Integrand size = 15, antiderivative size = 115 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=-\frac {32 \sqrt {d x}}{d}+\frac {16 \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}+\frac {16 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}+\frac {8 \sqrt {d x} \log \left (1-a x^2\right )}{d}+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d} \]
16*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/a^(1/4)/d^(1/2)+16*arctanh(a^(1/4)* (d*x)^(1/2)/d^(1/2))/a^(1/4)/d^(1/2)-32*(d*x)^(1/2)/d+8*ln(-a*x^2+1)*(d*x) ^(1/2)/d+2*polylog(2,a*x^2)*(d*x)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=\frac {5 x \operatorname {Gamma}\left (\frac {5}{4}\right ) \left (-16+16 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},a x^2\right )+4 \log \left (1-a x^2\right )+\operatorname {PolyLog}\left (2,a x^2\right )\right )}{2 \sqrt {d x} \operatorname {Gamma}\left (\frac {9}{4}\right )} \]
(5*x*Gamma[5/4]*(-16 + 16*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 4*Log[1 - a*x^2] + PolyLog[2, a*x^2]))/(2*Sqrt[d*x]*Gamma[9/4])
Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {7145, 25, 2905, 8, 262, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx\) |
| \(\Big \downarrow \) 7145 |
| \(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}-4 \int -\frac {\log \left (1-a x^2\right )}{\sqrt {d x}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \int \frac {\log \left (1-a x^2\right )}{\sqrt {d x}}dx+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle 4 \left (\frac {4 a \int \frac {x \sqrt {d x}}{1-a x^2}dx}{d}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 8 |
| \(\displaystyle 4 \left (\frac {4 a \int \frac {(d x)^{3/2}}{1-a x^2}dx}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle 4 \left (\frac {4 a \left (\frac {d^2 \int \frac {1}{\sqrt {d x} \left (1-a x^2\right )}dx}{a}-\frac {2 d \sqrt {d x}}{a}\right )}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle 4 \left (\frac {4 a \left (\frac {2 d \int \frac {1}{1-a x^2}d\sqrt {d x}}{a}-\frac {2 d \sqrt {d x}}{a}\right )}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle 4 \left (\frac {4 a \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}+\frac {1}{2} d \int \frac {1}{\sqrt {a} x d+d}d\sqrt {d x}\right )}{a}-\frac {2 d \sqrt {d x}}{a}\right )}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle 4 \left (\frac {4 a \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-\sqrt {a} d x}d\sqrt {d x}+\frac {\sqrt {d} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {2 d \sqrt {d x}}{a}\right )}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 4 \left (\frac {4 a \left (\frac {2 d \left (\frac {\sqrt {d} \arctan \left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {2 d \sqrt {d x}}{a}\right )}{d^2}+\frac {2 \sqrt {d x} \log \left (1-a x^2\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^2\right )}{d}\) |
4*((4*a*((-2*d*Sqrt[d*x])/a + (2*d*((Sqrt[d]*ArcTan[(a^(1/4)*Sqrt[d*x])/Sq rt[d]])/(2*a^(1/4)) + (Sqrt[d]*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(2*a^ (1/4))))/a))/d^2 + (2*Sqrt[d*x]*Log[1 - a*x^2])/d) + (2*Sqrt[d*x]*PolyLog[ 2, a*x^2])/d
3.1.74.3.1 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_) + (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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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[((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]
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 = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10
| method | result | size |
| meijerg | \(-\frac {\sqrt {x}\, \left (-\frac {64 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}}}{a}-\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{a \left (a \,x^{2}\right )^{\frac {1}{4}}}+\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \ln \left (-a \,x^{2}+1\right )}{a}+\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \operatorname {polylog}\left (2, a \,x^{2}\right )}{a}\right )}{2 \sqrt {d x}\, \left (-a \right )^{\frac {1}{4}}}\) | \(127\) |
| derivativedivides | \(\frac {2 \sqrt {d x}\, \operatorname {polylog}\left (2, a \,x^{2}\right )+8 \sqrt {d x}\, \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )+32 a \left (-\frac {\sqrt {d x}}{a}+\frac {\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{4 a}\right )}{d}\) | \(128\) |
| default | \(\frac {2 \sqrt {d x}\, \operatorname {polylog}\left (2, a \,x^{2}\right )+8 \sqrt {d x}\, \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )+32 a \left (-\frac {\sqrt {d x}}{a}+\frac {\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{4 a}\right )}{d}\) | \(128\) |
-1/2/(d*x)^(1/2)*x^(1/2)/(-a)^(1/4)*(-64*x^(1/2)*(-a)^(5/4)/a-16*x^(1/2)*( -a)^(5/4)/a/(a*x^2)^(1/4)*(ln(1-(a*x^2)^(1/4))-ln(1+(a*x^2)^(1/4))-2*arcta n((a*x^2)^(1/4)))+16*x^(1/2)*(-a)^(5/4)*ln(-a*x^2+1)/a+4*x^(1/2)*(-a)^(5/4 )/a*polylog(2,a*x^2))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.34 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=\frac {2 \, {\left (4 \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) + 4 i \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (i \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) - 4 i \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (-i \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) - 4 \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (-d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) + \sqrt {d x} {\left ({\rm Li}_2\left (a x^{2}\right ) + 4 \, \log \left (-a x^{2} + 1\right ) - 16\right )}\right )}}{d} \]
2*(4*d*(1/(a*d^2))^(1/4)*log(d*(1/(a*d^2))^(1/4) + sqrt(d*x)) + 4*I*d*(1/( a*d^2))^(1/4)*log(I*d*(1/(a*d^2))^(1/4) + sqrt(d*x)) - 4*I*d*(1/(a*d^2))^( 1/4)*log(-I*d*(1/(a*d^2))^(1/4) + sqrt(d*x)) - 4*d*(1/(a*d^2))^(1/4)*log(- d*(1/(a*d^2))^(1/4) + sqrt(d*x)) + sqrt(d*x)*(dilog(a*x^2) + 4*log(-a*x^2 + 1) - 16))/d
\[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=\int \frac {\operatorname {Li}_{2}\left (a x^{2}\right )}{\sqrt {d x}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=-\frac {2 \, {\left (8 \, \sqrt {d x} {\left (\log \left (d\right ) + 2\right )} - \frac {8 \, d \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}} - \sqrt {d x} {\rm Li}_2\left (a x^{2}\right ) - 4 \, \sqrt {d x} \log \left (-a d^{2} x^{2} + d^{2}\right ) + \frac {4 \, d \log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}}\right )}}{d} \]
-2*(8*sqrt(d*x)*(log(d) + 2) - 8*d*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d ))/sqrt(sqrt(a)*d) - sqrt(d*x)*dilog(a*x^2) - 4*sqrt(d*x)*log(-a*d^2*x^2 + d^2) + 4*d*log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/sqrt(sqrt(a)*d))/d
\[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=\int { \frac {{\rm Li}_2\left (a x^{2}\right )}{\sqrt {d x}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{\sqrt {d x}} \, dx=\int \frac {\mathrm {polylog}\left (2,a\,x^2\right )}{\sqrt {d\,x}} \,d x \]