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

Optimal. Leaf size=150 \[ -\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{2 e m \left (d+e x^m\right )^2}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{2 d^2 e m^2}+\frac{b n x^{1-m} \log (x) (f x)^{m-1}}{2 d^2 e m}+\frac{b n x^{1-m} (f x)^{m-1}}{2 d e m^2 \left (d+e x^m\right )} \]

[Out]

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

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Rubi [A]  time = 0.215251, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2339, 2338, 266, 44} \[ -\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{2 e m \left (d+e x^m\right )^2}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{2 d^2 e m^2}+\frac{b n x^{1-m} \log (x) (f x)^{m-1}}{2 d^2 e m}+\frac{b n x^{1-m} (f x)^{m-1}}{2 d e m^2 \left (d+e x^m\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.147099, size = 137, normalized size = 0.91 \[ \frac{x^{-m} (f x)^m \left (-a d^2 m-b d^2 m \log \left (c x^n\right )-b d^2 n \log \left (d+e x^m\right )+b d^2 n-b e^2 n x^{2 m} \log \left (d+e x^m\right )+b d e n x^m-2 b d e n x^m \log \left (d+e x^m\right )+b m n \log (x) \left (d+e x^m\right )^2\right )}{2 d^2 e f m^2 \left (d+e x^m\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.147, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{-1+m} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{ \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))/(d+e*x^m)^3,x)

[Out]

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

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Maxima [A]  time = 1.23201, size = 205, normalized size = 1.37 \begin{align*} \frac{1}{2} \, 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{b f^{m} \log \left (c x^{n}\right )}{2 \,{\left (e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m\right )}} - \frac{a 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))/(d+e*x^m)^3,x, algorithm="maxima")

[Out]

1/2*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*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*f^m/(e^3*f*m*x^(2*m) + 2*d*e^2*f*m*x^m +
d^2*e*f*m)

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Fricas [A]  time = 1.62013, size = 382, normalized size = 2.55 \begin{align*} \frac{b e^{2} f^{m - 1} m n x^{2 \, m} \log \left (x\right ) +{\left (2 \, b d e m n \log \left (x\right ) + b d e n\right )} f^{m - 1} x^{m} -{\left (b d^{2} m \log \left (c\right ) + a d^{2} m - b d^{2} n\right )} f^{m - 1} -{\left (b e^{2} f^{m - 1} n x^{2 \, m} + 2 \, b d e f^{m - 1} n x^{m} + b d^{2} f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{2 \,{\left (d^{2} e^{3} m^{2} x^{2 \, m} + 2 \, d^{3} e^{2} m^{2} x^{m} + d^{4} e m^{2}\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))/(d+e*x^m)^3,x, algorithm="fricas")

[Out]

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

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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))/(d+e*x**m)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.31322, size = 848, normalized size = 5.65 \begin{align*} \frac{b d f^{m} m n x^{2} x^{m} e \log \left (x\right )}{2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}} - \frac{b d f^{m} n x^{2} x^{m} e \log \left (x^{m} e + d\right )}{2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}} + \frac{b f^{m} m n x^{2} x^{2 \, m} e^{2} \log \left (x\right )}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} + \frac{b d f^{m} n x^{2} x^{m} e}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} - \frac{b d^{2} f^{m} n x^{2} \log \left (x^{m} e + d\right )}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} - \frac{b f^{m} n x^{2} x^{2 \, m} e^{2} \log \left (x^{m} e + d\right )}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} - \frac{b d^{2} f^{m} m x^{2} \log \left (c\right )}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} - \frac{a d^{2} f^{m} m x^{2}}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{3}\right )}} + \frac{b d^{2} f^{m} n x^{2}}{2 \,{\left (2 \, d^{3} f m^{2} x^{2} x^{m} e^{2} + d^{4} f m^{2} x^{2} e + d^{2} f m^{2} x^{2} x^{2 \, m} e^{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))/(d+e*x^m)^3,x, algorithm="giac")

[Out]

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