3.1.71 \(\int \frac {\text {PolyLog}(3,a x)}{(d x)^{7/2}} \, dx\) [71]

Optimal. Leaf size=125 \[ -\frac {16 a}{375 d^2 (d x)^{3/2}}-\frac {16 a^2}{125 d^3 \sqrt {d x}}+\frac {16 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {PolyLog}(2,a x)}{25 d (d x)^{5/2}}-\frac {2 \text {PolyLog}(3,a x)}{5 d (d x)^{5/2}} \]

[Out]

-16/375*a/d^2/(d*x)^(3/2)+16/125*a^(5/2)*arctanh(a^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+8/125*ln(-a*x+1)/d/(d*x)
^(5/2)-4/25*polylog(2,a*x)/d/(d*x)^(5/2)-2/5*polylog(3,a*x)/d/(d*x)^(5/2)-16/125*a^2/d^3/(d*x)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6726, 2442, 53, 65, 212} \begin {gather*} \frac {16 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}-\frac {16 a^2}{125 d^3 \sqrt {d x}}-\frac {16 a}{375 d^2 (d x)^{3/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x]/(d*x)^(7/2),x]

[Out]

(-16*a)/(375*d^2*(d*x)^(3/2)) - (16*a^2)/(125*d^3*Sqrt[d*x]) + (16*a^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]
])/(125*d^(7/2)) + (8*Log[1 - a*x])/(125*d*(d*x)^(5/2)) - (4*PolyLog[2, a*x])/(25*d*(d*x)^(5/2)) - (2*PolyLog[
3, a*x])/(5*d*(d*x)^(5/2))

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2442

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] - Dist[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 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}_3(a x)}{(d x)^{7/2}} \, dx &=-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {2}{5} \int \frac {\text {Li}_2(a x)}{(d x)^{7/2}} \, dx\\ &=-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}-\frac {4}{25} \int \frac {\log (1-a x)}{(d x)^{7/2}} \, dx\\ &=\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {(8 a) \int \frac {1}{(d x)^{5/2} (1-a x)} \, dx}{125 d}\\ &=-\frac {16 a}{375 d^2 (d x)^{3/2}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {\left (8 a^2\right ) \int \frac {1}{(d x)^{3/2} (1-a x)} \, dx}{125 d^2}\\ &=-\frac {16 a}{375 d^2 (d x)^{3/2}}-\frac {16 a^2}{125 d^3 \sqrt {d x}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {\left (8 a^3\right ) \int \frac {1}{\sqrt {d x} (1-a x)} \, dx}{125 d^3}\\ &=-\frac {16 a}{375 d^2 (d x)^{3/2}}-\frac {16 a^2}{125 d^3 \sqrt {d x}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}+\frac {\left (16 a^3\right ) \text {Subst}\left (\int \frac {1}{1-\frac {a x^2}{d}} \, dx,x,\sqrt {d x}\right )}{125 d^4}\\ &=-\frac {16 a}{375 d^2 (d x)^{3/2}}-\frac {16 a^2}{125 d^3 \sqrt {d x}}+\frac {16 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{125 d^{7/2}}+\frac {8 \log (1-a x)}{125 d (d x)^{5/2}}-\frac {4 \text {Li}_2(a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_3(a x)}{5 d (d x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 72, normalized size = 0.58 \begin {gather*} -\frac {2 x \left (8 a x+24 a^2 x^2-24 a^{5/2} x^{5/2} \tanh ^{-1}\left (\sqrt {a} \sqrt {x}\right )-12 \log (1-a x)+30 \text {PolyLog}(2,a x)+75 \text {PolyLog}(3,a x)\right )}{375 (d x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(8*a*x + 24*a^2*x^2 - 24*a^(5/2)*x^(5/2)*ArcTanh[Sqrt[a]*Sqrt[x]] - 12*Log[1 - a*x] + 30*PolyLog[2, a*x]
+ 75*PolyLog[3, a*x]))/(375*(d*x)^(7/2))

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Maple [A]
time = 0.09, size = 135, normalized size = 1.08

method	result	size
meijerg	\(\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {7}{2}} \left (-\frac {16}{375 x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2}}}-\frac {16 a}{125 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}}}-\frac {8 \sqrt {x}\, a^{2} \left (\ln \left (1-\sqrt {a x}\right )-\ln \left (1+\sqrt {a x}\right )\right )}{125 \left (-a \right )^{\frac {3}{2}} \sqrt {a x}}+\frac {8 \ln \left (-a x +1\right )}{125 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}-\frac {4 \polylog \left (2, a x \right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \polylog \left (3, a x \right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}\right )}{\left (d x \right )^{\frac {7}{2}} a}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(d*x)^(7/2)*x^(7/2)*(-a)^(7/2)/a*(-16/375/x^(3/2)/(-a)^(3/2)-16/125/x^(1/2)/(-a)^(3/2)*a-8/125*x^(1/2)/(-a)^
(3/2)*a^2/(a*x)^(1/2)*(ln(1-(a*x)^(1/2))-ln(1+(a*x)^(1/2)))+8/125/x^(5/2)/(-a)^(3/2)/a*ln(-a*x+1)-4/25/x^(5/2)
/(-a)^(3/2)/a*polylog(2,a*x)-2/5/x^(5/2)/(-a)^(3/2)/a*polylog(3,a*x))

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Maxima [A]
time = 0.48, size = 118, normalized size = 0.94 \begin {gather*} -\frac {2 \, {\left (\frac {12 \, 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 {24 \, a^{2} d^{2} x^{2} + 8 \, a d^{2} x + 30 \, d^{2} {\rm Li}_2\left (a x\right ) - 12 \, d^{2} \log \left (-a d x + d\right ) + 12 \, d^{2} \log \left (d\right ) + 75 \, d^{2} {\rm Li}_{3}(a x)}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{375 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/375*(12*a^3*log((sqrt(d*x)*a - sqrt(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/(sqrt(a*d)*d^2) + (24*a^2*d^2*x^2 + 8*
a*d^2*x + 30*d^2*dilog(a*x) - 12*d^2*log(-a*d*x + d) + 12*d^2*log(d) + 75*d^2*polylog(3, a*x))/((d*x)^(5/2)*d^
2))/d

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Fricas [A]
time = 0.43, size = 195, normalized size = 1.56 \begin {gather*} \left [\frac {2 \, {\left (12 \, 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 ) - 2 \, {\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} - 75 \, \sqrt {d x} {\rm polylog}\left (3, a x\right )\right )}}{375 \, d^{4} x^{3}}, -\frac {2 \, {\left (24 \, a^{2} d x^{3} \sqrt {-\frac {a}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {a}{d}}}{a x}\right ) + 2 \, {\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} + 75 \, \sqrt {d x} {\rm polylog}\left (3, a x\right )\right )}}{375 \, d^{4} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/375*(12*a^2*d*x^3*sqrt(a/d)*log((a*x + 2*sqrt(d*x)*sqrt(a/d) + 1)/(a*x - 1)) - 2*(12*a^2*x^2 + 4*a*x + 15*d
ilog(a*x) - 6*log(-a*x + 1))*sqrt(d*x) - 75*sqrt(d*x)*polylog(3, a*x))/(d^4*x^3), -2/375*(24*a^2*d*x^3*sqrt(-a
/d)*arctan(sqrt(d*x)*sqrt(-a/d)/(a*x)) + 2*(12*a^2*x^2 + 4*a*x + 15*dilog(a*x) - 6*log(-a*x + 1))*sqrt(d*x) +
75*sqrt(d*x)*polylog(3, a*x))/(d^4*x^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(polylog(3, a*x)/(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 (3,a\,x\right )}{{\left (d\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x)/(d*x)^(7/2),x)

[Out]

int(polylog(3, a*x)/(d*x)^(7/2), x)

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