3.358 \(\int \frac{(f x)^{-1+m} (a+b \log (c x^n))}{(d+e x^m)^4} \, dx\)

Optimal. Leaf size=188 \[ -\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}+\frac{b n x^{1-m} (f x)^{m-1}}{3 d^2 e m^2 \left (d+e x^m\right )}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{3 d^3 e m^2}+\frac{b n x^{1-m} \log (x) (f x)^{m-1}}{3 d^3 e m}+\frac{b n x^{1-m} (f x)^{m-1}}{6 d e m^2 \left (d+e x^m\right )^2} \]

[Out]

(b*n*x^(1 - m)*(f*x)^(-1 + m))/(6*d*e*m^2*(d + e*x^m)^2) + (b*n*x^(1 - m)*(f*x)^(-1 + m))/(3*d^2*e*m^2*(d + e*
x^m)) + (b*n*x^(1 - m)*(f*x)^(-1 + m)*Log[x])/(3*d^3*e*m) - (x^(1 - m)*(f*x)^(-1 + m)*(a + b*Log[c*x^n]))/(3*e
*m*(d + e*x^m)^3) - (b*n*x^(1 - m)*(f*x)^(-1 + m)*Log[d + e*x^m])/(3*d^3*e*m^2)

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Rubi [A]  time = 0.23405, antiderivative size = 188, 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} \[ -\frac{x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}+\frac{b n x^{1-m} (f x)^{m-1}}{3 d^2 e m^2 \left (d+e x^m\right )}-\frac{b n x^{1-m} (f x)^{m-1} \log \left (d+e x^m\right )}{3 d^3 e m^2}+\frac{b n x^{1-m} \log (x) (f x)^{m-1}}{3 d^3 e m}+\frac{b n x^{1-m} (f x)^{m-1}}{6 d e m^2 \left (d+e x^m\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^(-1 + m)*(a + b*Log[c*x^n]))/(d + e*x^m)^4,x]

[Out]

(b*n*x^(1 - m)*(f*x)^(-1 + m))/(6*d*e*m^2*(d + e*x^m)^2) + (b*n*x^(1 - m)*(f*x)^(-1 + m))/(3*d^2*e*m^2*(d + e*
x^m)) + (b*n*x^(1 - m)*(f*x)^(-1 + m)*Log[x])/(3*d^3*e*m) - (x^(1 - m)*(f*x)^(-1 + m)*(a + b*Log[c*x^n]))/(3*e
*m*(d + e*x^m)^3) - (b*n*x^(1 - m)*(f*x)^(-1 + m)*Log[d + e*x^m])/(3*d^3*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 )^4} \, 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 )^4} \, dx\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}+\frac{\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac{1}{x \left (d+e x^m\right )^3} \, dx}{3 e m}\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}+\frac{\left (b n x^{1-m} (f x)^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{1}{x (d+e x)^3} \, dx,x,x^m\right )}{3 e m^2}\\ &=-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}+\frac{\left (b n x^{1-m} (f x)^{-1+m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{d^3 x}-\frac{e}{d (d+e x)^3}-\frac{e}{d^2 (d+e x)^2}-\frac{e}{d^3 (d+e x)}\right ) \, dx,x,x^m\right )}{3 e m^2}\\ &=\frac{b n x^{1-m} (f x)^{-1+m}}{6 d e m^2 \left (d+e x^m\right )^2}+\frac{b n x^{1-m} (f x)^{-1+m}}{3 d^2 e m^2 \left (d+e x^m\right )}+\frac{b n x^{1-m} (f x)^{-1+m} \log (x)}{3 d^3 e m}-\frac{x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}-\frac{b n x^{1-m} (f x)^{-1+m} \log \left (d+e x^m\right )}{3 d^3 e m^2}\\ \end{align*}

Mathematica [A]  time = 0.153825, size = 178, normalized size = 0.95 \[ \frac{x^{-m} (f x)^m \left (-2 a d^3 m-2 b d^3 m \log \left (c x^n\right )+5 b d^2 e n x^m-6 b d^2 e n x^m \log \left (d+e x^m\right )-2 b d^3 n \log \left (d+e x^m\right )+3 b d^3 n+2 b d e^2 n x^{2 m}-6 b d e^2 n x^{2 m} \log \left (d+e x^m\right )-2 b e^3 n x^{3 m} \log \left (d+e x^m\right )+2 b m n \log (x) \left (d+e x^m\right )^3\right )}{6 d^3 e f m^2 \left (d+e x^m\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((f*x)^(-1 + m)*(a + b*Log[c*x^n]))/(d + e*x^m)^4,x]

[Out]

((f*x)^m*(-2*a*d^3*m + 3*b*d^3*n + 5*b*d^2*e*n*x^m + 2*b*d*e^2*n*x^(2*m) + 2*b*m*n*(d + e*x^m)^3*Log[x] - 2*b*
d^3*m*Log[c*x^n] - 2*b*d^3*n*Log[d + e*x^m] - 6*b*d^2*e*n*x^m*Log[d + e*x^m] - 6*b*d*e^2*n*x^(2*m)*Log[d + e*x
^m] - 2*b*e^3*n*x^(3*m)*Log[d + e*x^m]))/(6*d^3*e*f*m^2*x^m*(d + e*x^m)^3)

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Maple [F]  time = 0.951, 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 ) ^{4}}}\, 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)^4,x)

[Out]

int((f*x)^(-1+m)*(a+b*ln(c*x^n))/(d+e*x^m)^4,x)

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Maxima [A]  time = 1.22851, size = 284, normalized size = 1.51 \begin{align*} \frac{1}{6} \, b f^{m} n{\left (\frac{2 \, e x^{m} + 3 \, d}{{\left (d^{2} e^{3} f m x^{2 \, m} + 2 \, d^{3} e^{2} f m x^{m} + d^{4} e f m\right )} m} + \frac{2 \, \log \left (x\right )}{d^{3} e f m} - \frac{2 \, \log \left (e x^{m} + d\right )}{d^{3} e f m^{2}}\right )} - \frac{b f^{m} \log \left (c x^{n}\right )}{3 \,{\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} e f m\right )}} - \frac{a f^{m}}{3 \,{\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} 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)^4,x, algorithm="maxima")

[Out]

1/6*b*f^m*n*((2*e*x^m + 3*d)/((d^2*e^3*f*m*x^(2*m) + 2*d^3*e^2*f*m*x^m + d^4*e*f*m)*m) + 2*log(x)/(d^3*e*f*m)
- 2*log(e*x^m + d)/(d^3*e*f*m^2)) - 1/3*b*f^m*log(c*x^n)/(e^4*f*m*x^(3*m) + 3*d*e^3*f*m*x^(2*m) + 3*d^2*e^2*f*
m*x^m + d^3*e*f*m) - 1/3*a*f^m/(e^4*f*m*x^(3*m) + 3*d*e^3*f*m*x^(2*m) + 3*d^2*e^2*f*m*x^m + d^3*e*f*m)

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Fricas [A]  time = 1.63029, size = 559, normalized size = 2.97 \begin{align*} \frac{2 \, b e^{3} f^{m - 1} m n x^{3 \, m} \log \left (x\right ) + 2 \,{\left (3 \, b d e^{2} m n \log \left (x\right ) + b d e^{2} n\right )} f^{m - 1} x^{2 \, m} +{\left (6 \, b d^{2} e m n \log \left (x\right ) + 5 \, b d^{2} e n\right )} f^{m - 1} x^{m} -{\left (2 \, b d^{3} m \log \left (c\right ) + 2 \, a d^{3} m - 3 \, b d^{3} n\right )} f^{m - 1} - 2 \,{\left (b e^{3} f^{m - 1} n x^{3 \, m} + 3 \, b d e^{2} f^{m - 1} n x^{2 \, m} + 3 \, b d^{2} e f^{m - 1} n x^{m} + b d^{3} f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{6 \,{\left (d^{3} e^{4} m^{2} x^{3 \, m} + 3 \, d^{4} e^{3} m^{2} x^{2 \, m} + 3 \, d^{5} e^{2} m^{2} x^{m} + d^{6} 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)^4,x, algorithm="fricas")

[Out]

1/6*(2*b*e^3*f^(m - 1)*m*n*x^(3*m)*log(x) + 2*(3*b*d*e^2*m*n*log(x) + b*d*e^2*n)*f^(m - 1)*x^(2*m) + (6*b*d^2*
e*m*n*log(x) + 5*b*d^2*e*n)*f^(m - 1)*x^m - (2*b*d^3*m*log(c) + 2*a*d^3*m - 3*b*d^3*n)*f^(m - 1) - 2*(b*e^3*f^
(m - 1)*n*x^(3*m) + 3*b*d*e^2*f^(m - 1)*n*x^(2*m) + 3*b*d^2*e*f^(m - 1)*n*x^m + b*d^3*f^(m - 1)*n)*log(e*x^m +
d))/(d^3*e^4*m^2*x^(3*m) + 3*d^4*e^3*m^2*x^(2*m) + 3*d^5*e^2*m^2*x^m + d^6*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)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.42465, size = 1458, normalized size = 7.76 \begin{align*} \text{result too large to display} \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)^4,x, algorithm="giac")

[Out]

b*d^2*f^m*m*n*x^3*x^m*e*log(x)/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*
f*m^2*x^3*x^(3*m)*e^4) - b*d^2*f^m*n*x^3*x^m*e*log(x^m*e + d)/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d
^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) + b*d*f^m*m*n*x^3*x^(2*m)*e^2*log(x)/(3*d^5*f*m^2*x^3*x^
m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) + 5/6*b*d^2*f^m*n*x^3*x^m*e
/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) - 1/3*b
*d^3*f^m*n*x^3*log(x^m*e + d)/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f
*m^2*x^3*x^(3*m)*e^4) - b*d*f^m*n*x^3*x^(2*m)*e^2*log(x^m*e + d)/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e +
3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) - 1/3*b*d^3*f^m*m*x^3*log(c)/(3*d^5*f*m^2*x^3*x^m*e^2
+ d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) + 1/3*b*f^m*m*n*x^3*x^(3*m)*e^3*
log(x)/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) -
1/3*a*d^3*f^m*m*x^3/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*
x^(3*m)*e^4) + 1/2*b*d^3*f^m*n*x^3/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 +
d^3*f*m^2*x^3*x^(3*m)*e^4) + 1/3*b*d*f^m*n*x^3*x^(2*m)*e^2/(3*d^5*f*m^2*x^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*
f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4) - 1/3*b*f^m*n*x^3*x^(3*m)*e^3*log(x^m*e + d)/(3*d^5*f*m^2*x
^3*x^m*e^2 + d^6*f*m^2*x^3*e + 3*d^4*f*m^2*x^3*x^(2*m)*e^3 + d^3*f*m^2*x^3*x^(3*m)*e^4)