3.1.61 \(\int \frac {\text {PolyLog}(2,a x)}{\sqrt {d x}} \, dx\) [61]

Optimal. Leaf size=80 \[ -\frac {8 \sqrt {d x}}{d}+\frac {8 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {a} \sqrt {d}}+\frac {4 \sqrt {d x} \log (1-a x)}{d}+\frac {2 \sqrt {d x} \text {PolyLog}(2,a x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {2 \sqrt {d x} \text {Li}_2(a x)}{d}+\frac {4 \sqrt {d x} \log (1-a x)}{d}+\frac {8 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {a} \sqrt {d}}-\frac {8 \sqrt {d x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/Sqrt[d*x],x]

[Out]

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

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Mathematica [A]
time = 0.06, size = 63, normalized size = 0.79 \begin {gather*} \frac {8 \sqrt {x} \tanh ^{-1}\left (\sqrt {a} \sqrt {x}\right )+4 \sqrt {a} x (-2+\log (1-a x))+2 \sqrt {a} x \text {PolyLog}(2,a x)}{\sqrt {a} \sqrt {d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/Sqrt[d*x],x]

[Out]

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

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Maple [A]
time = 0.42, size = 74, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {2 \sqrt {d x}\, \polylog \left (2, a x \right )+4 \sqrt {d x}\, \ln \left (\frac {-a d x +d}{d}\right )+8 a \left (-\frac {\sqrt {d x}}{a}+\frac {d \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{a \sqrt {a d}}\right )}{d}\)	\(74\)
default	\(\frac {2 \sqrt {d x}\, \polylog \left (2, a x \right )+4 \sqrt {d x}\, \ln \left (\frac {-a d x +d}{d}\right )+8 a \left (-\frac {\sqrt {d x}}{a}+\frac {d \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{a \sqrt {a d}}\right )}{d}\)	\(74\)
meijerg	\(\frac {\sqrt {x}\, \sqrt {-a}\, \left (-\frac {8 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}}}{a}-\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}} \left (\ln \left (1-\sqrt {a x}\right )-\ln \left (1+\sqrt {a x}\right )\right )}{a \sqrt {a x}}+\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}} \ln \left (-a x +1\right )}{a}+\frac {2 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}} \polylog \left (2, a x \right )}{a}\right )}{\sqrt {d x}\, a}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x)/(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.46, size = 83, normalized size = 1.04 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {d x} {\left (\log \left (d\right ) + 2\right )} - \sqrt {d x} {\rm Li}_2\left (a x\right ) - 2 \, \sqrt {d x} \log \left (-a d x + d\right ) + \frac {2 \, d \log \left (\frac {\sqrt {d x} a - \sqrt {a d}}{\sqrt {d x} a + \sqrt {a d}}\right )}{\sqrt {a d}}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/(d*x)^(1/2),x, algorithm="maxima")

[Out]

-2*(2*sqrt(d*x)*(log(d) + 2) - sqrt(d*x)*dilog(a*x) - 2*sqrt(d*x)*log(-a*d*x + d) + 2*d*log((sqrt(d*x)*a - sqr
t(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/sqrt(a*d))/d

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Fricas [A]
time = 0.41, size = 135, normalized size = 1.69 \begin {gather*} \left [\frac {2 \, {\left (\sqrt {d x} {\left (a {\rm Li}_2\left (a x\right ) + 2 \, a \log \left (-a x + 1\right ) - 4 \, a\right )} + 2 \, \sqrt {a d} \log \left (\frac {a d x + 2 \, \sqrt {a d} \sqrt {d x} + d}{a x - 1}\right )\right )}}{a d}, \frac {2 \, {\left (\sqrt {d x} {\left (a {\rm Li}_2\left (a x\right ) + 2 \, a \log \left (-a x + 1\right ) - 4 \, a\right )} - 4 \, \sqrt {-a d} \arctan \left (\frac {\sqrt {-a d} \sqrt {d x}}{a d x}\right )\right )}}{a d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/(d*x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/(d*x)**(1/2),x)

[Out]

Integral(polylog(2, a*x)/sqrt(d*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(polylog(2,a*x)/(d*x)^(1/2),x, algorithm="giac")

[Out]

integrate(dilog(a*x)/sqrt(d*x), 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\right )}{\sqrt {d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/(d*x)^(1/2),x)

[Out]

int(polylog(2, a*x)/(d*x)^(1/2), x)

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