\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))}{(d+e x^m)^2} \, dx\) [356]

Optimal result

Integrand size = 27, antiderivative size = 69 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=\frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )}{d f m \left (d+e x^m\right )}-\frac {b n x^{-m} (f x)^m \log \left (d+e x^m\right )}{d e f m^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2373, 274, 266} \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=\frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )}{d f m \left (d+e x^m\right )}-\frac {b n x^{-m} (f x)^m \log \left (d+e x^m\right )}{d e f m^2} \]

[In]

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

[Out]

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

Rule 266

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]

Rule 274

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

Rule 2373

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]

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=-\frac {x^{-m} (f x)^m \left (a d m-b m n \left (d+e x^m\right ) \log (x)+b d m \log \left (c x^n\right )+b d n \log \left (d+e x^m\right )+b e n x^m \log \left (d+e x^m\right )\right )}{d e f m^2 \left (d+e x^m\right )} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (f x \right )^{m -1} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (d +e \,x^{m}\right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=\frac {b e f^{m - 1} m n x^{m} \log \left (x\right ) - {\left (b d m \log \left (c\right ) + a d m\right )} f^{m - 1} - {\left (b e f^{m - 1} n x^{m} + b d f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{d e^{2} m^{2} x^{m} + d^{2} e m^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.41 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=b f^{m} n {\left (\frac {\log \left (x\right )}{d e f m} - \frac {\log \left (e x^{m} + d\right )}{d e f m^{2}}\right )} - \frac {b f^{m} \log \left (c x^{n}\right )}{e^{2} f m x^{m} + d e f m} - \frac {a f^{m}}{e^{2} f m x^{m} + d e f m} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (69) = 138\).

Time = 0.35 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.93 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=\frac {b e f^{m} m n x x^{m} \log \left (x\right )}{d e^{2} f m^{2} x x^{m} + d^{2} e f m^{2} x} - \frac {b e f^{m} n x x^{m} \log \left (e x^{m} + d\right )}{d e^{2} f m^{2} x x^{m} + d^{2} e f m^{2} x} - \frac {b d f^{m} n x \log \left (e x^{m} + d\right )}{d e^{2} f m^{2} x x^{m} + d^{2} e f m^{2} x} - \frac {b d f^{m} m x \log \left (c\right )}{d e^{2} f m^{2} x x^{m} + d^{2} e f m^{2} x} - \frac {a d f^{m} m x}{d e^{2} f m^{2} x x^{m} + d^{2} e f m^{2} x} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^2} \, dx=\int \frac {{\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x^m\right )}^2} \,d x \]

[In]

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

[Out]

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