3.1.56 \(\int \frac {\log (d (\frac {1}{d}+f \sqrt {x})) (a+b \log (c x^n))^2}{x} \, dx\) [56]

Optimal. Leaf size=70 \[ -2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-d f \sqrt {x}\right )+8 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-d f \sqrt {x}\right )-16 b^2 n^2 \text {Li}_4\left (-d f \sqrt {x}\right ) \]

[Out]

-2*(a+b*ln(c*x^n))^2*polylog(2,-d*f*x^(1/2))+8*b*n*(a+b*ln(c*x^n))*polylog(3,-d*f*x^(1/2))-16*b^2*n^2*polylog(
4,-d*f*x^(1/2))

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Rubi [A]
time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2421, 2430, 6724} \begin {gather*} 8 b n \text {PolyLog}\left (3,-d f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )-2 \text {PolyLog}\left (2,-d f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )^2-16 b^2 n^2 \text {PolyLog}\left (4,-d f \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[d*(d^(-1) + f*Sqrt[x])]*(a + b*Log[c*x^n])^2)/x,x]

[Out]

-2*(a + b*Log[c*x^n])^2*PolyLog[2, -(d*f*Sqrt[x])] + 8*b*n*(a + b*Log[c*x^n])*PolyLog[3, -(d*f*Sqrt[x])] - 16*
b^2*n^2*PolyLog[4, -(d*f*Sqrt[x])]

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2430

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[PolyLo
g[k + 1, e*x^q]*((a + b*Log[c*x^n])^p/q), x] - Dist[b*n*(p/q), Int[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^(
p - 1)/x), x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\log \left (d \left (\frac {1}{d}+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx &=-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-d f \sqrt {x}\right )+(4 b n) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-d f \sqrt {x}\right )}{x} \, dx\\ &=-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-d f \sqrt {x}\right )+8 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-d f \sqrt {x}\right )-\left (8 b^2 n^2\right ) \int \frac {\text {Li}_3\left (-d f \sqrt {x}\right )}{x} \, dx\\ &=-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-d f \sqrt {x}\right )+8 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-d f \sqrt {x}\right )-16 b^2 n^2 \text {Li}_4\left (-d f \sqrt {x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 70, normalized size = 1.00 \begin {gather*} -2 \left (\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-d f \sqrt {x}\right )+4 b n \left (-\left (\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-d f \sqrt {x}\right )\right )+2 b n \text {Li}_4\left (-d f \sqrt {x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[d*(d^(-1) + f*Sqrt[x])]*(a + b*Log[c*x^n])^2)/x,x]

[Out]

-2*((a + b*Log[c*x^n])^2*PolyLog[2, -(d*f*Sqrt[x])] + 4*b*n*(-((a + b*Log[c*x^n])*PolyLog[3, -(d*f*Sqrt[x])])
+ 2*b*n*PolyLog[4, -(d*f*Sqrt[x])]))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right )^{2} \ln \left (d \left (\frac {1}{d}+f \sqrt {x}\right )\right )}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))^2*ln(d*(1/d+f*x^(1/2)))/x,x)

[Out]

int((a+b*ln(c*x^n))^2*ln(d*(1/d+f*x^(1/2)))/x,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(d*(1/d+f*x^(1/2)))/x,x, algorithm="maxima")

[Out]

integrate((b*log(c*x^n) + a)^2*log((f*sqrt(x) + 1/d)*d)/x, x)

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Fricas [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((a+b*log(c*x^n))^2*log(d*(1/d+f*x^(1/2)))/x,x, algorithm="fricas")

[Out]

integral((b^2*log(c*x^n)^2 + 2*a*b*log(c*x^n) + a^2)*log(d*f*sqrt(x) + 1)/x, x)

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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((a+b*ln(c*x**n))**2*ln(d*(1/d+f*x**(1/2)))/x,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((a+b*log(c*x^n))^2*log(d*(1/d+f*x^(1/2)))/x,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)^2*log((f*sqrt(x) + 1/d)*d)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (d\,\left (f\,\sqrt {x}+\frac {1}{d}\right )\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(d*(f*x^(1/2) + 1/d))*(a + b*log(c*x^n))^2)/x,x)

[Out]

int((log(d*(f*x^(1/2) + 1/d))*(a + b*log(c*x^n))^2)/x, x)

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