3.1.77 \(\int \frac {\text {PolyLog}(2,a x^2)}{(d x)^{7/2}} \, dx\) [77]

Optimal. Leaf size=126 \[ -\frac {32 a}{25 d^3 \sqrt {d x}}-\frac {16 a^{5/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {16 a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}} \]

[Out]

-16/25*a^(5/4)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+16/25*a^(5/4)*arctanh(a^(1/4)*(d*x)^(1/2)/d^(1/2))/
d^(7/2)+8/25*ln(-a*x^2+1)/d/(d*x)^(5/2)-2/5*polylog(2,a*x^2)/d/(d*x)^(5/2)-32/25*a/d^3/(d*x)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6726, 2505, 16, 331, 335, 304, 211, 214} \begin {gather*} -\frac {16 a^{5/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {16 a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}-\frac {32 a}{25 d^3 \sqrt {d x}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/(d*x)^(7/2),x]

[Out]

(-32*a)/(25*d^3*Sqrt[d*x]) - (16*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (16*a^(5/4)*ArcTa
nh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (8*Log[1 - a*x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[2, a*x^2])/
(5*d*(d*x)^(5/2))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 211

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

Rule 214

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2505

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] - Dist[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 6726

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[(d*x)^(m + 1)*(PolyLog[n
, a*(b*x^p)^q]/(d*(m + 1))), x] - Dist[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]

Rubi steps

\begin {align*} \int \frac {\text {Li}_2\left (a x^2\right )}{(d x)^{7/2}} \, dx &=-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}-\frac {4}{5} \int \frac {\log \left (1-a x^2\right )}{(d x)^{7/2}} \, dx\\ &=\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {(16 a) \int \frac {x}{(d x)^{5/2} \left (1-a x^2\right )} \, dx}{25 d}\\ &=\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {(16 a) \int \frac {1}{(d x)^{3/2} \left (1-a x^2\right )} \, dx}{25 d^2}\\ &=-\frac {32 a}{25 d^3 \sqrt {d x}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (16 a^2\right ) \int \frac {\sqrt {d x}}{1-a x^2} \, dx}{25 d^4}\\ &=-\frac {32 a}{25 d^3 \sqrt {d x}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (32 a^2\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{25 d^5}\\ &=-\frac {32 a}{25 d^3 \sqrt {d x}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (16 a^{3/2}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{25 d^3}-\frac {\left (16 a^{3/2}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{25 d^3}\\ &=-\frac {32 a}{25 d^3 \sqrt {d x}}-\frac {16 a^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {16 a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 70, normalized size = 0.56 \begin {gather*} -\frac {x \Gamma \left (-\frac {1}{4}\right ) \left (-48 a x^2+16 a^2 x^4 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};a x^2\right )+12 \log \left (1-a x^2\right )-15 \text {PolyLog}\left (2,a x^2\right )\right )}{150 (d x)^{7/2} \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/150*(x*Gamma[-1/4]*(-48*a*x^2 + 16*a^2*x^4*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 12*Log[1 - a*x^2] - 15*P
olyLog[2, a*x^2]))/((d*x)^(7/2)*Gamma[3/4])

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Maple [A]
time = 0.41, size = 131, normalized size = 1.04

method	result	size
meijerg	\(-\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {5}{4}} \left (-\frac {64}{25 \sqrt {x}\, \left (-a \right )^{\frac {1}{4}}}-\frac {16 x^{\frac {3}{2}} a \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{25 \left (-a \right )^{\frac {1}{4}} \left (a \,x^{2}\right )^{\frac {3}{4}}}+\frac {16 \ln \left (-a \,x^{2}+1\right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}-\frac {4 \polylog \left (2, a \,x^{2}\right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}\right )}{2 \left (d x \right )^{\frac {7}{2}}}\)	\(122\)
derivativedivides	\(\frac {-\frac {2 \polylog \left (2, a \,x^{2}\right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {32 a \left (-\frac {1}{d^{2} \sqrt {d x}}-\frac {2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{4 d^{2} \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{25}}{d}\)	\(131\)
default	\(\frac {-\frac {2 \polylog \left (2, a \,x^{2}\right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {32 a \left (-\frac {1}{d^{2} \sqrt {d x}}-\frac {2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{4 d^{2} \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{25}}{d}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/5*polylog(2,a*x^2)/(d*x)^(5/2)+4/25/(d*x)^(5/2)*ln((-a*d^2*x^2+d^2)/d^2)+16/25*a*(-1/d^2/(d*x)^(1/2)-1
/4/d^2/(d^2/a)^(1/4)*(2*arctan((d*x)^(1/2)/(d^2/a)^(1/4))-ln(((d*x)^(1/2)+(d^2/a)^(1/4))/((d*x)^(1/2)-(d^2/a)^
(1/4))))))

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Maxima [A]
time = 0.48, size = 151, normalized size = 1.20 \begin {gather*} -\frac {2 \, {\left (\frac {4 \, 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 {16 \, a d^{2} x^{2} + 5 \, d^{2} {\rm Li}_2\left (a x^{2}\right ) - 4 \, d^{2} \log \left (-a d^{2} x^{2} + d^{2}\right ) + 8 \, d^{2} \log \left (d\right )}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{25 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/25*(4*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 + (16*a*d^2*x^2 + 5*d^
2*dilog(a*x^2) - 4*d^2*log(-a*d^2*x^2 + d^2) + 8*d^2*log(d))/((d*x)^(5/2)*d^2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (89) = 178\).
time = 0.44, size = 212, normalized size = 1.68 \begin {gather*} \frac {2 \, {\left (16 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{4} d^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} - \sqrt {a^{5} d^{8} \sqrt {\frac {a^{5}}{d^{14}}} + a^{8} d x} d^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}}}{a^{5}}\right ) + 4 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (512 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 512 \, \sqrt {d x} a^{4}\right ) - 4 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (-512 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 512 \, \sqrt {d x} a^{4}\right ) - {\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}\right )}}{25 \, d^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(16*d^4*x^3*(a^5/d^14)^(1/4)*arctan(-(sqrt(d*x)*a^4*d^3*(a^5/d^14)^(1/4) - sqrt(a^5*d^8*sqrt(a^5/d^14) +
a^8*d*x)*d^3*(a^5/d^14)^(1/4))/a^5) + 4*d^4*x^3*(a^5/d^14)^(1/4)*log(512*d^11*(a^5/d^14)^(3/4) + 512*sqrt(d*x)
*a^4) - 4*d^4*x^3*(a^5/d^14)^(1/4)*log(-512*d^11*(a^5/d^14)^(3/4) + 512*sqrt(d*x)*a^4) - (16*a*x^2 + 5*dilog(a
*x^2) - 4*log(-a*x^2 + 1))*sqrt(d*x))/(d^4*x^3)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {polylog}\left (2,a\,x^2\right )}{{\left (d\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/(d*x)^(7/2),x)

[Out]

int(polylog(2, a*x^2)/(d*x)^(7/2), x)

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