3.4.58 \(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))}{(d+e x^m)^4} \, dx\) [358]

3.4.58.1 Optimal result

Integrand size = 27, antiderivative size = 188 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\frac {b n x^{1-m} (f x)^{-1+m}}{6 d e m^2 \left (d+e x^m\right )^2}+\frac {b n x^{1-m} (f x)^{-1+m}}{3 d^2 e m^2 \left (d+e x^m\right )}+\frac {b n x^{1-m} (f x)^{-1+m} \log (x)}{3 d^3 e m}-\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{3 e m \left (d+e x^m\right )^3}-\frac {b n x^{1-m} (f x)^{-1+m} \log \left (d+e x^m\right )}{3 d^3 e m^2} \]

output

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
 

3.4.58.2 Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.95 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\frac {x^{-m} (f x)^m \left (-2 a d^3 m+3 b d^3 n+5 b d^2 e n x^m+2 b d e^2 n x^{2 m}+2 b m n \left (d+e x^m\right )^3 \log (x)-2 b d^3 m \log \left (c x^n\right )-2 b d^3 n \log \left (d+e x^m\right )-6 b d^2 e n x^m \log \left (d+e x^m\right )-6 b d e^2 n x^{2 m} \log \left (d+e x^m\right )-2 b e^3 n x^{3 m} \log \left (d+e x^m\right )\right )}{6 d^3 e f m^2 \left (d+e x^m\right )^3} \]

input

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

output

((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)
 

3.4.58.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.57, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2777, 2776, 798, 54, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2777

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

\(\Big \downarrow \) 2776

\(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {b n \int \frac {1}{x \left (e x^m+d\right )^3}dx}{3 e m}-\frac {a+b \log \left (c x^n\right )}{3 e m \left (d+e x^m\right )^3}\right )\)

\(\Big \downarrow \) 798

\(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {b n \int \frac {x^{-m}}{\left (e x^m+d\right )^3}dx^m}{3 e m^2}-\frac {a+b \log \left (c x^n\right )}{3 e m \left (d+e x^m\right )^3}\right )\)

\(\Big \downarrow \) 54

\(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {b n \int \left (\frac {x^{-m}}{d^3}-\frac {e}{d^3 \left (e x^m+d\right )}-\frac {e}{d^2 \left (e x^m+d\right )^2}-\frac {e}{d \left (e x^m+d\right )^3}\right )dx^m}{3 e m^2}-\frac {a+b \log \left (c x^n\right )}{3 e m \left (d+e x^m\right )^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {b n \left (-\frac {\log \left (d+e x^m\right )}{d^3}+\frac {\log \left (x^m\right )}{d^3}+\frac {1}{d^2 \left (d+e x^m\right )}+\frac {1}{2 d \left (d+e x^m\right )^2}\right )}{3 e m^2}-\frac {a+b \log \left (c x^n\right )}{3 e m \left (d+e x^m\right )^3}\right )\)

input

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

output

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

3.4.58.3.1 Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2776

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*L 
og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[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] || G 
tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
 

rule 2777

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + ( 
e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[(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])
 

3.4.58.4 Maple [F]

\[\int \frac {\left (f x \right )^{m -1} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (d +e \,x^{m}\right )^{4}}d x\]

input

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

output

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

3.4.58.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.29 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\frac {2 \, b e^{3} f^{m - 1} m n x^{3 \, m} \log \left (x\right ) + 2 \, {\left (3 \, b d e^{2} m n \log \left (x\right ) + b d e^{2} n\right )} f^{m - 1} x^{2 \, m} + {\left (6 \, b d^{2} e m n \log \left (x\right ) + 5 \, b d^{2} e n\right )} f^{m - 1} x^{m} - {\left (2 \, b d^{3} m \log \left (c\right ) + 2 \, a d^{3} m - 3 \, b d^{3} n\right )} f^{m - 1} - 2 \, {\left (b e^{3} f^{m - 1} n x^{3 \, m} + 3 \, b d e^{2} f^{m - 1} n x^{2 \, m} + 3 \, b d^{2} e f^{m - 1} n x^{m} + b d^{3} f^{m - 1} n\right )} \log \left (e x^{m} + d\right )}{6 \, {\left (d^{3} e^{4} m^{2} x^{3 \, m} + 3 \, d^{4} e^{3} m^{2} x^{2 \, m} + 3 \, d^{5} e^{2} m^{2} x^{m} + d^{6} e m^{2}\right )}} \]

input

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))/(d+e*x^m)^4,x, algorithm="fricas")
 

output

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)
 

3.4.58.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.58.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.12 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\frac {1}{6} \, b f^{m} n {\left (\frac {2 \, e x^{m} + 3 \, d}{{\left (d^{2} e^{3} f m x^{2 \, m} + 2 \, d^{3} e^{2} f m x^{m} + d^{4} e f m\right )} m} + \frac {2 \, \log \left (x\right )}{d^{3} e f m} - \frac {2 \, \log \left (e x^{m} + d\right )}{d^{3} e f m^{2}}\right )} - \frac {b f^{m} \log \left (c x^{n}\right )}{3 \, {\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} e f m\right )}} - \frac {a f^{m}}{3 \, {\left (e^{4} f m x^{3 \, m} + 3 \, d e^{3} f m x^{2 \, m} + 3 \, d^{2} e^{2} f m x^{m} + d^{3} e f m\right )}} \]

input

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))/(d+e*x^m)^4,x, algorithm="maxima")
 

output

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)
 

3.4.58.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1102 vs. \(2 (178) = 356\).

Time = 0.37 (sec) , antiderivative size = 1102, normalized size of antiderivative = 5.86 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\frac {b e^{3} f^{m} m n x^{3} x^{3 \, m} \log \left (x\right )}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} + \frac {b d e^{2} f^{m} m n x^{3} x^{2 \, m} \log \left (x\right )}{d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}} + \frac {b d^{2} e f^{m} m n x^{3} x^{m} \log \left (x\right )}{d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}} - \frac {b e^{3} f^{m} n x^{3} x^{3 \, m} \log \left (e x^{m} + d\right )}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} - \frac {b d e^{2} f^{m} n x^{3} x^{2 \, m} \log \left (e x^{m} + d\right )}{d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}} - \frac {b d^{2} e f^{m} n x^{3} x^{m} \log \left (e x^{m} + d\right )}{d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}} + \frac {b d e^{2} f^{m} n x^{3} x^{2 \, m}}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} + \frac {5 \, b d^{2} e f^{m} n x^{3} x^{m}}{6 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} - \frac {b d^{3} f^{m} n x^{3} \log \left (e x^{m} + d\right )}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} - \frac {b d^{3} f^{m} m x^{3} \log \left (c\right )}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} - \frac {a d^{3} f^{m} m x^{3}}{3 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} + \frac {b d^{3} f^{m} n x^{3}}{2 \, {\left (d^{3} e^{4} f m^{2} x^{3} x^{3 \, m} + 3 \, d^{4} e^{3} f m^{2} x^{3} x^{2 \, m} + 3 \, d^{5} e^{2} f m^{2} x^{3} x^{m} + d^{6} e f m^{2} x^{3}\right )}} \]

input

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))/(d+e*x^m)^4,x, algorithm="giac")
 

output

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

3.4.58.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^m\right )^4} \, dx=\int \frac {{\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x^m\right )}^4} \,d x \]

input

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

output

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