3.364 \(\int \frac{(f x)^{-1+m} (a+b \log (c x^n))^2}{(d+e x^m)^2} \, dx\)

Optimal. Leaf size=138 \[ \frac{2 b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{d x^{-m}}{e}\right )}{d e m^3}-\frac{2 b n x^{1-m} (f x)^{m-1} \log \left (\frac{d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d e m^2}-\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )} \]

[Out]

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

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Rubi [A]  time = 0.340103, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2339, 2338, 2345, 2391} \[ \frac{2 b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{d x^{-m}}{e}\right )}{d e m^3}-\frac{2 b n x^{1-m} (f x)^{m-1} \log \left (\frac{d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d e m^2}-\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^(-1 + m)*(a + b*Log[c*x^n])^2)/(d + e*x^m)^2,x]

[Out]

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

Rule 2339

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

Rule 2338

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

Rule 2345

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> -Simp[(Log[1 +
d/(e*x^r)]*(a + b*Log[c*x^n])^p)/(d*r), x] + Dist[(b*n*p)/(d*r), Int[(Log[1 + d/(e*x^r)]*(a + b*Log[c*x^n])^(p
- 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx &=\left (x^{1-m} (f x)^{-1+m}\right ) \int \frac{x^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}+\frac{\left (2 b n x^{1-m} (f x)^{-1+m}\right ) \int \frac{a+b \log \left (c x^n\right )}{x \left (d+e x^m\right )} \, dx}{e m}\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac{2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{d x^{-m}}{e}\right )}{d e m^2}+\frac{\left (2 b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac{\log \left (1+\frac{d x^{-m}}{e}\right )}{x} \, dx}{d e m^2}\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac{2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{d x^{-m}}{e}\right )}{d e m^2}+\frac{2 b^2 n^2 x^{1-m} (f x)^{-1+m} \text{Li}_2\left (-\frac{d x^{-m}}{e}\right )}{d e m^3}\\ \end{align*}

Mathematica [A]  time = 0.445255, size = 157, normalized size = 1.14 \[ \frac{x^{-m} (f x)^m \left (\frac{2 b^2 n^2 \left (\text{PolyLog}\left (2,\frac{e x^m}{d}+1\right )+\left (\log \left (-\frac{e x^m}{d}\right )-m \log (x)\right ) \log \left (d+e x^m\right )+\frac{1}{2} m^2 \log ^2(x)\right )}{d}-\frac{m^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m}-\frac{2 a b m n \log \left (d-d x^m\right )}{d}+\frac{2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d}\right )}{e f m^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((f*x)^(-1 + m)*(a + b*Log[c*x^n])^2)/(d + e*x^m)^2,x]

[Out]

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

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Maple [F]  time = 0.911, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{-1+m} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{2}}{ \left ( d+e{x}^{m} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.35616, size = 618, normalized size = 4.48 \begin{align*} \frac{{\left (b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \,{\left (b^{2} e m^{2} n \log \left (c\right ) + a b e m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} x^{m} -{\left (b^{2} d m^{2} \log \left (c\right )^{2} + 2 \, a b d m^{2} \log \left (c\right ) + a^{2} d m^{2}\right )} f^{m - 1} - 2 \,{\left (b^{2} e f^{m - 1} n^{2} x^{m} + b^{2} d f^{m - 1} n^{2}\right )}{\rm Li}_2\left (-\frac{e x^{m} + d}{d} + 1\right ) - 2 \,{\left ({\left (b^{2} e m n \log \left (c\right ) + a b e m n\right )} f^{m - 1} x^{m} +{\left (b^{2} d m n \log \left (c\right ) + a b d m n\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \,{\left (b^{2} e f^{m - 1} m n^{2} x^{m} \log \left (x\right ) + b^{2} d f^{m - 1} m n^{2} \log \left (x\right )\right )} \log \left (\frac{e x^{m} + d}{d}\right )}{d e^{2} m^{3} x^{m} + d^{2} e m^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((f*x)**(-1+m)*(a+b*ln(c*x**n))**2/(d+e*x**m)**2,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} \left (f x\right )^{m - 1}}{{\left (e x^{m} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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