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

Optimal. Leaf size=70 \[ 8 b n \text{PolyLog}\left (3,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )-2 \text{PolyLog}\left (2,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )^2-16 b^2 n^2 \text{PolyLog}\left (4,-d f \sqrt{x}\right ) \]

[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])]

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Rubi [A]  time = 0.0670089, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2374, 2383, 6589} \[ 8 b n \text{PolyLog}\left (3,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )-2 \text{PolyLog}\left (2,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )^2-16 b^2 n^2 \text{PolyLog}\left (4,-d f \sqrt{x}\right ) \]

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 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(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*Log
[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 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[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 6589

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*}

Mathematica [A]  time = 0.14004, size = 70, normalized size = 1. \[ -2 \left (\text{PolyLog}\left (2,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )^2+4 b n \left (2 b n \text{PolyLog}\left (4,-d f \sqrt{x}\right )-\text{PolyLog}\left (3,-d f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )\right )\right ) \]

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. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{2}}{x}\ln \left ( d \left ({d}^{-1}+f\sqrt{x} \right ) \right ) }\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f \sqrt{x} + \frac{1}{d}\right )} d\right )}{x}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \log \left (c x^{n}\right )^{2} + 2 \, a b \log \left (c x^{n}\right ) + a^{2}\right )} \log \left (d f \sqrt{x} + 1\right )}{x}, x\right ) \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f \sqrt{x} + \frac{1}{d}\right )} d\right )}{x}\,{d x} \end{align*}

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)