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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 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])

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 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 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])

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 = 0.16, size = 178, normalized size = 0.95 \[ \frac {x^{-m} (f x)^m \left (-2 a d^3 m-2 b d^3 m \log \left (c x^n\right )-2 b d^3 n \log \left (d+e x^m\right )+3 b d^3 n+5 b d^2 e n x^m-6 b d^2 e n x^m \log \left (d+e x^m\right )-2 b e^3 n x^{3 m} \log \left (d+e x^m\right )+2 b d e^2 n x^{2 m}-6 b d e^2 n x^{2 m} \log \left (d+e x^m\right )+2 b m n \log (x) \left (d+e x^m\right )^3\right )}{6 d^3 e f m^2 \left (d+e x^m\right )^3} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 242, normalized size = 1.29 \[ \frac {2 \, b e^{3} f^{m - 1} m n x^{3 \, m} \log \relax (x) + 2 \, {\left (3 \, b d e^{2} m n \log \relax (x) + b d e^{2} n\right )} f^{m - 1} x^{2 \, m} + {\left (6 \, b d^{2} e m n \log \relax (x) + 5 \, b d^{2} e n\right )} f^{m - 1} x^{m} - {\left (2 \, b d^{3} m \log \relax (c) + 2 \, a d^{3} m - 3 \, b d^{3} n\right )} f^{m - 1} - 2 \, {\left (b e^{3} f^{m - 1} n x^{3 \, m} + 3 \, b d e^{2} f^{m - 1} n x^{2 \, m} + 3 \, b d^{2} e f^{m - 1} n x^{m} + b d^{3} f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{6 \, {\left (d^{3} e^{4} m^{2} x^{3 \, m} + 3 \, d^{4} e^{3} m^{2} x^{2 \, m} + 3 \, d^{5} e^{2} m^{2} x^{m} + d^{6} e m^{2}\right )}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.64, size = 1080, normalized size = 5.74 \[ \frac {b d^{2} f^{m} m n x^{3} x^{m} e \log \relax (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}} - \frac {b d^{2} f^{m} n x^{3} x^{m} e \log \left (x^{m} e + d\right )}{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}} + \frac {b d f^{m} m n x^{3} x^{2 \, m} e^{2} \log \relax (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}} + \frac {5 \, b d^{2} f^{m} n x^{3} x^{m} e}{6 \, {\left (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}\right )}} - \frac {b d^{3} f^{m} n x^{3} \log \left (x^{m} e + d\right )}{3 \, {\left (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}\right )}} - \frac {b d f^{m} n x^{3} x^{2 \, m} e^{2} \log \left (x^{m} e + d\right )}{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}} - \frac {b d^{3} f^{m} m x^{3} \log \relax (c)}{3 \, {\left (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}\right )}} + \frac {b f^{m} m n x^{3} x^{3 \, m} e^{3} \log \relax (x)}{3 \, {\left (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}\right )}} - \frac {a d^{3} f^{m} m x^{3}}{3 \, {\left (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}\right )}} + \frac {b d^{3} f^{m} n x^{3}}{2 \, {\left (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}\right )}} + \frac {b d f^{m} n x^{3} x^{2 \, m} e^{2}}{3 \, {\left (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}\right )}} - \frac {b f^{m} n x^{3} x^{3 \, m} e^{3} \log \left (x^{m} e + d\right )}{3 \, {\left (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}\right )}} \]

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)

________________________________________________________________________________________

maple [F]  time = 1.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \left (f x \right )^{m -1}}{\left (e \,x^{m}+d \right )^{4}}\, dx \]

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)^4,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.81, size = 210, normalized size = 1.12 \[ \frac {1}{6} \, b f^{m} n {\left (\frac {2 \, e x^{m} + 3 \, d}{{\left (d^{2} e^{3} f m x^{2 \, m} + 2 \, d^{3} e^{2} f m x^{m} + d^{4} e f m\right )} m} + \frac {2 \, \log \relax (x)}{d^{3} e f m} - \frac {2 \, \log \left (e x^{m} + d\right )}{d^{3} e f m^{2}}\right )} - \frac {b f^{m} \log \left (c x^{n}\right )}{3 \, {\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} e f m\right )}} - \frac {a f^{m}}{3 \, {\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} e f m\right )}} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x^m\right )}^4} \,d x \]

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)^4,x)

[Out]

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

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________