3.365 \(\int \frac{(f x)^{-1+m} (a+b \log (c x^n))^2}{(d+e x^m)^3} \, dx\)

Optimal. Leaf size=214 \[ \frac{b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{d x^{-m}}{e}\right )}{d^2 e m^3}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (\frac{d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e m^2}-\frac{b n x (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}+\frac{b^2 n^2 x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{d^2 e m^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.512463, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2339, 2338, 2349, 2345, 2391, 2335, 260} \[ \frac{b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{d x^{-m}}{e}\right )}{d^2 e m^3}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (\frac{d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e m^2}-\frac{b n x (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}+\frac{b^2 n^2 x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{d^2 e m^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n*x*(f*x)^(-1 + m)*(a + b*Log[c*x^n]))/(d^2*m^2*(d + e*x^m))) - (x^(1 - m)*(f*x)^(-1 + m)*(a + b*Log[c*x^
n])^2)/(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^2*e*m^2
) + (b^2*n^2*x^(1 - m)*(f*x)^(-1 + m)*Log[d + e*x^m])/(d^2*e*m^3) + (b^2*n^2*x^(1 - m)*(f*x)^(-1 + m)*PolyLog[
2, -(d/(e*x^m))])/(d^2*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 2349

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

Rule 2335

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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 )^3} \, 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 )^3} \, dx\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}+\frac{\left (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 )^2} \, dx}{e m}\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}-\frac{\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac{x^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx}{d m}+\frac{\left (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}{d e m}\\ &=-\frac{b n x (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}-\frac{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^2 e m^2}+\frac{\left (b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac{x^{-1+m}}{d+e x^m} \, dx}{d^2 m^2}+\frac{\left (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^2 e m^2}\\ &=-\frac{b n x (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d^2 m^2 \left (d+e x^m\right )}-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{2 e m \left (d+e x^m\right )^2}-\frac{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^2 e m^2}+\frac{b^2 n^2 x^{1-m} (f x)^{-1+m} \log \left (d+e x^m\right )}{d^2 e m^3}+\frac{b^2 n^2 x^{1-m} (f x)^{-1+m} \text{Li}_2\left (-\frac{d x^{-m}}{e}\right )}{d^2 e m^3}\\ \end{align*}

Mathematica [A]  time = 0.358638, size = 207, normalized size = 0.97 \[ \frac{x^{-m} (f x)^m \left (\frac{2 b^2 n^2 \left (\text{PolyLog}\left (2,\frac{e x^m}{d}+1\right )+\left (\log \left (-\frac{e x^m}{d}\right )-m \log (x)\right ) \log \left (d+e x^m\right )+\frac{1}{2} m^2 \log ^2(x)\right )}{d^2}-\frac{m^2 \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2}+\frac{2 b m n \left (a+b \log \left (c x^n\right )\right )}{d \left (d+e x^m\right )}-\frac{2 a b m n \log \left (d-d x^m\right )}{d^2}+\frac{2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d^2}+\frac{2 b^2 n^2 \log \left (d-d x^m\right )}{d^2}\right )}{2 e f m^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.922, 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 ) ^{3}}}\, 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)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a b f^{m} n{\left (\frac{1}{{\left (d e^{2} f m x^{m} + d^{2} e f m\right )} m} + \frac{\log \left (x\right )}{d^{2} e f m} - \frac{\log \left (e x^{m} + d\right )}{d^{2} e f m^{2}}\right )} - \frac{1}{2} \,{\left (\frac{f^{m} \log \left (x^{n}\right )^{2}}{e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m} - 2 \, \int \frac{e f^{m} m x^{m} \log \left (c\right )^{2} +{\left (d f^{m} n +{\left (2 \, e f^{m} m \log \left (c\right ) + e f^{m} n\right )} x^{m}\right )} \log \left (x^{n}\right )}{e^{4} f m x x^{3 \, m} + 3 \, d e^{3} f m x x^{2 \, m} + 3 \, d^{2} e^{2} f m x x^{m} + d^{3} e f m x}\,{d x}\right )} b^{2} - \frac{a b f^{m} \log \left (c x^{n}\right )}{e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m} - \frac{a^{2} f^{m}}{2 \,{\left (e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m\right )}} \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)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.38674, size = 1204, normalized size = 5.63 \begin{align*} \frac{{\left (b^{2} e^{2} m^{2} n^{2} \log \left (x\right )^{2} + 2 \,{\left (b^{2} e^{2} m^{2} n \log \left (c\right ) + a b e^{2} m^{2} n - b^{2} e^{2} m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{2 \, m} + 2 \,{\left (b^{2} d e m^{2} n^{2} \log \left (x\right )^{2} + b^{2} d e m n \log \left (c\right ) + a b d e m n +{\left (2 \, b^{2} d e m^{2} n \log \left (c\right ) + 2 \, a b d e m^{2} n - b^{2} d e m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{m} -{\left (b^{2} d^{2} m^{2} \log \left (c\right )^{2} + a^{2} d^{2} m^{2} - 2 \, a b d^{2} m n + 2 \,{\left (a b d^{2} m^{2} - b^{2} d^{2} m n\right )} \log \left (c\right )\right )} f^{m - 1} - 2 \,{\left (b^{2} e^{2} f^{m - 1} n^{2} x^{2 \, m} + 2 \, b^{2} d e f^{m - 1} n^{2} x^{m} + b^{2} d^{2} f^{m - 1} n^{2}\right )}{\rm Li}_2\left (-\frac{e x^{m} + d}{d} + 1\right ) - 2 \,{\left ({\left (b^{2} e^{2} m n \log \left (c\right ) + a b e^{2} m n - b^{2} e^{2} n^{2}\right )} f^{m - 1} x^{2 \, m} + 2 \,{\left (b^{2} d e m n \log \left (c\right ) + a b d e m n - b^{2} d e n^{2}\right )} f^{m - 1} x^{m} +{\left (b^{2} d^{2} m n \log \left (c\right ) + a b d^{2} m n - b^{2} d^{2} n^{2}\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \,{\left (b^{2} e^{2} f^{m - 1} m n^{2} x^{2 \, m} \log \left (x\right ) + 2 \, b^{2} d e f^{m - 1} m n^{2} x^{m} \log \left (x\right ) + b^{2} d^{2} f^{m - 1} m n^{2} \log \left (x\right )\right )} \log \left (\frac{e x^{m} + d}{d}\right )}{2 \,{\left (d^{2} e^{3} m^{3} x^{2 \, m} + 2 \, d^{3} e^{2} m^{3} x^{m} + d^{4} e m^{3}\right )}} \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)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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