3.1.64 \(\int \frac {\text {PolyLog}(2,a x)}{(d x)^{7/2}} \, dx\) [64]

Optimal. Leaf size=106 \[ -\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}+\frac {8 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {PolyLog}(2,a x)}{5 d (d x)^{5/2}} \]

[Out]

-8/75*a/d^2/(d*x)^(3/2)+8/25*a^(5/2)*arctanh(a^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)+4/25*ln(-a*x+1)/d/(d*x)^(5/2
)-2/5*polylog(2,a*x)/d/(d*x)^(5/2)-8/25*a^2/d^3/(d*x)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {8 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}-\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/(d*x)^(7/2),x]

[Out]

(-8*a)/(75*d^2*(d*x)^(3/2)) - (8*a^2)/(25*d^3*Sqrt[d*x]) + (8*a^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]])/(2
5*d^(7/2)) + (4*Log[1 - a*x])/(25*d*(d*x)^(5/2)) - (2*PolyLog[2, 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}_2(a x)}{(d x)^{7/2}} \, dx &=-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}-\frac {2}{5} \int \frac {\log (1-a x)}{(d x)^{7/2}} \, dx\\ &=\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}+\frac {(4 a) \int \frac {1}{(d x)^{5/2} (1-a x)} \, dx}{25 d}\\ &=-\frac {8 a}{75 d^2 (d x)^{3/2}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}+\frac {\left (4 a^2\right ) \int \frac {1}{(d x)^{3/2} (1-a x)} \, dx}{25 d^2}\\ &=-\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}+\frac {\left (4 a^3\right ) \int \frac {1}{\sqrt {d x} (1-a x)} \, dx}{25 d^3}\\ &=-\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}+\frac {\left (8 a^3\right ) \text {Subst}\left (\int \frac {1}{1-\frac {a x^2}{d}} \, dx,x,\sqrt {d x}\right )}{25 d^4}\\ &=-\frac {8 a}{75 d^2 (d x)^{3/2}}-\frac {8 a^2}{25 d^3 \sqrt {d x}}+\frac {8 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d x}}{\sqrt {d}}\right )}{25 d^{7/2}}+\frac {4 \log (1-a x)}{25 d (d x)^{5/2}}-\frac {2 \text {Li}_2(a x)}{5 d (d x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 65, normalized size = 0.61 \begin {gather*} -\frac {2 x \left (4 a x+12 a^2 x^2-12 a^{5/2} x^{5/2} \tanh ^{-1}\left (\sqrt {a} \sqrt {x}\right )-6 \log (1-a x)+15 \text {PolyLog}(2,a x)\right )}{75 (d x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(4*a*x + 12*a^2*x^2 - 12*a^(5/2)*x^(5/2)*ArcTanh[Sqrt[a]*Sqrt[x]] - 6*Log[1 - a*x] + 15*PolyLog[2, a*x])
)/(75*(d*x)^(7/2))

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Maple [A]
time = 0.37, size = 88, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {2 \polylog \left (2, a x \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {4 \ln \left (\frac {-a d x +d}{d}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {8 a \left (-\frac {1}{3 d \left (d x \right )^{\frac {3}{2}}}-\frac {a}{d^{2} \sqrt {d x}}+\frac {a^{2} \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{d^{2} \sqrt {a d}}\right )}{25}}{d}\)	\(88\)
default	\(\frac {-\frac {2 \polylog \left (2, a x \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {4 \ln \left (\frac {-a d x +d}{d}\right )}{25 \left (d x \right )^{\frac {5}{2}}}+\frac {8 a \left (-\frac {1}{3 d \left (d x \right )^{\frac {3}{2}}}-\frac {a}{d^{2} \sqrt {d x}}+\frac {a^{2} \arctanh \left (\frac {a \sqrt {d x}}{\sqrt {a d}}\right )}{d^{2} \sqrt {a d}}\right )}{25}}{d}\)	\(88\)
meijerg	\(\frac {x^{\frac {7}{2}} \left (-a \right )^{\frac {7}{2}} \left (-\frac {8}{75 x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2}}}-\frac {8 a}{25 \sqrt {x}\, \left (-a \right )^{\frac {3}{2}}}-\frac {4 \sqrt {x}\, a^{2} \left (\ln \left (1-\sqrt {a x}\right )-\ln \left (1+\sqrt {a x}\right )\right )}{25 \left (-a \right )^{\frac {3}{2}} \sqrt {a x}}+\frac {4 \ln \left (-a x +1\right )}{25 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \polylog \left (2, a x \right )}{5 x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{2}} a}\right )}{\left (d x \right )^{\frac {7}{2}} a}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 108, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (\frac {6 \, 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 {12 \, a^{2} d^{2} x^{2} + 4 \, a d^{2} x + 15 \, d^{2} {\rm Li}_2\left (a x\right ) - 6 \, d^{2} \log \left (-a d x + d\right ) + 6 \, d^{2} \log \left (d\right )}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{75 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/75*(6*a^3*log((sqrt(d*x)*a - sqrt(a*d))/(sqrt(d*x)*a + sqrt(a*d)))/(sqrt(a*d)*d^2) + (12*a^2*d^2*x^2 + 4*a*
d^2*x + 15*d^2*dilog(a*x) - 6*d^2*log(-a*d*x + d) + 6*d^2*log(d))/((d*x)^(5/2)*d^2))/d

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Fricas [A]
time = 0.39, size = 170, normalized size = 1.60 \begin {gather*} \left [\frac {2 \, {\left (6 \, 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 ) - {\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}\right )}}{75 \, d^{4} x^{3}}, -\frac {2 \, {\left (12 \, a^{2} d x^{3} \sqrt {-\frac {a}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {a}{d}}}{a x}\right ) + {\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}\right )}}{75 \, d^{4} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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)^(7/2),x, algorithm="giac")

[Out]

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

[Out]

int(polylog(2, a*x)/(d*x)^(7/2), x)

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