\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))}{(d+e x^m)^3} \, dx\) [357]

Optimal result

Integrand size = 27, antiderivative size = 150 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\frac {b n x^{1-m} (f x)^{-1+m}}{2 d e m^2 \left (d+e x^m\right )}+\frac {b n x^{1-m} (f x)^{-1+m} \log (x)}{2 d^2 e m}-\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{2 e m \left (d+e x^m\right )^2}-\frac {b n x^{1-m} (f x)^{-1+m} \log \left (d+e x^m\right )}{2 d^2 e m^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2377, 2376, 272, 46} \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=-\frac {x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{2 e m \left (d+e x^m\right )^2}-\frac {b n x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{2 d^2 e m^2}+\frac {b n x^{1-m} \log (x) (f x)^{m-1}}{2 d^2 e m}+\frac {b n x^{1-m} (f x)^{m-1}}{2 d e m^2 \left (d+e x^m\right )} \]

[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 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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 2376

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 2377

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])

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.91 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\frac {x^{-m} (f x)^m \left (-a d^2 m+b d^2 n+b d e n x^m+b m n \left (d+e x^m\right )^2 \log (x)-b d^2 m \log \left (c x^n\right )-b d^2 n \log \left (d+e x^m\right )-2 b d e n x^m \log \left (d+e x^m\right )-b e^2 n x^{2 m} \log \left (d+e x^m\right )\right )}{2 d^2 e f m^2 \left (d+e x^m\right )^2} \]

[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)

Maple [F]

\[\int \frac {\left (f x \right )^{m -1} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (d +e \,x^{m}\right )^{3}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\frac {b e^{2} f^{m - 1} m n x^{2 \, m} \log \left (x\right ) + {\left (2 \, b d e m n \log \left (x\right ) + b d e n\right )} f^{m - 1} x^{m} - {\left (b d^{2} m \log \left (c\right ) + a d^{2} m - b d^{2} n\right )} f^{m - 1} - {\left (b e^{2} f^{m - 1} n x^{2 \, m} + 2 \, b d e f^{m - 1} n x^{m} + b d^{2} f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{2 \, {\left (d^{2} e^{3} m^{2} x^{2 \, m} + 2 \, d^{3} e^{2} m^{2} x^{m} + d^{4} e m^{2}\right )}} \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\frac {1}{2} \, b f^{m} n {\left (\frac {1}{{\left (d e^{2} f m x^{m} + d^{2} e f m\right )} m} + \frac {\log \left (x\right )}{d^{2} e f m} - \frac {\log \left (e x^{m} + d\right )}{d^{2} e f m^{2}}\right )} - \frac {b f^{m} \log \left (c x^{n}\right )}{2 \, {\left (e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m\right )}} - \frac {a f^{m}}{2 \, {\left (e^{3} f m x^{2 \, m} + 2 \, d e^{2} f m x^{m} + d^{2} e f m\right )}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 633 vs. \(2 (142) = 284\).

Time = 0.36 (sec) , antiderivative size = 633, normalized size of antiderivative = 4.22 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\frac {b e^{2} f^{m} m n x^{2} x^{2 \, m} \log \left (x\right )}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} + \frac {b d e f^{m} m n x^{2} x^{m} \log \left (x\right )}{d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}} - \frac {b e^{2} f^{m} n x^{2} x^{2 \, m} \log \left (e x^{m} + d\right )}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} - \frac {b d e f^{m} n x^{2} x^{m} \log \left (e x^{m} + d\right )}{d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}} + \frac {b d e f^{m} n x^{2} x^{m}}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} - \frac {b d^{2} f^{m} n x^{2} \log \left (e x^{m} + d\right )}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} - \frac {b d^{2} f^{m} m x^{2} \log \left (c\right )}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} - \frac {a d^{2} f^{m} m x^{2}}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} + \frac {b d^{2} f^{m} n x^{2}}{2 \, {\left (d^{2} e^{3} f m^{2} x^{2} x^{2 \, m} + 2 \, d^{3} e^{2} f m^{2} x^{2} x^{m} + d^{4} e f m^{2} x^{2}\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^3} \, dx=\int \frac {{\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x^m\right )}^3} \,d x \]

[In]

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

[Out]

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