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

Optimal. Leaf size=69 \[ \frac{(f x)^m \left (a+b \log \left (c x^n\right )\right )}{d f m \left (d+e x^m\right )}-\frac{b n x^{-m} (f x)^m \log \left (d+e x^m\right )}{d e f m^2} \]

[Out]

((f*x)^m*(a + b*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)

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Rubi [A]  time = 0.107119, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2335, 268, 260} \[ \frac{(f x)^m \left (a+b \log \left (c x^n\right )\right )}{d f m \left (d+e x^m\right )}-\frac{b n x^{-m} (f x)^m \log \left (d+e x^m\right )}{d e f m^2} \]

Antiderivative was successfully verified.

[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 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 268

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracP
art[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 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 )}{\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) \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]  time = 0.121435, size = 89, normalized size = 1.29 \[ -\frac{x^{-m} (f x)^m \left (a d m+b d m \log \left (c x^n\right )+b e n x^m \log \left (d+e x^m\right )-b m n \log (x) \left (d+e x^m\right )+b d n \log \left (d+e x^m\right )\right )}{d e f m^2 \left (d+e x^m\right )} \]

Antiderivative was successfully verified.

[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)))

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Maple [F]  time = 1.158, 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 ) ^{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))/(d+e*x^m)^2,x)

[Out]

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

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Maxima [A]  time = 1.19838, size = 131, normalized size = 1.9 \begin{align*} 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} \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)^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)

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Fricas [A]  time = 1.36141, size = 205, normalized size = 2.97 \begin{align*} \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}} \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)^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)

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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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.36732, size = 278, normalized size = 4.03 \begin{align*} \frac{b f^{m} m n x x^{m} e \log \left (x\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac{b f^{m} n x x^{m} e \log \left (x^{m} e + d\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac{b d f^{m} n x \log \left (x^{m} e + d\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac{b d f^{m} m x \log \left (c\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac{a d f^{m} m x}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} \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)^2,x, algorithm="giac")

[Out]

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