\(\int \frac {\log (c (a+\frac {b}{x})^p)}{(d+e x)^4} \, dx\) [204]

Optimal result

Integrand size = 20, antiderivative size = 175 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\frac {b p}{6 d (a d-b e) (d+e x)^2}+\frac {b (2 a d-b e) p}{3 d^2 (a d-b e)^2 (d+e x)}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e (d+e x)^3}-\frac {p \log (x)}{3 d^3 e}+\frac {a^3 p \log (b+a x)}{3 e (a d-b e)^3}-\frac {b \left (3 a^2 d^2-3 a b d e+b^2 e^2\right ) p \log (d+e x)}{3 d^3 (a d-b e)^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.94 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\frac {\frac {b e p}{2 d (a d-b e) (d+e x)^2}+\frac {b e (2 a d-b e) p}{d^2 (a d-b e)^2 (d+e x)}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3}-\frac {p \log (x)}{d^3}+\frac {a^3 p \log (b+a x)}{(a d-b e)^3}-\frac {b e \left (3 a^2 d^2-3 a b d e+b^2 e^2\right ) p \log (d+e x)}{d^3 (a d-b e)^3}}{3 e} \] Input:

Integrate[Log[c*(a + b/x)^p]/(d + e*x)^4,x]
 

Output:

((b*e*p)/(2*d*(a*d - b*e)*(d + e*x)^2) + (b*e*(2*a*d - b*e)*p)/(d^2*(a*d - 
 b*e)^2*(d + e*x)) - Log[c*(a + b/x)^p]/(d + e*x)^3 - (p*Log[x])/d^3 + (a^ 
3*p*Log[b + a*x])/(a*d - b*e)^3 - (b*e*(3*a^2*d^2 - 3*a*b*d*e + b^2*e^2)*p 
*Log[d + e*x])/(d^3*(a*d - b*e)^3))/(3*e)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2913, 1016, 93, 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.

\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx\)

\(\Big \downarrow \) 2913

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

\(\Big \downarrow \) 1016

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

\(\Big \downarrow \) 93

\(\displaystyle -\frac {b p \int \left (\frac {a^4}{b (b e-a d)^3 (b+a x)}+\frac {1}{b d^3 x}+\frac {e^2 \left (3 a^2 d^2-3 a b e d+b^2 e^2\right )}{d^3 (a d-b e)^3 (d+e x)}+\frac {e^2 (2 a d-b e)}{d^2 (a d-b e)^2 (d+e x)^2}+\frac {e^2}{d (a d-b e) (d+e x)^3}\right )dx}{3 e}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Log[c*(a + b/x)^p]/(d + e*x)^4,x]
 

Output:

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

Defintions of rubi rules used

rule 93

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre 
eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
 

rule 1016

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^( 
p_.), x_Symbol] :> Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ 
[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !I 
ntegerQ[p])
 

rule 2009

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

rule 2913

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. 
)*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n 
)^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1)))   Int[x^(n - 1)*((f + g 
*x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x 
] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
 

Maple [A] (verified)

Time = 2.63 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.94

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{3 e \left (e x +d \right )^{3}}-\frac {p b \left (-\frac {a^{3} \ln \left (a x +b \right )}{b \left (d a -b e \right )^{3}}-\frac {e}{2 d \left (d a -b e \right ) \left (e x +d \right )^{2}}-\frac {e \left (2 d a -b e \right )}{d^{2} \left (d a -b e \right )^{2} \left (e x +d \right )}+\frac {e \left (3 a^{2} d^{2}-3 d e a b +b^{2} e^{2}\right ) \ln \left (e x +d \right )}{d^{3} \left (d a -b e \right )^{3}}+\frac {\ln \left (x \right )}{b \,d^{3}}\right )}{3 e}\)	\(165\)
parallelrisch	\(-\frac {3 x^{3} b^{4} e^{5} p^{2}-6 \ln \left (x \right ) x^{2} b^{4} d \,e^{4} p^{2}+6 \ln \left (e x +d \right ) x^{2} b^{4} d \,e^{4} p^{2}-6 \ln \left (x \right ) x \,b^{4} d^{2} e^{3} p^{2}+6 \ln \left (e x +d \right ) x \,b^{4} d^{2} e^{3} p^{2}+6 \ln \left (x \right ) a \,b^{3} d^{4} e \,p^{2}-6 \ln \left (e x +d \right ) a \,b^{3} d^{4} e \,p^{2}-2 x^{3} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} b \,d^{3} e^{2} p +18 \ln \left (e x +d \right ) x^{2} a^{2} b^{2} d^{3} e^{2} p^{2}-18 \ln \left (e x +d \right ) x^{2} a \,b^{3} d^{2} e^{3} p^{2}-18 \ln \left (x \right ) x \,a^{2} b^{2} d^{4} e \,p^{2}+18 \ln \left (x \right ) x a \,b^{3} d^{3} e^{2} p^{2}+18 \ln \left (e x +d \right ) x \,a^{2} b^{2} d^{4} e \,p^{2}-18 \ln \left (e x +d \right ) x a \,b^{3} d^{3} e^{2} p^{2}+6 \ln \left (e x +d \right ) a^{2} b^{2} d^{5} p^{2}+2 \ln \left (e x +d \right ) b^{4} d^{3} e^{2} p^{2}+7 x^{2} b^{4} d \,e^{4} p^{2}+4 x \,b^{4} d^{2} e^{3} p^{2}-6 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{2} b^{2} d^{5} p -2 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{4} d^{3} e^{2} p -2 \ln \left (x \right ) x^{3} b^{4} e^{5} p^{2}+2 \ln \left (e x +d \right ) x^{3} b^{4} e^{5} p^{2}-6 \ln \left (x \right ) a^{2} b^{2} d^{5} p^{2}-2 \ln \left (x \right ) b^{4} d^{3} e^{2} p^{2}-6 x^{2} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} b \,d^{4} e p -6 \ln \left (x \right ) x^{3} a^{2} b^{2} d^{2} e^{3} p^{2}+6 \ln \left (x \right ) x^{3} a \,b^{3} d \,e^{4} p^{2}+6 \ln \left (e x +d \right ) x^{3} a^{2} b^{2} d^{2} e^{3} p^{2}-6 \ln \left (e x +d \right ) x^{3} a \,b^{3} d \,e^{4} p^{2}-18 \ln \left (x \right ) x^{2} a^{2} b^{2} d^{3} e^{2} p^{2}+18 \ln \left (x \right ) x^{2} a \,b^{3} d^{2} e^{3} p^{2}-8 x^{3} a \,b^{3} d \,e^{4} p^{2}+11 x^{2} a^{2} b^{2} d^{3} e^{2} p^{2}-18 x^{2} a \,b^{3} d^{2} e^{3} p^{2}-6 x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} b \,d^{5} p +6 x \,a^{2} b^{2} d^{4} e \,p^{2}-10 x a \,b^{3} d^{3} e^{2} p^{2}+6 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a \,b^{3} d^{4} e p +5 x^{3} a^{2} b^{2} d^{2} e^{3} p^{2}}{6 \left (e x +d \right )^{3} \left (a^{3} d^{3}-3 a^{2} b \,d^{2} e +3 b^{2} d \,e^{2} a -b^{3} e^{3}\right ) b \,d^{3} p}\)	\(846\)

Input:

int(ln(c*(a+b/x)^p)/(e*x+d)^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*ln(c*(a+b/x)^p)/e/(e*x+d)^3-1/3*p*b/e*(-1/b*a^3/(a*d-b*e)^3*ln(a*x+b) 
-1/2*e/d/(a*d-b*e)/(e*x+d)^2-e*(2*a*d-b*e)/d^2/(a*d-b*e)^2/(e*x+d)+e*(3*a^ 
2*d^2-3*a*b*d*e+b^2*e^2)/d^3/(a*d-b*e)^3*ln(e*x+d)+1/b/d^3*ln(x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 818 vs. \(2 (163) = 326\).

Time = 3.47 (sec) , antiderivative size = 818, normalized size of antiderivative = 4.67 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:

integrate(log(c*(a+b/x)^p)/(e*x+d)^4,x, algorithm="fricas")
 

Output:

1/6*(2*(2*a^2*b*d^3*e^3 - 3*a*b^2*d^2*e^4 + b^3*d*e^5)*p*x^2 + (9*a^2*b*d^ 
4*e^2 - 14*a*b^2*d^3*e^3 + 5*b^3*d^2*e^4)*p*x - 2*(a^3*d^6 - 3*a^2*b*d^5*e 
 + 3*a*b^2*d^4*e^2 - b^3*d^3*e^3)*p*log((a*x + b)/x) + (5*a^2*b*d^5*e - 8* 
a*b^2*d^4*e^2 + 3*b^3*d^3*e^3)*p + 2*(a^3*d^3*e^3*p*x^3 + 3*a^3*d^4*e^2*p* 
x^2 + 3*a^3*d^5*e*p*x + a^3*d^6*p)*log(a*x + b) - 2*((3*a^2*b*d^2*e^4 - 3* 
a*b^2*d*e^5 + b^3*e^6)*p*x^3 + 3*(3*a^2*b*d^3*e^3 - 3*a*b^2*d^2*e^4 + b^3* 
d*e^5)*p*x^2 + 3*(3*a^2*b*d^4*e^2 - 3*a*b^2*d^3*e^3 + b^3*d^2*e^4)*p*x + ( 
3*a^2*b*d^5*e - 3*a*b^2*d^4*e^2 + b^3*d^3*e^3)*p)*log(e*x + d) - 2*(a^3*d^ 
6 - 3*a^2*b*d^5*e + 3*a*b^2*d^4*e^2 - b^3*d^3*e^3)*log(c) - 2*((a^3*d^3*e^ 
3 - 3*a^2*b*d^2*e^4 + 3*a*b^2*d*e^5 - b^3*e^6)*p*x^3 + 3*(a^3*d^4*e^2 - 3* 
a^2*b*d^3*e^3 + 3*a*b^2*d^2*e^4 - b^3*d*e^5)*p*x^2 + 3*(a^3*d^5*e - 3*a^2* 
b*d^4*e^2 + 3*a*b^2*d^3*e^3 - b^3*d^2*e^4)*p*x + (a^3*d^6 - 3*a^2*b*d^5*e 
+ 3*a*b^2*d^4*e^2 - b^3*d^3*e^3)*p)*log(x))/(a^3*d^9*e - 3*a^2*b*d^8*e^2 + 
 3*a*b^2*d^7*e^3 - b^3*d^6*e^4 + (a^3*d^6*e^4 - 3*a^2*b*d^5*e^5 + 3*a*b^2* 
d^4*e^6 - b^3*d^3*e^7)*x^3 + 3*(a^3*d^7*e^3 - 3*a^2*b*d^6*e^4 + 3*a*b^2*d^ 
5*e^5 - b^3*d^4*e^6)*x^2 + 3*(a^3*d^8*e^2 - 3*a^2*b*d^7*e^3 + 3*a*b^2*d^6* 
e^4 - b^3*d^5*e^5)*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\text {Timed out} \] Input:

integrate(ln(c*(a+b/x)**p)/(e*x+d)**4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.71 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\frac {{\left (\frac {2 \, a^{3} \log \left (a x + b\right )}{a^{3} b d^{3} - 3 \, a^{2} b^{2} d^{2} e + 3 \, a b^{3} d e^{2} - b^{4} e^{3}} - \frac {2 \, {\left (3 \, a^{2} d^{2} e - 3 \, a b d e^{2} + b^{2} e^{3}\right )} \log \left (e x + d\right )}{a^{3} d^{6} - 3 \, a^{2} b d^{5} e + 3 \, a b^{2} d^{4} e^{2} - b^{3} d^{3} e^{3}} + \frac {5 \, a d^{2} e - 3 \, b d e^{2} + 2 \, {\left (2 \, a d e^{2} - b e^{3}\right )} x}{a^{2} d^{6} - 2 \, a b d^{5} e + b^{2} d^{4} e^{2} + {\left (a^{2} d^{4} e^{2} - 2 \, a b d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a^{2} d^{5} e - 2 \, a b d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x} - \frac {2 \, \log \left (x\right )}{b d^{3}}\right )} b p}{6 \, e} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{3 \, {\left (e x + d\right )}^{3} e} \] Input:

integrate(log(c*(a+b/x)^p)/(e*x+d)^4,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1097 vs. \(2 (163) = 326\).

Time = 0.14 (sec) , antiderivative size = 1097, normalized size of antiderivative = 6.27 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:

integrate(log(c*(a+b/x)^p)/(e*x+d)^4,x, algorithm="giac")
 

Output:

-1/6*(2*(3*a^2*b^2*d^2*p - 3*a*b^3*d*e*p + b^4*e^2*p)*log(-a*d + b*e + (a* 
x + b)*d/x)/(a^3*d^6 - 3*a^2*b*d^5*e + 3*a*b^2*d^4*e^2 - b^3*d^3*e^3) + 2* 
(3*a^2*b^2*d^2*p - 3*a*b^3*d*e*p + b^4*e^2*p - 6*(a*x + b)*a*b^2*d^2*p/x + 
 3*(a*x + b)*b^3*d*e*p/x + 3*(a*x + b)^2*b^2*d^2*p/x^2)*log((a*x + b)/x)/( 
a^3*d^6 - 3*a^2*b*d^5*e + 3*a*b^2*d^4*e^2 - b^3*d^3*e^3 - 3*(a*x + b)*a^2* 
d^6/x + 6*(a*x + b)*a*b*d^5*e/x - 3*(a*x + b)*b^2*d^4*e^2/x + 3*(a*x + b)^ 
2*a*d^6/x^2 - 3*(a*x + b)^2*b*d^5*e/x^2 - (a*x + b)^3*d^6/x^3) - 2*(3*a^2* 
b^2*d^2*p - 3*a*b^3*d*e*p + b^4*e^2*p)*log((a*x + b)/x)/(a^3*d^6 - 3*a^2*b 
*d^5*e + 3*a*b^2*d^4*e^2 - b^3*d^3*e^3) - (6*a^3*b^3*d^3*e*p - 15*a^2*b^4* 
d^2*e^2*p + 12*a*b^5*d*e^3*p - 3*b^6*e^4*p - 6*a^4*b^2*d^4*log(c) + 18*a^3 
*b^3*d^3*e*log(c) - 20*a^2*b^4*d^2*e^2*log(c) + 10*a*b^5*d*e^3*log(c) - 2* 
b^6*e^4*log(c) - 12*(a*x + b)*a^2*b^3*d^3*e*p/x + 19*(a*x + b)*a*b^4*d^2*e 
^2*p/x - 7*(a*x + b)*b^5*d*e^3*p/x + 12*(a*x + b)*a^3*b^2*d^4*log(c)/x - 3 
0*(a*x + b)*a^2*b^3*d^3*e*log(c)/x + 24*(a*x + b)*a*b^4*d^2*e^2*log(c)/x - 
 6*(a*x + b)*b^5*d*e^3*log(c)/x + 6*(a*x + b)^2*a*b^3*d^3*e*p/x^2 - 4*(a*x 
 + b)^2*b^4*d^2*e^2*p/x^2 - 6*(a*x + b)^2*a^2*b^2*d^4*log(c)/x^2 + 12*(a*x 
 + b)^2*a*b^3*d^3*e*log(c)/x^2 - 6*(a*x + b)^2*b^4*d^2*e^2*log(c)/x^2)/(a^ 
5*d^8 - 5*a^4*b*d^7*e + 10*a^3*b^2*d^6*e^2 - 10*a^2*b^3*d^5*e^3 + 5*a*b^4* 
d^4*e^4 - b^5*d^3*e^5 - 3*(a*x + b)*a^4*d^8/x + 12*(a*x + b)*a^3*b*d^7*e/x 
 - 18*(a*x + b)*a^2*b^2*d^6*e^2/x + 12*(a*x + b)*a*b^3*d^5*e^3/x - 3*(a...
 

Mupad [B] (verification not implemented)

Time = 27.22 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.78 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx=\frac {p\,\ln \left (d+e\,x\right )}{3\,d^3\,e}-\frac {3\,b^2\,e^2\,p}{2\,\left (3\,a^2\,d^5\,e+6\,a^2\,d^4\,e^2\,x+3\,a^2\,d^3\,e^3\,x^2-6\,a\,b\,d^4\,e^2-12\,a\,b\,d^3\,e^3\,x-6\,a\,b\,d^2\,e^4\,x^2+3\,b^2\,d^3\,e^3+6\,b^2\,d^2\,e^4\,x+3\,b^2\,d\,e^5\,x^2\right )}-\frac {p\,\ln \left (x\right )}{3\,d^3\,e}-\frac {a^3\,p\,\ln \left (b+a\,x\right )}{-3\,a^3\,d^3\,e+9\,a^2\,b\,d^2\,e^2-9\,a\,b^2\,d\,e^3+3\,b^3\,e^4}-\frac {\ln \left (c\,{\left (\frac {b+a\,x}{x}\right )}^p\right )}{3\,\left (d^3\,e+3\,d^2\,e^2\,x+3\,d\,e^3\,x^2+e^4\,x^3\right )}-\frac {b^2\,e^3\,p\,x}{3\,a^2\,d^6\,e+6\,a^2\,d^5\,e^2\,x+3\,a^2\,d^4\,e^3\,x^2-6\,a\,b\,d^5\,e^2-12\,a\,b\,d^4\,e^3\,x-6\,a\,b\,d^3\,e^4\,x^2+3\,b^2\,d^4\,e^3+6\,b^2\,d^3\,e^4\,x+3\,b^2\,d^2\,e^5\,x^2}-\frac {a^3\,d^3\,p\,\ln \left (d+e\,x\right )}{3\,a^3\,d^6\,e-9\,a^2\,b\,d^5\,e^2+9\,a\,b^2\,d^4\,e^3-3\,b^3\,d^3\,e^4}+\frac {5\,a\,b\,d\,e\,p}{2\,\left (3\,a^2\,d^5\,e+6\,a^2\,d^4\,e^2\,x+3\,a^2\,d^3\,e^3\,x^2-6\,a\,b\,d^4\,e^2-12\,a\,b\,d^3\,e^3\,x-6\,a\,b\,d^2\,e^4\,x^2+3\,b^2\,d^3\,e^3+6\,b^2\,d^2\,e^4\,x+3\,b^2\,d\,e^5\,x^2\right )}+\frac {2\,a\,b\,d\,e^2\,p\,x}{3\,a^2\,d^6\,e+6\,a^2\,d^5\,e^2\,x+3\,a^2\,d^4\,e^3\,x^2-6\,a\,b\,d^5\,e^2-12\,a\,b\,d^4\,e^3\,x-6\,a\,b\,d^3\,e^4\,x^2+3\,b^2\,d^4\,e^3+6\,b^2\,d^3\,e^4\,x+3\,b^2\,d^2\,e^5\,x^2} \] Input:

int(log(c*(a + b/x)^p)/(d + e*x)^4,x)
 

Output:

(p*log(d + e*x))/(3*d^3*e) - (3*b^2*e^2*p)/(2*(3*a^2*d^5*e + 3*b^2*d^3*e^3 
 + 6*a^2*d^4*e^2*x + 6*b^2*d^2*e^4*x + 3*b^2*d*e^5*x^2 + 3*a^2*d^3*e^3*x^2 
 - 6*a*b*d^4*e^2 - 12*a*b*d^3*e^3*x - 6*a*b*d^2*e^4*x^2)) - (p*log(x))/(3* 
d^3*e) - (a^3*p*log(b + a*x))/(3*b^3*e^4 - 3*a^3*d^3*e + 9*a^2*b*d^2*e^2 - 
 9*a*b^2*d*e^3) - log(c*((b + a*x)/x)^p)/(3*(d^3*e + e^4*x^3 + 3*d^2*e^2*x 
 + 3*d*e^3*x^2)) - (b^2*e^3*p*x)/(3*a^2*d^6*e + 3*b^2*d^4*e^3 + 6*a^2*d^5* 
e^2*x + 6*b^2*d^3*e^4*x + 3*a^2*d^4*e^3*x^2 + 3*b^2*d^2*e^5*x^2 - 6*a*b*d^ 
5*e^2 - 12*a*b*d^4*e^3*x - 6*a*b*d^3*e^4*x^2) - (a^3*d^3*p*log(d + e*x))/( 
3*a^3*d^6*e - 3*b^3*d^3*e^4 + 9*a*b^2*d^4*e^3 - 9*a^2*b*d^5*e^2) + (5*a*b* 
d*e*p)/(2*(3*a^2*d^5*e + 3*b^2*d^3*e^3 + 6*a^2*d^4*e^2*x + 6*b^2*d^2*e^4*x 
 + 3*b^2*d*e^5*x^2 + 3*a^2*d^3*e^3*x^2 - 6*a*b*d^4*e^2 - 12*a*b*d^3*e^3*x 
- 6*a*b*d^2*e^4*x^2)) + (2*a*b*d*e^2*p*x)/(3*a^2*d^6*e + 3*b^2*d^4*e^3 + 6 
*a^2*d^5*e^2*x + 6*b^2*d^3*e^4*x + 3*a^2*d^4*e^3*x^2 + 3*b^2*d^2*e^5*x^2 - 
 6*a*b*d^5*e^2 - 12*a*b*d^4*e^3*x - 6*a*b*d^3*e^4*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 984, normalized size of antiderivative = 5.62 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:

int(log(c*(a+b/x)^p)/(e*x+d)^4,x)
 

Output:

(6*log(a*x + b)*a**3*d**6*p + 18*log(a*x + b)*a**3*d**5*e*p*x + 18*log(a*x 
 + b)*a**3*d**4*e**2*p*x**2 + 6*log(a*x + b)*a**3*d**3*e**3*p*x**3 - 18*lo 
g(d + e*x)*a**2*b*d**5*e*p - 54*log(d + e*x)*a**2*b*d**4*e**2*p*x - 54*log 
(d + e*x)*a**2*b*d**3*e**3*p*x**2 - 18*log(d + e*x)*a**2*b*d**2*e**4*p*x** 
3 + 18*log(d + e*x)*a*b**2*d**4*e**2*p + 54*log(d + e*x)*a*b**2*d**3*e**3* 
p*x + 54*log(d + e*x)*a*b**2*d**2*e**4*p*x**2 + 18*log(d + e*x)*a*b**2*d*e 
**5*p*x**3 - 6*log(d + e*x)*b**3*d**3*e**3*p - 18*log(d + e*x)*b**3*d**2*e 
**4*p*x - 18*log(d + e*x)*b**3*d*e**5*p*x**2 - 6*log(d + e*x)*b**3*e**6*p* 
x**3 - 6*log(((a*x + b)**p*c)/x**p)*a**3*d**6 + 18*log(((a*x + b)**p*c)/x* 
*p)*a**2*b*d**5*e - 18*log(((a*x + b)**p*c)/x**p)*a*b**2*d**4*e**2 + 6*log 
(((a*x + b)**p*c)/x**p)*b**3*d**3*e**3 - 6*log(x)*a**3*d**6*p - 18*log(x)* 
a**3*d**5*e*p*x - 18*log(x)*a**3*d**4*e**2*p*x**2 - 6*log(x)*a**3*d**3*e** 
3*p*x**3 + 18*log(x)*a**2*b*d**5*e*p + 54*log(x)*a**2*b*d**4*e**2*p*x + 54 
*log(x)*a**2*b*d**3*e**3*p*x**2 + 18*log(x)*a**2*b*d**2*e**4*p*x**3 - 18*l 
og(x)*a*b**2*d**4*e**2*p - 54*log(x)*a*b**2*d**3*e**3*p*x - 54*log(x)*a*b* 
*2*d**2*e**4*p*x**2 - 18*log(x)*a*b**2*d*e**5*p*x**3 + 6*log(x)*b**3*d**3* 
e**3*p + 18*log(x)*b**3*d**2*e**4*p*x + 18*log(x)*b**3*d*e**5*p*x**2 + 6*l 
og(x)*b**3*e**6*p*x**3 + 11*a**2*b*d**5*e*p + 15*a**2*b*d**4*e**2*p*x - 4* 
a**2*b*d**2*e**4*p*x**3 - 18*a*b**2*d**4*e**2*p - 24*a*b**2*d**3*e**3*p*x 
+ 6*a*b**2*d*e**5*p*x**3 + 7*b**3*d**3*e**3*p + 9*b**3*d**2*e**4*p*x - ...