Integrand size = 15, antiderivative size = 147 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=-\frac {128 a}{125 d^3 \sqrt {d x}}-\frac {64 a^{5/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {64 a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \operatorname {PolyLog}\left (2,a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}} \]
-64/125*a^(5/4)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+64/125*a^(5/4) *arctanh(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+32/125*ln(-a*x^2+1)/d/(d*x)^ (5/2)-8/25*polylog(2,a*x^2)/d/(d*x)^(5/2)-2/5*polylog(3,a*x^2)/d/(d*x)^(5/ 2)-128/125*a/d^3/(d*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.54 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=-\frac {x \operatorname {Gamma}\left (-\frac {1}{4}\right ) \left (-192 a x^2+64 a^2 x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},a x^2\right )+48 \log \left (1-a x^2\right )-60 \operatorname {PolyLog}\left (2,a x^2\right )-75 \operatorname {PolyLog}\left (3,a x^2\right )\right )}{750 (d x)^{7/2} \operatorname {Gamma}\left (\frac {3}{4}\right )} \]
-1/750*(x*Gamma[-1/4]*(-192*a*x^2 + 64*a^2*x^4*Hypergeometric2F1[3/4, 1, 7 /4, a*x^2] + 48*Log[1 - a*x^2] - 60*PolyLog[2, a*x^2] - 75*PolyLog[3, a*x^ 2]))/((d*x)^(7/2)*Gamma[3/4])
Time = 0.39 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {7145, 7145, 25, 2905, 8, 264, 266, 27, 827, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {4}{5} \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{(d x)^{7/2}}dx-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {4}{5} \left (\frac {4}{5} \int -\frac {\log \left (1-a x^2\right )}{(d x)^{7/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \int \frac {\log \left (1-a x^2\right )}{(d x)^{7/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \int \frac {x}{(d x)^{5/2} \left (1-a x^2\right )}dx}{5 d}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \int \frac {1}{(d x)^{3/2} \left (1-a x^2\right )}dx}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {a \int \frac {\sqrt {d x}}{1-a x^2}dx}{d^2}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {2 a \int \frac {d^3 x}{d^2-a d^2 x^2}d\sqrt {d x}}{d^3}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {2 a \int \frac {d x}{d^2-a d^2 x^2}d\sqrt {d x}}{d}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {2 a \left (\frac {\int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} x d+d}d\sqrt {d x}}{2 \sqrt {a}}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {2 a \left (\frac {\int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 a^{3/4} \sqrt {d}}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {4}{5} \left (-\frac {4}{5} \left (-\frac {4 a \left (\frac {2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 a^{3/4} \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 a^{3/4} \sqrt {d}}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{5 d^2}-\frac {2 \log \left (1-a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}\) |
(4*((-4*((-4*a*(-2/(d*Sqrt[d*x]) + (2*a*(-1/2*ArcTan[(a^(1/4)*Sqrt[d*x])/S qrt[d]]/(a^(3/4)*Sqrt[d]) + ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]]/(2*a^(3/4 )*Sqrt[d])))/d))/(5*d^2) - (2*Log[1 - a*x^2])/(5*d*(d*x)^(5/2))))/5 - (2*P olyLog[2, a*x^2])/(5*d*(d*x)^(5/2))))/5 - (2*PolyLog[3, a*x^2])/(5*d*(d*x) ^(5/2))
3.1.84.3.1 Defintions of rubi rules used
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]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
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]
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]
Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97
method | result | size |
meijerg | \(-\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {5}{4}} \left (-\frac {256}{125 \sqrt {x}\, \left (-a \right )^{\frac {1}{4}}}-\frac {64 x^{\frac {3}{2}} a \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{125 \left (-a \right )^{\frac {1}{4}} \left (a \,x^{2}\right )^{\frac {3}{4}}}+\frac {64 \ln \left (-a \,x^{2}+1\right )}{125 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}-\frac {16 \operatorname {polylog}\left (2, a \,x^{2}\right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}-\frac {4 \operatorname {polylog}\left (3, a \,x^{2}\right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}\right )}{2 \left (d x \right )^{\frac {7}{2}}}\) | \(142\) |
-1/2/(d*x)^(7/2)*x^(7/2)*(-a)^(5/4)*(-256/125/x^(1/2)/(-a)^(1/4)-64/125*x^ (3/2)/(-a)^(1/4)*a/(a*x^2)^(3/4)*(ln(1-(a*x^2)^(1/4))-ln(1+(a*x^2)^(1/4))+ 2*arctan((a*x^2)^(1/4)))+64/125/x^(5/2)/(-a)^(1/4)*ln(-a*x^2+1)/a-16/25/x^ (5/2)/(-a)^(1/4)/a*polylog(2,a*x^2)-4/5/x^(5/2)/(-a)^(1/4)/a*polylog(3,a*x ^2))
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.55 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (16 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (32768 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) - 16 i \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (32768 i \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) + 16 i \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (-32768 i \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) - 16 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (-32768 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) - 4 \, {\left (16 \, a x^{2} + 5 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )} \sqrt {d x} - 25 \, \sqrt {d x} {\rm polylog}\left (3, a x^{2}\right )\right )}}{125 \, d^{4} x^{3}} \]
2/125*(16*d^4*x^3*(a^5/d^14)^(1/4)*log(32768*d^11*(a^5/d^14)^(3/4) + 32768 *sqrt(d*x)*a^4) - 16*I*d^4*x^3*(a^5/d^14)^(1/4)*log(32768*I*d^11*(a^5/d^14 )^(3/4) + 32768*sqrt(d*x)*a^4) + 16*I*d^4*x^3*(a^5/d^14)^(1/4)*log(-32768* I*d^11*(a^5/d^14)^(3/4) + 32768*sqrt(d*x)*a^4) - 16*d^4*x^3*(a^5/d^14)^(1/ 4)*log(-32768*d^11*(a^5/d^14)^(3/4) + 32768*sqrt(d*x)*a^4) - 4*(16*a*x^2 + 5*dilog(a*x^2) - 4*log(-a*x^2 + 1))*sqrt(d*x) - 25*sqrt(d*x)*polylog(3, a *x^2))/(d^4*x^3)
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{\left (d x\right )^{\frac {7}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.11 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {16 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}} + \frac {\log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}}\right )}}{d^{2}} + \frac {64 \, a d^{2} x^{2} + 20 \, d^{2} {\rm Li}_2\left (a x^{2}\right ) - 16 \, d^{2} \log \left (-a d^{2} x^{2} + d^{2}\right ) + 32 \, d^{2} \log \left (d\right ) + 25 \, d^{2} {\rm Li}_{3}(a x^{2})}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{125 \, d} \]
-2/125*(16*a^2*(2*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)* d)*sqrt(a)) + log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/(sqrt(sqrt(a)*d)*sqrt(a)))/d^2 + (64*a*d^2*x^2 + 20*d ^2*dilog(a*x^2) - 16*d^2*log(-a*d^2*x^2 + d^2) + 32*d^2*log(d) + 25*d^2*po lylog(3, a*x^2))/((d*x)^(5/2)*d^2))/d
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{2})}{\left (d x\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{(d x)^{7/2}} \, dx=\int \frac {\mathrm {polylog}\left (3,a\,x^2\right )}{{\left (d\,x\right )}^{7/2}} \,d x \]