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*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)

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Rubi [A]  time = 0.11, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.12, 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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fricas [A]  time = 0.41, size = 89, normalized size = 1.29 \[ \frac {b e f^{m - 1} m n x^{m} \log \relax (x) - {\left (b d m \log \relax (c) + a d m\right )} f^{m - 1} - {\left (b e f^{m - 1} n x^{m} + b d f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{d e^{2} m^{2} x^{m} + d^{2} e m^{2}} \]

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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giac [B]  time = 0.45, size = 206, normalized size = 2.99 \[ \frac {b f^{m} m n x x^{m} e \log \relax (x)}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac {b f^{m} n x x^{m} e \log \left (x^{m} e + d\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac {b d f^{m} n x \log \left (x^{m} e + d\right )}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac {b d f^{m} m x \log \relax (c)}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} - \frac {a d f^{m} m x}{d f m^{2} x x^{m} e^{2} + d^{2} f m^{2} x e} \]

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)

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maple [F]  time = 1.27, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \left (f x \right )^{m -1}}{\left (e \,x^{m}+d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.72, size = 97, normalized size = 1.41 \[ b f^{m} n {\left (\frac {\log \relax (x)}{d e f m} - \frac {\log \left (e x^{m} + d\right )}{d e f m^{2}}\right )} - \frac {b f^{m} \log \left (c x^{n}\right )}{e^{2} f m x^{m} + d e f m} - \frac {a f^{m}}{e^{2} f m x^{m} + d e f m} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x^m\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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