Integrand size = 27, antiderivative size = 74 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=-\frac {b e n \log (d+e x)}{f (d f-e g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{f (g+f x)}+\frac {b e n \log (g+f x)}{f (d f-e g)} \]
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=\frac {-\frac {a+b \log \left (c (d+e x)^n\right )}{g+f x}+\frac {b e n (\log (d+e x)-\log (g+f x))}{-d f+e g}}{f} \]
(-((a + b*Log[c*(d + e*x)^n])/(g + f*x)) + (b*e*n*(Log[d + e*x] - Log[g + f*x]))/(-(d*f) + e*g))/f
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2005, 2842, 47, 16}
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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+\frac {g}{x}\right )^2} \, dx\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f x+g)^2}dx\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {b e n \int \frac {1}{(d+e x) (g+f x)}dx}{f}-\frac {a+b \log \left (c (d+e x)^n\right )}{f (f x+g)}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {b e n \left (\frac {f \int \frac {1}{g+f x}dx}{d f-e g}-\frac {e \int \frac {1}{d+e x}dx}{d f-e g}\right )}{f}-\frac {a+b \log \left (c (d+e x)^n\right )}{f (f x+g)}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {b e n \left (\frac {\log (f x+g)}{d f-e g}-\frac {\log (d+e x)}{d f-e g}\right )}{f}-\frac {a+b \log \left (c (d+e x)^n\right )}{f (f x+g)}\) |
-((a + b*Log[c*(d + e*x)^n])/(f*(g + f*x))) + (b*e*n*(-(Log[d + e*x]/(d*f - e*g)) + Log[g + f*x]/(d*f - e*g)))/f
3.4.7.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Time = 0.35 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.72
method | result | size |
parallelrisch | \(-\frac {\ln \left (e x +d \right ) x b \,e^{2} f n -\ln \left (f x +g \right ) x b \,e^{2} f n +\ln \left (e x +d \right ) b \,e^{2} g n -\ln \left (f x +g \right ) b \,e^{2} g n +\ln \left (c \left (e x +d \right )^{n}\right ) b d e f -\ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{2} g +a d e f -a \,e^{2} g}{\left (d f -e g \right ) \left (f x +g \right ) e f}\) | \(127\) |
risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f \left (f x +g \right )}-\frac {-i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b d f -i \pi b e g \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b d f -i \pi b e g \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) b d f +i \pi b e g \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b d f +i \pi b e g \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}-2 \ln \left (-f x -g \right ) b e f n x +2 \ln \left (e x +d \right ) b e f n x -2 \ln \left (-f x -g \right ) b e g n +2 \ln \left (e x +d \right ) b e g n +2 \ln \left (c \right ) b d f -2 \ln \left (c \right ) b e g +2 a d f -2 a e g}{2 \left (f x +g \right ) f \left (d f -e g \right )}\) | \(354\) |
-(ln(e*x+d)*x*b*e^2*f*n-ln(f*x+g)*x*b*e^2*f*n+ln(e*x+d)*b*e^2*g*n-ln(f*x+g )*b*e^2*g*n+ln(c*(e*x+d)^n)*b*d*e*f-ln(c*(e*x+d)^n)*b*e^2*g+a*d*e*f-a*e^2* g)/(d*f-e*g)/(f*x+g)/e/f
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=-\frac {a d f - a e g + {\left (b e f n x + b d f n\right )} \log \left (e x + d\right ) - {\left (b e f n x + b e g n\right )} \log \left (f x + g\right ) + {\left (b d f - b e g\right )} \log \left (c\right )}{d f^{2} g - e f g^{2} + {\left (d f^{3} - e f^{2} g\right )} x} \]
-(a*d*f - a*e*g + (b*e*f*n*x + b*d*f*n)*log(e*x + d) - (b*e*f*n*x + b*e*g* n)*log(f*x + g) + (b*d*f - b*e*g)*log(c))/(d*f^2*g - e*f*g^2 + (d*f^3 - e* f^2*g)*x)
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (61) = 122\).
Time = 34.40 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.50 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=\begin {cases} \frac {a x + \frac {b d \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - b n x + b x \log {\left (c \left (d + e x\right )^{n} \right )}}{g^{2}} & \text {for}\: f = 0 \\- \frac {a}{f^{2} x + f g} - \frac {b n}{f^{2} x + f g} - \frac {b \log {\left (c \left (e x + \frac {e g}{f}\right )^{n} \right )}}{f^{2} x + f g} & \text {for}\: d = \frac {e g}{f} \\- \frac {a d f}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} + \frac {a e g}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} - \frac {b d f \log {\left (c \left (d + e x\right )^{n} \right )}}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} + \frac {b e f n x \log {\left (x + \frac {g}{f} \right )}}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} - \frac {b e f x \log {\left (c \left (d + e x\right )^{n} \right )}}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} + \frac {b e g n \log {\left (x + \frac {g}{f} \right )}}{d f^{3} x + d f^{2} g - e f^{2} g x - e f g^{2}} & \text {otherwise} \end {cases} \]
Piecewise(((a*x + b*d*log(c*(d + e*x)**n)/e - b*n*x + b*x*log(c*(d + e*x)* *n))/g**2, Eq(f, 0)), (-a/(f**2*x + f*g) - b*n/(f**2*x + f*g) - b*log(c*(e *x + e*g/f)**n)/(f**2*x + f*g), Eq(d, e*g/f)), (-a*d*f/(d*f**3*x + d*f**2* g - e*f**2*g*x - e*f*g**2) + a*e*g/(d*f**3*x + d*f**2*g - e*f**2*g*x - e*f *g**2) - b*d*f*log(c*(d + e*x)**n)/(d*f**3*x + d*f**2*g - e*f**2*g*x - e*f *g**2) + b*e*f*n*x*log(x + g/f)/(d*f**3*x + d*f**2*g - e*f**2*g*x - e*f*g* *2) - b*e*f*x*log(c*(d + e*x)**n)/(d*f**3*x + d*f**2*g - e*f**2*g*x - e*f* g**2) + b*e*g*n*log(x + g/f)/(d*f**3*x + d*f**2*g - e*f**2*g*x - e*f*g**2) , True))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=-b e n {\left (\frac {\log \left (e x + d\right )}{d f^{2} - e f g} - \frac {\log \left (f x + g\right )}{d f^{2} - e f g}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{f^{2} x + f g} - \frac {a}{f^{2} x + f g} \]
-b*e*n*(log(e*x + d)/(d*f^2 - e*f*g) - log(f*x + g)/(d*f^2 - e*f*g)) - b*l og((e*x + d)^n*c)/(f^2*x + f*g) - a/(f^2*x + f*g)
Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=-\frac {b e n \log \left (e x + d\right )}{d f^{2} - e f g} + \frac {b e n \log \left (f x + g\right )}{d f^{2} - e f g} - \frac {b n \log \left (e x + d\right )}{f^{2} x + f g} - \frac {b \log \left (c\right ) + a}{f^{2} x + f g} \]
-b*e*n*log(e*x + d)/(d*f^2 - e*f*g) + b*e*n*log(f*x + g)/(d*f^2 - e*f*g) - b*n*log(e*x + d)/(f^2*x + f*g) - (b*log(c) + a)/(f^2*x + f*g)
Time = 1.99 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+\frac {g}{x}\right )^2 x^2} \, dx=-\frac {a}{x\,f^2+g\,f}-\frac {b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{f\,\left (g+f\,x\right )}+\frac {b\,e\,n\,\mathrm {atan}\left (\frac {e\,g\,2{}\mathrm {i}+e\,f\,x\,2{}\mathrm {i}}{d\,f-e\,g}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{f\,\left (d\,f-e\,g\right )} \]