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

Optimal. Leaf size=145 \[ \frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-2 \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2+8 b n \text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {\log \left (\frac {f \sqrt {x}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-16 b^2 n^2 \text {Li}_4\left (-\frac {f \sqrt {x}}{e}\right ) \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2375, 2337, 2374, 2383, 6589} \[ -2 \text {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2+8 b n \text {PolyLog}\left (3,-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-16 b^2 n^2 \text {PolyLog}\left (4,-\frac {f \sqrt {x}}{e}\right )+\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {\log \left (\frac {f \sqrt {x}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[d*(e + f*Sqrt[x])]*(a + b*Log[c*x^n])^3)/(3*b*n) - (Log[1 + (f*Sqrt[x])/e]*(a + b*Log[c*x^n])^3)/(3*b*n)
- 2*(a + b*Log[c*x^n])^2*PolyLog[2, -((f*Sqrt[x])/e)] + 8*b*n*(a + b*Log[c*x^n])*PolyLog[3, -((f*Sqrt[x])/e)]
- 16*b^2*n^2*PolyLog[4, -((f*Sqrt[x])/e)]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

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 2375

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :
> Simp[(Log[d*(e + f*x^m)^r]*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] - Dist[(f*m*r)/(b*n*(p + 1)), Int[(
x^(m - 1)*(a + b*Log[c*x^n])^(p + 1))/(e + f*x^m), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p,
0] && NeQ[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 (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx &=\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {f \int \frac {\left (a+b \log \left (c x^n\right )\right )^3}{\left (e+f \sqrt {x}\right ) \sqrt {x}} \, dx}{6 b n}\\ &=\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {\log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}+\int \frac {\log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx\\ &=\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {\log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )+(4 b n) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )}{x} \, dx\\ &=\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {\log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )+8 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right )-\left (8 b^2 n^2\right ) \int \frac {\text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right )}{x} \, dx\\ &=\frac {\log \left (d \left (e+f \sqrt {x}\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {\log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )+8 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right )-16 b^2 n^2 \text {Li}_4\left (-\frac {f \sqrt {x}}{e}\right )\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 263, normalized size = 1.81 \[ \frac {1}{3} \left (\log (x) \log \left (d \left (e+f \sqrt {x}\right )\right ) \left (-3 b n \log (x) \left (a+b \log \left (c x^n\right )\right )+3 \left (a+b \log \left (c x^n\right )\right )^2+b^2 n^2 \log ^2(x)\right )-3 b n \left (-8 \text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right )+4 \log (x) \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )+\log ^2(x) \log \left (\frac {f \sqrt {x}}{e}+1\right )\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-3 \left (2 \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )+\log (x) \log \left (\frac {f \sqrt {x}}{e}+1\right )\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )^2-b^2 n^2 \left (48 \text {Li}_4\left (-\frac {f \sqrt {x}}{e}\right )+6 \log ^2(x) \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )-24 \log (x) \text {Li}_3\left (-\frac {f \sqrt {x}}{e}\right )+\log ^3(x) \log \left (\frac {f \sqrt {x}}{e}+1\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[d*(e + f*Sqrt[x])]*Log[x]*(b^2*n^2*Log[x]^2 - 3*b*n*Log[x]*(a + b*Log[c*x^n]) + 3*(a + b*Log[c*x^n])^2) -
3*(a - b*n*Log[x] + b*Log[c*x^n])^2*(Log[1 + (f*Sqrt[x])/e]*Log[x] + 2*PolyLog[2, -((f*Sqrt[x])/e)]) - 3*b*n*
(a - b*n*Log[x] + b*Log[c*x^n])*(Log[1 + (f*Sqrt[x])/e]*Log[x]^2 + 4*Log[x]*PolyLog[2, -((f*Sqrt[x])/e)] - 8*P
olyLog[3, -((f*Sqrt[x])/e)]) - b^2*n^2*(Log[1 + (f*Sqrt[x])/e]*Log[x]^3 + 6*Log[x]^2*PolyLog[2, -((f*Sqrt[x])/
e)] - 24*Log[x]*PolyLog[3, -((f*Sqrt[x])/e)] + 48*PolyLog[4, -((f*Sqrt[x])/e)]))/3

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \log \left (c x^{n}\right )^{2} + 2 \, a b \log \left (c x^{n}\right ) + a^{2}\right )} \log \left (d f \sqrt {x} + d e\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f \sqrt {x} + e\right )} d\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{2} \ln \left (\left (f \sqrt {x}+e \right ) d \right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f \sqrt {x} + e\right )} d\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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