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

Optimal. Leaf size=607 \[ -\frac {2 b n \left (e x \log \left (-\frac {e x}{d}\right )-(d+e x) \log (d+e x)\right ) \left (m \log (x)-\log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{d x}-\frac {\left (\log \left (f x^m\right )+m (-\log (x))+m\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2}{x}+\frac {b m n \left (e x \left (\log ^2(x)-2 \left (\text {Li}_2\left (-\frac {e x}{d}\right )+\log (x) \log \left (\frac {e x}{d}+1\right )\right )\right )+2 e x \log \left (-\frac {e x}{d}\right )-2 (d+e x) \log (d+e x)-2 d \log (x) \log (d+e x)\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{d x}-\frac {m \log (x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2}{x}-\frac {2 b^2 e n^2 \text {Li}_2\left (-\frac {e x}{d}\right ) \log \left (f x^m\right )}{d}-\frac {b^2 e n^2 \log ^2(d+e x) \log \left (f x^m\right )}{d}-\frac {b^2 n^2 \log ^2(d+e x) \log \left (f x^m\right )}{x}+\frac {2 b^2 e n^2 \log (x) \log (d+e x) \log \left (f x^m\right )}{d}-\frac {2 b^2 e n^2 \log (x) \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{d}+\frac {2 b^2 e m n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{d}-\frac {2 b^2 e m n^2 \text {Li}_3\left (\frac {e x}{d}+1\right )}{d}+\frac {2 b^2 e m n^2 \text {Li}_2\left (\frac {e x}{d}+1\right ) (\log (d+e x)+1)}{d}-\frac {b^2 e m n^2 \log ^2(d+e x)}{d}+\frac {b^2 e m n^2 \log \left (-\frac {e x}{d}\right ) \log ^2(d+e x)}{d}-\frac {b^2 m n^2 \log ^2(d+e x)}{x}-\frac {b^2 e m n^2 \log ^2(x) \log (d+e x)}{d}+\frac {b^2 e m n^2 \log ^2(x) \log \left (\frac {e x}{d}+1\right )}{d}+\frac {2 b^2 e m n^2 \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d} \]

[Out]

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

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Rubi [F]  time = 0.03, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x^2} \, dx &=\int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 513, normalized size = 0.85 \[ \frac {2 b n \left ((d+e x) \log (d+e x)-e x \log \left (-\frac {e x}{d}\right )\right ) \left (m \log (x)-\log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+d \left (-\log \left (f x^m\right )+m \log (x)-m\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2-b m n \left (-e x \left (\log ^2(x)-2 \left (\text {Li}_2\left (-\frac {e x}{d}\right )+\log (x) \log \left (\frac {e x}{d}+1\right )\right )\right )-2 e x \log \left (-\frac {e x}{d}\right )+2 (d+e x) \log (d+e x)+2 d \log (x) \log (d+e x)\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )-d m \log (x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+b^2 n^2 \left (2 e x \text {Li}_2\left (\frac {e x}{d}+1\right ) \left (m \log (d+e x)+\log \left (f x^m\right )+m (-\log (x))+m\right )-d \log ^2(d+e x) \log \left (f x^m\right )-e x \log ^2(d+e x) \log \left (f x^m\right )+2 e x \log \left (-\frac {e x}{d}\right ) \log (d+e x) \log \left (f x^m\right )+2 e m x \text {Li}_3\left (-\frac {e x}{d}\right )-2 e m x \text {Li}_3\left (\frac {e x}{d}+1\right )-2 e m x \log (x) \text {Li}_2\left (-\frac {e x}{d}\right )+e m x \log ^2(x) \log (d+e x)-e m x \log ^2(x) \log \left (\frac {e x}{d}+1\right )-d m \log ^2(d+e x)-e m x \log ^2(d+e x)+e m x \log \left (-\frac {e x}{d}\right ) \log ^2(d+e x)-2 e m x \log (x) \log \left (-\frac {e x}{d}\right ) \log (d+e x)+2 e m x \log \left (-\frac {e x}{d}\right ) \log (d+e x)\right )}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*n*(m*Log[x] - Log[f*x^m])*(-(e*x*Log[-((e*x)/d)]) + (d + e*x)*Log[d + e*x])*(a - b*n*Log[d + e*x] + b*Log
[c*(d + e*x)^n]) - d*m*Log[x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 + d*(-m + m*Log[x] - Log[f*x^m])
*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - b*m*n*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*(-2*e*x
*Log[-((e*x)/d)] + 2*(d + e*x)*Log[d + e*x] + 2*d*Log[x]*Log[d + e*x] - e*x*(Log[x]^2 - 2*(Log[x]*Log[1 + (e*x
)/d] + PolyLog[2, -((e*x)/d)]))) + b^2*n^2*(e*m*x*Log[x]^2*Log[d + e*x] + 2*e*m*x*Log[-((e*x)/d)]*Log[d + e*x]
- 2*e*m*x*Log[x]*Log[-((e*x)/d)]*Log[d + e*x] + 2*e*x*Log[-((e*x)/d)]*Log[f*x^m]*Log[d + e*x] - d*m*Log[d + e
*x]^2 - e*m*x*Log[d + e*x]^2 + e*m*x*Log[-((e*x)/d)]*Log[d + e*x]^2 - d*Log[f*x^m]*Log[d + e*x]^2 - e*x*Log[f*
x^m]*Log[d + e*x]^2 - e*m*x*Log[x]^2*Log[1 + (e*x)/d] - 2*e*m*x*Log[x]*PolyLog[2, -((e*x)/d)] + 2*e*x*(m - m*L
og[x] + Log[f*x^m] + m*Log[d + e*x])*PolyLog[2, 1 + (e*x)/d] + 2*e*m*x*PolyLog[3, -((e*x)/d)] - 2*e*m*x*PolyLo
g[3, 1 + (e*x)/d]))/(d*x)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} \log \left (f x^{m}\right ) + 2 \, a b \log \left ({\left (e x + d\right )}^{n} c\right ) \log \left (f x^{m}\right ) + a^{2} \log \left (f x^{m}\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.73, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2} \ln \left (f \,x^{m}\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (b^{2} {\left (m + \log \relax (f)\right )} + b^{2} \log \left (x^{m}\right )\right )} \log \left ({\left (e x + d\right )}^{n}\right )^{2}}{x} + \int \frac {b^{2} d \log \relax (c)^{2} \log \relax (f) + 2 \, a b d \log \relax (c) \log \relax (f) + a^{2} d \log \relax (f) + {\left (b^{2} e \log \relax (c)^{2} \log \relax (f) + 2 \, a b e \log \relax (c) \log \relax (f) + a^{2} e \log \relax (f)\right )} x + 2 \, {\left (b^{2} d \log \relax (c) \log \relax (f) + a b d \log \relax (f) + {\left (a b e \log \relax (f) + {\left (e \log \relax (c) \log \relax (f) + {\left (m n + n \log \relax (f)\right )} e\right )} b^{2}\right )} x + {\left (b^{2} d \log \relax (c) + a b d + {\left ({\left (e n + e \log \relax (c)\right )} b^{2} + a b e\right )} x\right )} \log \left (x^{m}\right )\right )} \log \left ({\left (e x + d\right )}^{n}\right ) + {\left (b^{2} d \log \relax (c)^{2} + 2 \, a b d \log \relax (c) + a^{2} d + {\left (b^{2} e \log \relax (c)^{2} + 2 \, a b e \log \relax (c) + a^{2} e\right )} x\right )} \log \left (x^{m}\right )}{e x^{3} + d x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*(m + log(f)) + b^2*log(x^m))*log((e*x + d)^n)^2/x + integrate((b^2*d*log(c)^2*log(f) + 2*a*b*d*log(c)*lo
g(f) + a^2*d*log(f) + (b^2*e*log(c)^2*log(f) + 2*a*b*e*log(c)*log(f) + a^2*e*log(f))*x + 2*(b^2*d*log(c)*log(f
) + a*b*d*log(f) + (a*b*e*log(f) + (e*log(c)*log(f) + (m*n + n*log(f))*e)*b^2)*x + (b^2*d*log(c) + a*b*d + ((e
*n + e*log(c))*b^2 + a*b*e)*x)*log(x^m))*log((e*x + d)^n) + (b^2*d*log(c)^2 + 2*a*b*d*log(c) + a^2*d + (b^2*e*
log(c)^2 + 2*a*b*e*log(c) + a^2*e)*x)*log(x^m))/(e*x^3 + d*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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