3.372 \(\int \frac{\log (f x^m) (a+b \log (c (d+e x)^n))^2}{x^3} \, dx\)

Optimal. Leaf size=939 \[ \text{result too large to display} \]

[Out]

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

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Rubi [F]  time = 0.0295654, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n])^2)/x^3,x]

[Out]

Defer[Int][(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n])^2)/x^3, 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^3} \, dx &=\int \frac{\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x^3} \, dx\\ \end{align*}

Mathematica [A]  time = 1.17526, size = 781, normalized size = 0.83 \[ \frac{-2 b m n \left (e x \left (-2 e x \left (\text{PolyLog}\left (2,-\frac{e x}{d}\right )+\log (x) \log \left (\frac{e x}{d}+1\right )\right )+2 d (\log (x)+1)+e x \log ^2(x)\right )+\left (d^2-e^2 x^2\right ) \log (d+e x)+2 d^2 \log (x) \log (d+e x)+e^2 x^2 \log \left (-\frac{e x}{d}\right )+e x (d+e x)\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+b^2 n^2 \left (-2 e^2 x^2 \text{PolyLog}\left (2,\frac{e x}{d}+1\right ) \left (2 m \log (d+e x)+2 \log \left (f x^m\right )-2 m \log (x)+m\right )-4 e^2 m x^2 \text{PolyLog}\left (3,-\frac{e x}{d}\right )+4 e^2 m x^2 \text{PolyLog}\left (3,\frac{e x}{d}+1\right )+4 e^2 m x^2 (\log (x)-1) \text{PolyLog}\left (2,-\frac{e x}{d}\right )-2 d^2 \log ^2(d+e x) \log \left (f x^m\right )-d^2 m \log ^2(d+e x)+2 e^2 x^2 \log ^2(d+e x) \log \left (f x^m\right )-4 e^2 x^2 \log (d+e x) \log \left (f x^m\right )-4 e^2 x^2 \log \left (-\frac{e x}{d}\right ) \log (d+e x) \log \left (f x^m\right )+e^2 m x^2 \log ^2(d+e x)-2 e^2 m x^2 \log \left (-\frac{e x}{d}\right ) \log ^2(d+e x)-2 e^2 m x^2 \log ^2(x) \log (d+e x)+2 e^2 m x^2 \log ^2(x) \log \left (\frac{e x}{d}+1\right )+2 e^2 m x^2 \log \left (-\frac{e x}{d}\right )-6 e^2 m x^2 \log (d+e x)+4 e^2 m x^2 \log (x) \log (d+e x)-2 e^2 m x^2 \log \left (-\frac{e x}{d}\right ) \log (d+e x)+4 e^2 m x^2 \log (x) \log \left (-\frac{e x}{d}\right ) \log (d+e x)-4 e^2 m x^2 \log (x) \log \left (\frac{e x}{d}+1\right )-4 d e x \log (d+e x) \log \left (f x^m\right )-6 d e m x \log (d+e x)+4 e^2 x^2 \log (x) \log \left (f x^m\right )-2 e^2 m x^2 \log ^2(x)+4 e^2 m x^2 \log (x)\right )+d^2 \left (-2 \log \left (f x^m\right )+2 m \log (x)-m\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2-2 d^2 m \log (x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+4 b n \left (e^2 x^2 \log \left (-\frac{e x}{d}\right )+(d+e x) ((d-e x) \log (d+e x)+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 )}{4 d^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n])^2)/x^3,x]

[Out]

(4*b*n*(m*Log[x] - Log[f*x^m])*(e^2*x^2*Log[-((e*x)/d)] + (d + e*x)*(e*x + (d - e*x)*Log[d + e*x]))*(a - b*n*L
og[d + e*x] + b*Log[c*(d + e*x)^n]) - 2*d^2*m*Log[x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 + d^2*(-m
+ 2*m*Log[x] - 2*Log[f*x^m])*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - 2*b*m*n*(a - b*n*Log[d + e*x]
+ b*Log[c*(d + e*x)^n])*(e*x*(d + e*x) + e^2*x^2*Log[-((e*x)/d)] + (d^2 - e^2*x^2)*Log[d + e*x] + 2*d^2*Log[x]
*Log[d + e*x] + e*x*(e*x*Log[x]^2 + 2*d*(1 + Log[x]) - 2*e*x*(Log[x]*Log[1 + (e*x)/d] + PolyLog[2, -((e*x)/d)]
))) + b^2*n^2*(4*e^2*m*x^2*Log[x] - 2*e^2*m*x^2*Log[x]^2 + 2*e^2*m*x^2*Log[-((e*x)/d)] + 4*e^2*x^2*Log[x]*Log[
f*x^m] - 6*d*e*m*x*Log[d + e*x] - 6*e^2*m*x^2*Log[d + e*x] + 4*e^2*m*x^2*Log[x]*Log[d + e*x] - 2*e^2*m*x^2*Log
[x]^2*Log[d + e*x] - 2*e^2*m*x^2*Log[-((e*x)/d)]*Log[d + e*x] + 4*e^2*m*x^2*Log[x]*Log[-((e*x)/d)]*Log[d + e*x
] - 4*d*e*x*Log[f*x^m]*Log[d + e*x] - 4*e^2*x^2*Log[f*x^m]*Log[d + e*x] - 4*e^2*x^2*Log[-((e*x)/d)]*Log[f*x^m]
*Log[d + e*x] - d^2*m*Log[d + e*x]^2 + e^2*m*x^2*Log[d + e*x]^2 - 2*e^2*m*x^2*Log[-((e*x)/d)]*Log[d + e*x]^2 -
2*d^2*Log[f*x^m]*Log[d + e*x]^2 + 2*e^2*x^2*Log[f*x^m]*Log[d + e*x]^2 - 4*e^2*m*x^2*Log[x]*Log[1 + (e*x)/d] +
2*e^2*m*x^2*Log[x]^2*Log[1 + (e*x)/d] + 4*e^2*m*x^2*(-1 + Log[x])*PolyLog[2, -((e*x)/d)] - 2*e^2*x^2*(m - 2*m
*Log[x] + 2*Log[f*x^m] + 2*m*Log[d + e*x])*PolyLog[2, 1 + (e*x)/d] - 4*e^2*m*x^2*PolyLog[3, -((e*x)/d)] + 4*e^
2*m*x^2*PolyLog[3, 1 + (e*x)/d]))/(4*d^2*x^2)

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Maple [F]  time = 1.454, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( f{x}^{m} \right ) \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (b^{2}{\left (m + 2 \, \log \left (f\right )\right )} + 2 \, b^{2} \log \left (x^{m}\right )\right )} \log \left ({\left (e x + d\right )}^{n}\right )^{2}}{4 \, x^{2}} + \int \frac{2 \, b^{2} d \log \left (c\right )^{2} \log \left (f\right ) + 4 \, a b d \log \left (c\right ) \log \left (f\right ) + 2 \, a^{2} d \log \left (f\right ) + 2 \,{\left (b^{2} e \log \left (c\right )^{2} \log \left (f\right ) + 2 \, a b e \log \left (c\right ) \log \left (f\right ) + a^{2} e \log \left (f\right )\right )} x +{\left (4 \, b^{2} d \log \left (c\right ) \log \left (f\right ) + 4 \, a b d \log \left (f\right ) +{\left (4 \, a b e \log \left (f\right ) +{\left (4 \, e \log \left (c\right ) \log \left (f\right ) +{\left (m n + 2 \, n \log \left (f\right )\right )} e\right )} b^{2}\right )} x + 2 \,{\left (2 \, b^{2} d \log \left (c\right ) + 2 \, a b d +{\left ({\left (e n + 2 \, e \log \left (c\right )\right )} b^{2} + 2 \, a b e\right )} x\right )} \log \left (x^{m}\right )\right )} \log \left ({\left (e x + d\right )}^{n}\right ) + 2 \,{\left (b^{2} d \log \left (c\right )^{2} + 2 \, a b d \log \left (c\right ) + a^{2} d +{\left (b^{2} e \log \left (c\right )^{2} + 2 \, a b e \log \left (c\right ) + a^{2} e\right )} x\right )} \log \left (x^{m}\right )}{2 \,{\left (e x^{4} + d x^{3}\right )}}\,{d x} \end{align*}

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^3,x, algorithm="maxima")

[Out]

-1/4*(b^2*(m + 2*log(f)) + 2*b^2*log(x^m))*log((e*x + d)^n)^2/x^2 + integrate(1/2*(2*b^2*d*log(c)^2*log(f) + 4
*a*b*d*log(c)*log(f) + 2*a^2*d*log(f) + 2*(b^2*e*log(c)^2*log(f) + 2*a*b*e*log(c)*log(f) + a^2*e*log(f))*x + (
4*b^2*d*log(c)*log(f) + 4*a*b*d*log(f) + (4*a*b*e*log(f) + (4*e*log(c)*log(f) + (m*n + 2*n*log(f))*e)*b^2)*x +
2*(2*b^2*d*log(c) + 2*a*b*d + ((e*n + 2*e*log(c))*b^2 + 2*a*b*e)*x)*log(x^m))*log((e*x + d)^n) + 2*(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^4 + d*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3}}, x\right ) \end{align*}

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^3,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^3, 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(ln(f*x**m)*(a+b*ln(c*(e*x+d)**n))**2/x**3,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 ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} \log \left (f x^{m}\right )}{x^{3}}\,{d x} \end{align*}

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^3,x, algorithm="giac")

[Out]

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