3.3.3 \(\int \frac {\log (c (a+\frac {b}{x})^p)}{(d+e x)^3} \, dx\) [203]

3.3.3.1 Optimal result

Integrand size = 20, antiderivative size = 127 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\frac {b p}{2 d (a d-b e) (d+e x)}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e (d+e x)^2}-\frac {p \log (x)}{2 d^2 e}+\frac {a^2 p \log (b+a x)}{2 e (a d-b e)^2}-\frac {b (2 a d-b e) p \log (d+e x)}{2 d^2 (a d-b e)^2} \]

output

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

3.3.3.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.89 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\frac {\frac {b e p}{d (a d-b e) (d+e x)}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2}-\frac {p \log (x)}{d^2}+\frac {a^2 p \log (b+a x)}{(a d-b e)^2}+\frac {b e (-2 a d+b e) p \log (d+e x)}{d^2 (a d-b e)^2}}{2 e} \]

input

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

output

((b*e*p)/(d*(a*d - b*e)*(d + e*x)) - Log[c*(a + b/x)^p]/(d + e*x)^2 - (p*L 
og[x])/d^2 + (a^2*p*Log[b + a*x])/(a*d - b*e)^2 + (b*e*(-2*a*d + b*e)*p*Lo 
g[d + e*x])/(d^2*(a*d - b*e)^2))/(2*e)
 

3.3.3.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 123, 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.

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

\(\Big \downarrow \) 2913

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

\(\Big \downarrow \) 1016

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

\(\Big \downarrow \) 93

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.3.3.3.1 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]
 

3.3.3.4 Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 e \left (e x +d \right )^{2}}-\frac {p b \left (\frac {\ln \left (x \right )}{b \,d^{2}}-\frac {e}{d \left (a d -b e \right ) \left (e x +d \right )}+\frac {e \left (2 a d -b e \right ) \ln \left (e x +d \right )}{d^{2} \left (a d -b e \right )^{2}}-\frac {a^{2} \ln \left (a x +b \right )}{b \left (a d -b e \right )^{2}}\right )}{2 e}\)	\(120\)
parallelrisch	\(-\frac {-2 \ln \left (x \right ) x^{2} a^{2} b d \,e^{4} p^{2}+2 \ln \left (e x +d \right ) x^{2} a^{2} b d \,e^{4} p^{2}-4 \ln \left (x \right ) x \,a^{2} b \,d^{2} e^{3} p^{2}+2 \ln \left (x \right ) x a \,b^{2} d \,e^{4} p^{2}+4 \ln \left (e x +d \right ) x \,a^{2} b \,d^{2} e^{3} p^{2}-2 \ln \left (e x +d \right ) x a \,b^{2} d \,e^{4} p^{2}-a^{2} b \,d^{3} e^{2} p^{2}+a \,b^{2} d^{2} e^{3} p^{2}+\ln \left (x \right ) x^{2} a \,b^{2} e^{5} p^{2}-\ln \left (e x +d \right ) x^{2} a \,b^{2} e^{5} p^{2}-2 \ln \left (x \right ) a^{2} b \,d^{3} e^{2} p^{2}+\ln \left (x \right ) a \,b^{2} d^{2} e^{3} p^{2}+2 \ln \left (e x +d \right ) a^{2} b \,d^{3} e^{2} p^{2}-\ln \left (e x +d \right ) a \,b^{2} d^{2} e^{3} p^{2}+x a \,b^{2} d \,e^{4} p^{2}-2 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{2} b \,d^{3} e^{2} p -x \,a^{2} b \,d^{2} e^{3} p^{2}-2 x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} d^{3} e^{2} p -x^{2} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} d^{2} e^{3} p +\ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a \,b^{2} d^{2} e^{3} p}{2 \left (e x +d \right )^{2} d^{2} \left (a d -b e \right )^{2} a \,e^{2} p}\)	\(428\)

input

int(ln(c*(a+b/x)^p)/(e*x+d)^3,x,method=_RETURNVERBOSE)
 

output

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

3.3.3.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (117) = 234\).

Time = 0.78 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.37 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\frac {{\left (a b d^{2} e^{2} - b^{2} d e^{3}\right )} p x - {\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} p \log \left (\frac {a x + b}{x}\right ) + {\left (a b d^{3} e - b^{2} d^{2} e^{2}\right )} p + {\left (a^{2} d^{2} e^{2} p x^{2} + 2 \, a^{2} d^{3} e p x + a^{2} d^{4} p\right )} \log \left (a x + b\right ) - {\left ({\left (2 \, a b d e^{3} - b^{2} e^{4}\right )} p x^{2} + 2 \, {\left (2 \, a b d^{2} e^{2} - b^{2} d e^{3}\right )} p x + {\left (2 \, a b d^{3} e - b^{2} d^{2} e^{2}\right )} p\right )} \log \left (e x + d\right ) - {\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} \log \left (c\right ) - {\left ({\left (a^{2} d^{2} e^{2} - 2 \, a b d e^{3} + b^{2} e^{4}\right )} p x^{2} + 2 \, {\left (a^{2} d^{3} e - 2 \, a b d^{2} e^{2} + b^{2} d e^{3}\right )} p x + {\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} p\right )} \log \left (x\right )}{2 \, {\left (a^{2} d^{6} e - 2 \, a b d^{5} e^{2} + b^{2} d^{4} e^{3} + {\left (a^{2} d^{4} e^{3} - 2 \, a b d^{3} e^{4} + b^{2} d^{2} e^{5}\right )} x^{2} + 2 \, {\left (a^{2} d^{5} e^{2} - 2 \, a b d^{4} e^{3} + b^{2} d^{3} e^{4}\right )} x\right )}} \]

input

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="fricas")
 

output

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

3.3.3.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\text {Timed out} \]

input

integrate(ln(c*(a+b/x)**p)/(e*x+d)**3,x)
 

output

Timed out
 

3.3.3.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.26 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\frac {{\left (\frac {a^{2} \log \left (a x + b\right )}{a^{2} b d^{2} - 2 \, a b^{2} d e + b^{3} e^{2}} - \frac {{\left (2 \, a d e - b e^{2}\right )} \log \left (e x + d\right )}{a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}} + \frac {e}{a d^{3} - b d^{2} e + {\left (a d^{2} e - b d e^{2}\right )} x} - \frac {\log \left (x\right )}{b d^{2}}\right )} b p}{2 \, e} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{2 \, {\left (e x + d\right )}^{2} e} \]

input

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="maxima")
 

output

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

3.3.3.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (117) = 234\).

Time = 0.32 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.70 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=-\frac {\frac {{\left (2 \, a b^{2} d p - b^{3} e p\right )} \log \left (-a d + b e + \frac {{\left (a x + b\right )} d}{x}\right )}{a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}} + \frac {{\left (2 \, a b^{2} d p - b^{3} e p - \frac {2 \, {\left (a x + b\right )} b^{2} d p}{x}\right )} \log \left (\frac {a x + b}{x}\right )}{a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2} - \frac {2 \, {\left (a x + b\right )} a d^{4}}{x} + \frac {2 \, {\left (a x + b\right )} b d^{3} e}{x} + \frac {{\left (a x + b\right )}^{2} d^{4}}{x^{2}}} - \frac {{\left (2 \, a b^{2} d p - b^{3} e p\right )} \log \left (\frac {a x + b}{x}\right )}{a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}} - \frac {a b^{3} d e p - b^{4} e^{2} p - 2 \, a^{2} b^{2} d^{2} \log \left (c\right ) + 3 \, a b^{3} d e \log \left (c\right ) - b^{4} e^{2} \log \left (c\right ) - \frac {{\left (a x + b\right )} b^{3} d e p}{x} + \frac {2 \, {\left (a x + b\right )} a b^{2} d^{2} \log \left (c\right )}{x} - \frac {2 \, {\left (a x + b\right )} b^{3} d e \log \left (c\right )}{x}}{a^{3} d^{5} - 3 \, a^{2} b d^{4} e + 3 \, a b^{2} d^{3} e^{2} - b^{3} d^{2} e^{3} - \frac {2 \, {\left (a x + b\right )} a^{2} d^{5}}{x} + \frac {4 \, {\left (a x + b\right )} a b d^{4} e}{x} - \frac {2 \, {\left (a x + b\right )} b^{2} d^{3} e^{2}}{x} + \frac {{\left (a x + b\right )}^{2} a d^{5}}{x^{2}} - \frac {{\left (a x + b\right )}^{2} b d^{4} e}{x^{2}}}}{2 \, b} \]

input

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="giac")
 

output

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

3.3.3.9 Mupad [B] (verification not implemented)

Time = 2.11 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.71 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^3} \, dx=\frac {a^2\,p\,\ln \left (b+a\,x\right )}{2\,a^2\,d^2\,e-4\,a\,b\,d\,e^2+2\,b^2\,e^3}-\frac {\ln \left (c\,{\left (\frac {b+a\,x}{x}\right )}^p\right )}{2\,\left (d^2\,e+2\,d\,e^2\,x+e^3\,x^2\right )}-\frac {p\,\ln \left (x\right )}{2\,d^2\,e}-\frac {b\,e\,p}{2\,b\,d^2\,e^2-2\,a\,d^3\,e+2\,b\,d\,e^3\,x-2\,a\,d^2\,e^2\,x}+\frac {b^2\,e\,p\,\ln \left (d+e\,x\right )}{2\,a^2\,d^4-4\,a\,b\,d^3\,e+2\,b^2\,d^2\,e^2}-\frac {2\,a\,b\,d\,p\,\ln \left (d+e\,x\right )}{2\,a^2\,d^4-4\,a\,b\,d^3\,e+2\,b^2\,d^2\,e^2} \]

input

int(log(c*(a + b/x)^p)/(d + e*x)^3,x)
 

output

(a^2*p*log(b + a*x))/(2*b^2*e^3 + 2*a^2*d^2*e - 4*a*b*d*e^2) - log(c*((b + 
 a*x)/x)^p)/(2*(d^2*e + e^3*x^2 + 2*d*e^2*x)) - (p*log(x))/(2*d^2*e) - (b* 
e*p)/(2*b*d^2*e^2 - 2*a*d^3*e + 2*b*d*e^3*x - 2*a*d^2*e^2*x) + (b^2*e*p*lo 
g(d + e*x))/(2*a^2*d^4 + 2*b^2*d^2*e^2 - 4*a*b*d^3*e) - (2*a*b*d*p*log(d + 
 e*x))/(2*a^2*d^4 + 2*b^2*d^2*e^2 - 4*a*b*d^3*e)