3.4.72 \(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))^2}{x^3} \, dx\) [372]

Optimal. Leaf size=939 \[ \frac {b^2 e^2 m n^2 \log (x)}{d^2}-\frac {b^2 e^2 m n^2 \log ^2(x)}{2 d^2}+\frac {b^2 e^2 m n^2 \log \left (-\frac {e x}{d}\right )}{2 d^2}+\frac {b^2 e^2 n^2 \log (x) \log \left (f x^m\right )}{d^2}-\frac {3 b^2 e^2 m n^2 \log (d+e x)}{2 d^2}-\frac {3 b^2 e m n^2 \log (d+e x)}{2 d x}+\frac {b^2 e^2 m n^2 \log (x) \log (d+e x)}{d^2}+\frac {b^2 e^2 m n^2 \log ^2(x) \log (d+e x)}{2 d^2}-\frac {b^2 e^2 m n^2 \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{2 d^2}-\frac {b^2 e^2 n^2 \log \left (f x^m\right ) \log (d+e x)}{d^2}-\frac {b^2 e n^2 \log \left (f x^m\right ) \log (d+e x)}{d x}-\frac {b^2 e^2 n^2 \log (x) \log \left (f x^m\right ) \log (d+e x)}{d^2}+\frac {b^2 e^2 m n^2 \log ^2(d+e x)}{4 d^2}-\frac {b^2 m n^2 \log ^2(d+e x)}{4 x^2}-\frac {b^2 e^2 m n^2 \log \left (-\frac {e x}{d}\right ) \log ^2(d+e x)}{2 d^2}+\frac {b^2 e^2 n^2 \log \left (f x^m\right ) \log ^2(d+e x)}{2 d^2}-\frac {b^2 n^2 \log \left (f x^m\right ) \log ^2(d+e x)}{2 x^2}+\frac {b n \left (m \log (x)-\log \left (f x^m\right )\right ) \left (e^2 x^2 \log \left (-\frac {e x}{d}\right )+(d+e x) (e x+(d-e x) \log (d+e x))\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{d^2 x^2}-\frac {m \log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2}{2 x^2}-\frac {\left (m-2 m \log (x)+2 \log \left (f x^m\right )\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2}{4 x^2}-\frac {b^2 e^2 m n^2 \log (x) \log \left (1+\frac {e x}{d}\right )}{d^2}-\frac {b^2 e^2 m n^2 \log ^2(x) \log \left (1+\frac {e x}{d}\right )}{2 d^2}+\frac {b^2 e^2 n^2 \log (x) \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{d^2}-\frac {b^2 e^2 n^2 \left (m-\log \left (f x^m\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}-\frac {b m n \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (e x (d+e x)+e^2 x^2 \log \left (-\frac {e x}{d}\right )+\left (d^2-e^2 x^2\right ) \log (d+e x)+2 d^2 \log (x) \log (d+e x)+e x \left (e x \log ^2(x)+2 d (1+\log (x))-2 e x \left (\log (x) \log \left (1+\frac {e x}{d}\right )+\text {Li}_2\left (-\frac {e x}{d}\right )\right )\right )\right )}{2 d^2 x^2}-\frac {b^2 e^2 m n^2 (1+2 \log (d+e x)) \text {Li}_2\left (1+\frac {e x}{d}\right )}{2 d^2}-\frac {b^2 e^2 m n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{d^2}+\frac {b^2 e^2 m n^2 \text {Li}_3\left (1+\frac {e x}{d}\right )}{d^2} \]

[Out]

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

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Rubi [F]
time = 0.02, 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 = {} \begin {gather*} \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 {gather*}

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*}

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Mathematica [A]
time = 0.64, size = 781, normalized size = 0.83 \begin {gather*} \frac {4 b n \left (m \log (x)-\log \left (f x^m\right )\right ) \left (e^2 x^2 \log \left (-\frac {e x}{d}\right )+(d+e x) (e x+(d-e x) \log (d+e x))\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )-2 d^2 m \log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2+d^2 \left (-m+2 m \log (x)-2 \log \left (f x^m\right )\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2-2 b m n \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (e x (d+e x)+e^2 x^2 \log \left (-\frac {e x}{d}\right )+\left (d^2-e^2 x^2\right ) \log (d+e x)+2 d^2 \log (x) \log (d+e x)+e x \left (e x \log ^2(x)+2 d (1+\log (x))-2 e x \left (\log (x) \log \left (1+\frac {e x}{d}\right )+\text {Li}_2\left (-\frac {e x}{d}\right )\right )\right )\right )+b^2 n^2 \left (4 e^2 m x^2 \log (x)-2 e^2 m x^2 \log ^2(x)+2 e^2 m x^2 \log \left (-\frac {e x}{d}\right )+4 e^2 x^2 \log (x) \log \left (f x^m\right )-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 ^2(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 d e x \log \left (f x^m\right ) \log (d+e x)-4 e^2 x^2 \log \left (f x^m\right ) \log (d+e x)-4 e^2 x^2 \log \left (-\frac {e x}{d}\right ) \log \left (f x^m\right ) \log (d+e x)-d^2 m \log ^2(d+e x)+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 d^2 \log \left (f x^m\right ) \log ^2(d+e x)+2 e^2 x^2 \log \left (f x^m\right ) \log ^2(d+e x)-4 e^2 m x^2 \log (x) \log \left (1+\frac {e x}{d}\right )+2 e^2 m x^2 \log ^2(x) \log \left (1+\frac {e x}{d}\right )+4 e^2 m x^2 (-1+\log (x)) \text {Li}_2\left (-\frac {e x}{d}\right )-2 e^2 x^2 \left (m-2 m \log (x)+2 \log \left (f x^m\right )+2 m \log (d+e x)\right ) \text {Li}_2\left (1+\frac {e x}{d}\right )-4 e^2 m x^2 \text {Li}_3\left (-\frac {e x}{d}\right )+4 e^2 m x^2 \text {Li}_3\left (1+\frac {e x}{d}\right )\right )}{4 d^2 x^2} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (f \,x^{m}\right ) \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )^{2}}{x^{3}}\, dx\]

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 = 0.00, 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(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((x*e + 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*log(c)^2*log(f) + 2*a*b*log(c)*log(f) + a^2*log(f))*x*e + (4*b^
2*d*log(c)*log(f) + 4*a*b*d*log(f) + ((m*n + 2*n*log(f) + 4*log(c)*log(f))*b^2 + 4*a*b*log(f))*x*e + 2*(2*b^2*
d*log(c) + 2*a*b*d + (b^2*(n + 2*log(c)) + 2*a*b)*x*e)*log(x^m))*log((x*e + d)^n) + 2*(b^2*d*log(c)^2 + 2*a*b*
d*log(c) + a^2*d + (b^2*log(c)^2 + 2*a*b*log(c) + a^2)*x*e)*log(x^m))/(x^4*e + d*x^3), 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(log(f*x^m)*(a+b*log(c*(e*x+d)^n))^2/x^3,x, algorithm="fricas")

[Out]

integral((b^2*log((x*e + d)^n*c)^2*log(f*x^m) + 2*a*b*log((x*e + d)^n*c)*log(f*x^m) + a^2*log(f*x^m))/x^3, 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(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 = 0.00, 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(log(f*x^m)*(a+b*log(c*(e*x+d)^n))^2/x^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (f\,x^m\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{x^3} \,d x \end {gather*}

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^3,x)

[Out]

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

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