3/(d*x)^(1/2)+8/25*a^(5/2)*arctanh(a^(1 
/2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+4/25*ln(-a*x+1)/d/(d*x)^(5/2)-2/5*polylog 
(2,a*x)/d/(d*x)^(5/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=-\frac {2 x \left (4 a x+12 a^2 x^2-12 a^{5/2} x^{5/2} \text {arctanh}\left (\sqrt {a} \sqrt {x}\right )-6 \log (1-a x)+15 \operatorname {PolyLog}(2,a x)\right )}{75 (d x)^{7/2}} \] Input:

Integrate[PolyLog[2, a*x]/(d*x)^(7/2),x]
 

Output:

(-2*x*(4*a*x + 12*a^2*x^2 - 12*a^(5/2)*x^(5/2)*ArcTanh[Sqrt[a]*Sqrt[x]] - 
6*Log[1 - a*x] + 15*PolyLog[2, a*x]))/(75*(d*x)^(7/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {7145, 25, 2842, 61, 61, 73, 219}

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.

\(\displaystyle \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 7145

\(\displaystyle \frac {2}{5} \int -\frac {\log (1-a x)}{(d x)^{7/2}}dx-\frac {2 \operatorname {PolyLog}(2,a x)}{5 d (d x)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2}{5} \int \frac {\log (1-a x)}{(d x)^{7/2}}dx-\frac {2 \operatorname {PolyLog}(2,a x)}{5 d (d x)^{5/2}}\)

\(\Big \downarrow \) 2842

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

Int[PolyLog[2, a*x]/(d*x)^(7/2),x]
 

Output:

(-2*((-2*a*(-2/(3*d*(d*x)^(3/2)) + (a*(-2/(d*Sqrt[d*x]) + (2*Sqrt[a]*ArcTa 
nh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]])/d^(3/2)))/d))/(5*d) - (2*Log[1 - a*x])/(5 
*d*(d*x)^(5/2))))/5 - (2*PolyLog[2, a*x])/(5*d*(d*x)^(5/2))
 

Defintions of rubi rules used

rule 25

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

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 2842

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ 
))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( 
g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1)))   Int[(f + g*x)^(q + 1)/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && 
NeQ[q, -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 [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {2 \operatorname {polylog}\left (2, x a \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {4 \ln \left (\frac {-d x a +d}{d}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {8 a \left (-\frac {1}{3 d \left (d x \right )^{\frac {3}{2}}}-\frac {a}{d^{2} \sqrt {d x}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x}\, a}{\sqrt {a d}}\right )}{d^{2} \sqrt {a d}}\right )}{25}}{d}\)	\(88\)
default	\(\frac {-\frac {2 \operatorname {polylog}\left (2, x a \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {4 \ln \left (\frac {-d x a +d}{d}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {8 a \left (-\frac {1}{3 d \left (d x \right )^{\frac {3}{2}}}-\frac {a}{d^{2} \sqrt {d x}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x}\, a}{\sqrt {a d}}\right )}{d^{2} \sqrt {a d}}\right )}{25}}{d}\)	\(88\)
meijerg	\(\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {7}{2}} \left (-\frac {8}{75 x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2}}}-\frac {8 a}{25 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}}}-\frac {4 \sqrt {x}\, a^{2} \left (\ln \left (1-\sqrt {x a}\right )-\ln \left (1+\sqrt {x a}\right )\right )}{25 \left (-a \right )^{\frac {3}{2}} \sqrt {x a}}+\frac {4 \ln \left (-x a +1\right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \operatorname {polylog}\left (2, x a \right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}\right )}{\left (d x \right )^{\frac {7}{2}} a}\)	\(117\)

Input:

int(polylog(2,x*a)/(d*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=\left [\frac {2 \, {\left (6 \, a^{2} d x^{3} \sqrt {\frac {a}{d}} \log \left (\frac {a x + 2 \, \sqrt {d x} \sqrt {\frac {a}{d}} + 1}{a x - 1}\right ) - {\left (12 \, a^{2} x^{2} + 4 \, a x + 15 \, {\rm Li}_2\left (a x\right ) - 6 \, \log \left (-a x + 1\right )\right )} \sqrt {d x}\right )}}{75 \, d^{4} x^{3}}, -\frac {2 \, {\left (12 \, a^{2} d x^{3} \sqrt {-\frac {a}{d}} \arctan \left (\sqrt {d x} \sqrt {-\frac {a}{d}}\right ) + {\left (12 \, a^{2} x^{2} + 4 \, a x + 15 \, {\rm Li}_2\left (a x\right ) - 6 \, \log \left (-a x + 1\right )\right )} \sqrt {d x}\right )}}{75 \, d^{4} x^{3}}\right ] \] Input:

integrate(polylog(2,a*x)/(d*x)^(7/2),x, algorithm="fricas")
 

Output:

[2/75*(6*a^2*d*x^3*sqrt(a/d)*log((a*x + 2*sqrt(d*x)*sqrt(a/d) + 1)/(a*x - 
1)) - (12*a^2*x^2 + 4*a*x + 15*dilog(a*x) - 6*log(-a*x + 1))*sqrt(d*x))/(d 
^4*x^3), -2/75*(12*a^2*d*x^3*sqrt(-a/d)*arctan(sqrt(d*x)*sqrt(-a/d)) + (12 
*a^2*x^2 + 4*a*x + 15*dilog(a*x) - 6*log(-a*x + 1))*sqrt(d*x))/(d^4*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(polylog(2,a*x)/(d*x)**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {6 \, a^{3} \log \left (\frac {\sqrt {d x} a - \sqrt {a d}}{\sqrt {d x} a + \sqrt {a d}}\right )}{\sqrt {a d} d^{2}} + \frac {12 \, a^{2} d^{2} x^{2} + 4 \, a d^{2} x + 15 \, d^{2} {\rm Li}_2\left (a x\right ) - 6 \, d^{2} \log \left (-a d x + d\right ) + 6 \, d^{2} \log \left (d\right )}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{75 \, d} \] Input:

integrate(polylog(2,a*x)/(d*x)^(7/2),x, algorithm="maxima")
 

Output:

-2/75*(6*a^3*log((sqrt(d*x)*a - sqrt(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/(sqr 
t(a*d)*d^2) + (12*a^2*d^2*x^2 + 4*a*d^2*x + 15*d^2*dilog(a*x) - 6*d^2*log( 
-a*d*x + d) + 6*d^2*log(d))/((d*x)^(5/2)*d^2))/d
 

Giac [F]

\[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=\int { \frac {{\rm Li}_2\left (a x\right )}{\left (d x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(polylog(2,a*x)/(d*x)^(7/2),x, algorithm="giac")
 

Output:

integrate(dilog(a*x)/(d*x)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=\int \frac {\mathrm {polylog}\left (2,a\,x\right )}{{\left (d\,x\right )}^{7/2}} \,d x \] Input:

int(polylog(2, a*x)/(d*x)^(7/2),x)
 

Output:

int(polylog(2, a*x)/(d*x)^(7/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\operatorname {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx=\frac {\int \frac {\mathit {polylog}\left (2, a x \right )}{\sqrt {x}\, x^{3}}d x}{\sqrt {d}\, d^{3}} \] Input:

int(polylog(2,a*x)/(d*x)^(7/2),x)
 

Output:

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