3.1.59 \(\int (d x)^{3/2} \text {PolyLog}(2,a x) \, dx\) [59]

Optimal. Leaf size=117 \[ -\frac {8 d \sqrt {d x}}{25 a^2}-\frac {8 (d x)^{3/2}}{75 a}-\frac {8 (d x)^{5/2}}{125 d}+\frac {8 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/2}}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {PolyLog}(2,a x)}{5 d} \]

[Out]

-8/75*(d*x)^(3/2)/a-8/125*(d*x)^(5/2)/d+8/25*d^(3/2)*arctanh(a^(1/2)*(d*x)^(1/2)/d^(1/2))/a^(5/2)+4/25*(d*x)^(
5/2)*ln(-a*x+1)/d+2/5*(d*x)^(5/2)*polylog(2,a*x)/d-8/25*d*(d*x)^(1/2)/a^2

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Rubi [A]
time = 0.05, antiderivative size = 117, 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, 52, 65, 212} \begin {gather*} \frac {8 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/2}}-\frac {8 d \sqrt {d x}}{25 a^2}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}-\frac {8 (d x)^{3/2}}{75 a}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}-\frac {8 (d x)^{5/2}}{125 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[2, a*x],x]

[Out]

(-8*d*Sqrt[d*x])/(25*a^2) - (8*(d*x)^(3/2))/(75*a) - (8*(d*x)^(5/2))/(125*d) + (8*d^(3/2)*ArcTanh[(Sqrt[a]*Sqr
t[d*x])/Sqrt[d]])/(25*a^(5/2)) + (4*(d*x)^(5/2)*Log[1 - a*x])/(25*d) + (2*(d*x)^(5/2)*PolyLog[2, a*x])/(5*d)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 (d x)^{3/2} \text {Li}_2(a x) \, dx &=\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {2}{5} \int (d x)^{3/2} \log (1-a x) \, dx\\ &=\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {(4 a) \int \frac {(d x)^{5/2}}{1-a x} \, dx}{25 d}\\ &=-\frac {8 (d x)^{5/2}}{125 d}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {4}{25} \int \frac {(d x)^{3/2}}{1-a x} \, dx\\ &=-\frac {8 (d x)^{3/2}}{75 a}-\frac {8 (d x)^{5/2}}{125 d}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {(4 d) \int \frac {\sqrt {d x}}{1-a x} \, dx}{25 a}\\ &=-\frac {8 d \sqrt {d x}}{25 a^2}-\frac {8 (d x)^{3/2}}{75 a}-\frac {8 (d x)^{5/2}}{125 d}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {\left (4 d^2\right ) \int \frac {1}{\sqrt {d x} (1-a x)} \, dx}{25 a^2}\\ &=-\frac {8 d \sqrt {d x}}{25 a^2}-\frac {8 (d x)^{3/2}}{75 a}-\frac {8 (d x)^{5/2}}{125 d}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}+\frac {(8 d) \text {Subst}\left (\int \frac {1}{1-\frac {a x^2}{d}} \, dx,x,\sqrt {d x}\right )}{25 a^2}\\ &=-\frac {8 d \sqrt {d x}}{25 a^2}-\frac {8 (d x)^{3/2}}{75 a}-\frac {8 (d x)^{5/2}}{125 d}+\frac {8 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/2}}+\frac {4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2(a x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 90, normalized size = 0.77 \begin {gather*} \frac {2 (d x)^{3/2} \left (\frac {4 \tanh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{5 a^{5/2}}+\frac {2}{75} \sqrt {x} \left (-\frac {2 \left (15+5 a x+3 a^2 x^2\right )}{a^2}+15 x^2 \log (1-a x)\right )+x^{5/2} \text {PolyLog}(2,a x)\right )}{5 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*((4*ArcTanh[Sqrt[a]*Sqrt[x]])/(5*a^(5/2)) + (2*Sqrt[x]*((-2*(15 + 5*a*x + 3*a^2*x^2))/a^2 + 15*
x^2*Log[1 - a*x]))/75 + x^(5/2)*PolyLog[2, a*x]))/(5*x^(3/2))

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Maple [A]
time = 0.39, size = 101, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} \polylog \left (2, a x \right )}{5}+\frac {4 \left (d x \right )^{\frac {5}{2}} \ln \left (\frac {-a d x +d}{d}\right )}{25}+\frac {8 a \left (-\frac {\frac {\left (d x \right )^{\frac {5}{2}} a^{2}}{5}+\frac {d \left (d x \right )^{\frac {3}{2}} a}{3}+d^{2} \sqrt {d x}}{a^{3}}+\frac {d^{3} \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{a^{3} \sqrt {a d}}\right )}{25}}{d}\)	\(101\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} \polylog \left (2, a x \right )}{5}+\frac {4 \left (d x \right )^{\frac {5}{2}} \ln \left (\frac {-a d x +d}{d}\right )}{25}+\frac {8 a \left (-\frac {\frac {\left (d x \right )^{\frac {5}{2}} a^{2}}{5}+\frac {d \left (d x \right )^{\frac {3}{2}} a}{3}+d^{2} \sqrt {d x}}{a^{3}}+\frac {d^{3} \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{a^{3} \sqrt {a d}}\right )}{25}}{d}\)	\(101\)
meijerg	\(\frac {\left (d x \right )^{\frac {3}{2}} \left (-\frac {2 \sqrt {x}\, \left (-a \right )^{\frac {7}{2}} \left (84 a^{2} x^{2}+140 a x +420\right )}{2625 a^{3}}-\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {7}{2}} \left (\ln \left (1-\sqrt {a x}\right )-\ln \left (1+\sqrt {a x}\right )\right )}{25 a^{3} \sqrt {a x}}+\frac {4 x^{\frac {5}{2}} \left (-a \right )^{\frac {7}{2}} \ln \left (-a x +1\right )}{25 a}+\frac {2 x^{\frac {5}{2}} \left (-a \right )^{\frac {7}{2}} \polylog \left (2, a x \right )}{5 a}\right )}{x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2}} a}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 128, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (\frac {30 \, d^{3} \log \left (\frac {\sqrt {d x} a - \sqrt {a d}}{\sqrt {d x} a + \sqrt {a d}}\right )}{\sqrt {a d} a^{2}} - \frac {75 \, \left (d x\right )^{\frac {5}{2}} a^{2} {\rm Li}_2\left (a x\right ) + 30 \, \left (d x\right )^{\frac {5}{2}} a^{2} \log \left (-a d x + d\right ) - 6 \, {\left (5 \, a^{2} \log \left (d\right ) + 2 \, a^{2}\right )} \left (d x\right )^{\frac {5}{2}} - 20 \, \left (d x\right )^{\frac {3}{2}} a d - 60 \, \sqrt {d x} d^{2}}{a^{2}}\right )}}{375 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/375*(30*d^3*log((sqrt(d*x)*a - sqrt(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/(sqrt(a*d)*a^2) - (75*(d*x)^(5/2)*a^2*
dilog(a*x) + 30*(d*x)^(5/2)*a^2*log(-a*d*x + d) - 6*(5*a^2*log(d) + 2*a^2)*(d*x)^(5/2) - 20*(d*x)^(3/2)*a*d -
60*sqrt(d*x)*d^2)/a^2)/d

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Fricas [A]
time = 0.39, size = 190, normalized size = 1.62 \begin {gather*} \left [\frac {2 \, {\left (30 \, d \sqrt {\frac {d}{a}} \log \left (\frac {a d x + 2 \, \sqrt {d x} a \sqrt {\frac {d}{a}} + d}{a x - 1}\right ) + {\left (75 \, a^{2} d x^{2} {\rm Li}_2\left (a x\right ) + 30 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 12 \, a^{2} d x^{2} - 20 \, a d x - 60 \, d\right )} \sqrt {d x}\right )}}{375 \, a^{2}}, -\frac {2 \, {\left (60 \, d \sqrt {-\frac {d}{a}} \arctan \left (\frac {\sqrt {d x} a \sqrt {-\frac {d}{a}}}{d}\right ) - {\left (75 \, a^{2} d x^{2} {\rm Li}_2\left (a x\right ) + 30 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 12 \, a^{2} d x^{2} - 20 \, a d x - 60 \, d\right )} \sqrt {d x}\right )}}{375 \, a^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/375*(30*d*sqrt(d/a)*log((a*d*x + 2*sqrt(d*x)*a*sqrt(d/a) + d)/(a*x - 1)) + (75*a^2*d*x^2*dilog(a*x) + 30*a^
2*d*x^2*log(-a*x + 1) - 12*a^2*d*x^2 - 20*a*d*x - 60*d)*sqrt(d*x))/a^2, -2/375*(60*d*sqrt(-d/a)*arctan(sqrt(d*
x)*a*sqrt(-d/a)/d) - (75*a^2*d*x^2*dilog(a*x) + 30*a^2*d*x^2*log(-a*x + 1) - 12*a^2*d*x^2 - 20*a*d*x - 60*d)*s
qrt(d*x))/a^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((d*x)^(3/2)*dilog(a*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^{3/2}\,\mathrm {polylog}\left (2,a\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*polylog(2, a*x),x)

[Out]

int((d*x)^(3/2)*polylog(2, a*x), x)

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