3.84 \(\int \frac {\text {Li}_3(a x^2)}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac {64 a^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {64 a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}-\frac {128 a}{125 d^3 \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}} \]

[Out]

-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)

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Rubi [A]  time = 0.11, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6591, 2455, 16, 325, 329, 298, 205, 208} \[ -\frac {8 \text {PolyLog}\left (2,a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}-\frac {64 a^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {64 a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}-\frac {128 a}{125 d^3 \sqrt {d x}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(7/2),x]

[Out]

(-128*a)/(125*d^3*Sqrt[d*x]) - (64*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (64*a^(5/4)*Ar
cTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (32*Log[1 - a*x^2])/(125*d*(d*x)^(5/2)) - (8*PolyLog[2, a*
x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[3, 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 205

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

Rule 208

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

Rule 298

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 325

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 329

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 2455

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 6591

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}_3\left (a x^2\right )}{(d x)^{7/2}} \, dx &=-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {4}{5} \int \frac {\text {Li}_2\left (a x^2\right )}{(d x)^{7/2}} \, dx\\ &=-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}-\frac {16}{25} \int \frac {\log \left (1-a x^2\right )}{(d x)^{7/2}} \, dx\\ &=\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {(64 a) \int \frac {x}{(d x)^{5/2} \left (1-a x^2\right )} \, dx}{125 d}\\ &=\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {(64 a) \int \frac {1}{(d x)^{3/2} \left (1-a x^2\right )} \, dx}{125 d^2}\\ &=-\frac {128 a}{125 d^3 \sqrt {d x}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (64 a^2\right ) \int \frac {\sqrt {d x}}{1-a x^2} \, dx}{125 d^4}\\ &=-\frac {128 a}{125 d^3 \sqrt {d x}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (128 a^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{125 d^5}\\ &=-\frac {128 a}{125 d^3 \sqrt {d x}}+\frac {32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac {\left (64 a^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{125 d^3}-\frac {\left (64 a^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{125 d^3}\\ &=-\frac {128 a}{125 d^3 \sqrt {d x}}-\frac {64 a^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {64 a^{5/4} \tanh ^{-1}\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 \text {Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 79, normalized size = 0.54 \[ -\frac {x \Gamma \left (-\frac {1}{4}\right ) \left (64 a^2 x^4 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};a x^2\right )-60 \text {Li}_2\left (a x^2\right )-75 \text {Li}_3\left (a x^2\right )-192 a x^2+48 \log \left (1-a x^2\right )\right )}{750 \Gamma \left (\frac {3}{4}\right ) (d x)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^2]/(d*x)^(7/2),x]

[Out]

-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])

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fricas [C]  time = 0.96, size = 226, normalized size = 1.54 \[ \frac {2 \, {\left (64 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{4} d^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} - \sqrt {a^{5} d^{8} \sqrt {\frac {a^{5}}{d^{14}}} + a^{8} d x} d^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}}}{a^{5}}\right ) + 16 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (32768 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) - 16 \, d^{4} x^{3} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {1}{4}} \log \left (-32768 \, d^{11} \left (\frac {a^{5}}{d^{14}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a^{4}\right ) - 4 \, {\left (16 \, a x^{2} + 5 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )} \sqrt {d x} - 25 \, \sqrt {d x} {\rm polylog}\left (3, a x^{2}\right )\right )}}{125 \, d^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/125*(64*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) + 16*d^4*x^3*(a^5/d^14)^(1/4)*log(32768*d^11*(a^5/d^14)^(3/4) + 32768*sqr
t(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}(a x^{2})}{\left (d x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/(d*x)^(7/2), x)

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maple [A]  time = 0.18, size = 142, normalized size = 0.97 \[ -\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {5}{4}} \left (-\frac {256}{125 \sqrt {x}\, \left (-a \right )^{\frac {1}{4}}}-\frac {64 x^{\frac {3}{2}} a \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{125 \left (-a \right )^{\frac {1}{4}} \left (a \,x^{2}\right )^{\frac {3}{4}}}+\frac {64 \ln \left (-a \,x^{2}+1\right )}{125 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}-\frac {16 \polylog \left (2, a \,x^{2}\right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}-\frac {4 \polylog \left (3, a \,x^{2}\right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {1}{4}} a}\right )}{2 \left (d x \right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/(d*x)^(7/2),x)

[Out]

-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)/a*ln(-a*x^2+1)-16/25/
x^(5/2)/(-a)^(1/4)*polylog(2,a*x^2)/a-4/5/x^(5/2)/(-a)^(1/4)/a*polylog(3,a*x^2))

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maxima [A]  time = 0.49, size = 163, normalized size = 1.11 \[ -\frac {2 \, {\left (\frac {16 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}} + \frac {\log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}}\right )}}{d^{2}} + \frac {64 \, a d^{2} x^{2} + 20 \, d^{2} {\rm Li}_2\left (a x^{2}\right ) - 16 \, d^{2} \log \left (-a d^{2} x^{2} + d^{2}\right ) + 32 \, d^{2} \log \relax (d) + 25 \, d^{2} {\rm Li}_{3}(a x^{2})}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{125 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(7/2),x, algorithm="maxima")

[Out]

-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*polylog(3, a*x^2))/((d*x)^(5/2)*d^2)
)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {polylog}\left (3,a\,x^2\right )}{{\left (d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/(d*x)^(7/2),x)

[Out]

int(polylog(3, a*x^2)/(d*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{\left (d x\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/(d*x)**(7/2),x)

[Out]

Integral(polylog(3, a*x**2)/(d*x)**(7/2), x)

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