3.1.69 \(\int \frac {\text {PolyLog}(3,a x)}{(d x)^{3/2}} \, dx\) [69]

Optimal. Leaf size=85 \[ \frac {16 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {8 \log (1-a x)}{d \sqrt {d x}}-\frac {4 \text {PolyLog}(2,a x)}{d \sqrt {d x}}-\frac {2 \text {PolyLog}(3,a x)}{d \sqrt {d x}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6726, 2442, 65, 212} \begin {gather*} \frac {16 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {4 \text {Li}_2(a x)}{d \sqrt {d x}}-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}+\frac {8 \log (1-a x)}{d \sqrt {d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x]/(d*x)^(3/2),x]

[Out]

(16*Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]])/d^(3/2) + (8*Log[1 - a*x])/(d*Sqrt[d*x]) - (4*PolyLog[2, a*x
])/(d*Sqrt[d*x]) - (2*PolyLog[3, a*x])/(d*Sqrt[d*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)^{3/2}} \, dx &=-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}+2 \int \frac {\text {Li}_2(a x)}{(d x)^{3/2}} \, dx\\ &=-\frac {4 \text {Li}_2(a x)}{d \sqrt {d x}}-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}-4 \int \frac {\log (1-a x)}{(d x)^{3/2}} \, dx\\ &=\frac {8 \log (1-a x)}{d \sqrt {d x}}-\frac {4 \text {Li}_2(a x)}{d \sqrt {d x}}-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}+\frac {(8 a) \int \frac {1}{\sqrt {d x} (1-a x)} \, dx}{d}\\ &=\frac {8 \log (1-a x)}{d \sqrt {d x}}-\frac {4 \text {Li}_2(a x)}{d \sqrt {d x}}-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}+\frac {(16 a) \text {Subst}\left (\int \frac {1}{1-\frac {a x^2}{d}} \, dx,x,\sqrt {d x}\right )}{d^2}\\ &=\frac {16 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {8 \log (1-a x)}{d \sqrt {d x}}-\frac {4 \text {Li}_2(a x)}{d \sqrt {d x}}-\frac {2 \text {Li}_3(a x)}{d \sqrt {d x}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 58, normalized size = 0.68 \begin {gather*} \frac {2 x \left (8 \sqrt {a} \sqrt {x} \tanh ^{-1}\left (\sqrt {a} \sqrt {x}\right )+4 \log (1-a x)-2 \text {PolyLog}(2,a x)-\text {PolyLog}(3,a x)\right )}{(d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(8*Sqrt[a]*Sqrt[x]*ArcTanh[Sqrt[a]*Sqrt[x]] + 4*Log[1 - a*x] - 2*PolyLog[2, a*x] - PolyLog[3, a*x]))/(d*x
)^(3/2)

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Maple [A]
time = 0.07, size = 111, normalized size = 1.31

method	result	size
meijerg	\(\frac {x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2}} \left (-\frac {8 \sqrt {x}\, \sqrt {-a}\, \left (\ln \left (1-\sqrt {a x}\right )-\ln \left (1+\sqrt {a x}\right )\right )}{\sqrt {a x}}+\frac {8 \sqrt {-a}\, \ln \left (-a x +1\right )}{\sqrt {x}\, a}-\frac {4 \sqrt {-a}\, \polylog \left (2, a x \right )}{\sqrt {x}\, a}-\frac {2 \sqrt {-a}\, \polylog \left (3, a x \right )}{\sqrt {x}\, a}\right )}{\left (d x \right )^{\frac {3}{2}} a}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(d*x)^(3/2)*x^(3/2)*(-a)^(3/2)/a*(-8*x^(1/2)*(-a)^(1/2)/(a*x)^(1/2)*(ln(1-(a*x)^(1/2))-ln(1+(a*x)^(1/2)))+8/
x^(1/2)*(-a)^(1/2)/a*ln(-a*x+1)-4/x^(1/2)*(-a)^(1/2)/a*polylog(2,a*x)-2/x^(1/2)*(-a)^(1/2)/a*polylog(3,a*x))

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Maxima [A]
time = 0.47, size = 78, normalized size = 0.92 \begin {gather*} -\frac {2 \, {\left (\frac {4 \, a \log \left (\frac {\sqrt {d x} a - \sqrt {a d}}{\sqrt {d x} a + \sqrt {a d}}\right )}{\sqrt {a d}} + \frac {2 \, {\rm Li}_2\left (a x\right ) - 4 \, \log \left (-a d x + d\right ) + 4 \, \log \left (d\right ) + {\rm Li}_{3}(a x)}{\sqrt {d x}}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(4*a*log((sqrt(d*x)*a - sqrt(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/sqrt(a*d) + (2*dilog(a*x) - 4*log(-a*d*x + d)
+ 4*log(d) + polylog(3, a*x))/sqrt(d*x))/d

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Fricas [A]
time = 0.39, size = 156, normalized size = 1.84 \begin {gather*} \left [\frac {2 \, {\left (4 \, d x \sqrt {\frac {a}{d}} \log \left (\frac {a x + 2 \, \sqrt {d x} \sqrt {\frac {a}{d}} + 1}{a x - 1}\right ) - 2 \, \sqrt {d x} {\left ({\rm Li}_2\left (a x\right ) - 2 \, \log \left (-a x + 1\right )\right )} - \sqrt {d x} {\rm polylog}\left (3, a x\right )\right )}}{d^{2} x}, -\frac {2 \, {\left (8 \, d x \sqrt {-\frac {a}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {a}{d}}}{a x}\right ) + 2 \, \sqrt {d x} {\left ({\rm Li}_2\left (a x\right ) - 2 \, \log \left (-a x + 1\right )\right )} + \sqrt {d x} {\rm polylog}\left (3, a x\right )\right )}}{d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*(4*d*x*sqrt(a/d)*log((a*x + 2*sqrt(d*x)*sqrt(a/d) + 1)/(a*x - 1)) - 2*sqrt(d*x)*(dilog(a*x) - 2*log(-a*x +
1)) - sqrt(d*x)*polylog(3, a*x))/(d^2*x), -2*(8*d*x*sqrt(-a/d)*arctan(sqrt(d*x)*sqrt(-a/d)/(a*x)) + 2*sqrt(d*x
)*(dilog(a*x) - 2*log(-a*x + 1)) + sqrt(d*x)*polylog(3, a*x))/(d^2*x)]

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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 {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x)/(d*x)^(3/2),x)

[Out]

int(polylog(3, a*x)/(d*x)^(3/2), x)

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