Integrand size = 22, antiderivative size = 74 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=\frac {b e n \log (d+e x)}{g (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{g (f+g x)}-\frac {b e n \log (f+g x)}{g (e f-d g)} \]
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=\frac {-\frac {a+b \log \left (c (d+e x)^n\right )}{f+g x}+\frac {b e n (\log (d+e x)-\log (f+g x))}{e f-d g}}{g} \]
(-((a + b*Log[c*(d + e*x)^n])/(f + g*x)) + (b*e*n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g))/g
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {b e n \int \frac {1}{(d+e x) (f+g x)}dx}{g}-\frac {a+b \log \left (c (d+e x)^n\right )}{g (f+g x)}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {b e n \left (\frac {e \int \frac {1}{d+e x}dx}{e f-d g}-\frac {g \int \frac {1}{f+g x}dx}{e f-d g}\right )}{g}-\frac {a+b \log \left (c (d+e x)^n\right )}{g (f+g x)}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {b e n \left (\frac {\log (d+e x)}{e f-d g}-\frac {\log (f+g x)}{e f-d g}\right )}{g}-\frac {a+b \log \left (c (d+e x)^n\right )}{g (f+g x)}\) |
-((a + b*Log[c*(d + e*x)^n])/(g*(f + g*x))) + (b*e*n*(Log[d + e*x]/(e*f - d*g) - Log[f + g*x]/(e*f - d*g)))/g
3.3.52.3.1 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[((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 = 0.62 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.72
method | result | size |
parallelrisch | \(-\frac {\ln \left (e x +d \right ) x b \,e^{2} g n -\ln \left (g x +f \right ) x b \,e^{2} g n +\ln \left (e x +d \right ) b \,e^{2} f n -\ln \left (g x +f \right ) b \,e^{2} f n +\ln \left (c \left (e x +d \right )^{n}\right ) b d e g -\ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{2} f +a d e g -a \,e^{2} f}{\left (d g -e f \right ) \left (g x +f \right ) e g}\) | \(127\) |
risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{g \left (g x +f \right )}-\frac {i \pi b e f \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b d g \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b e f \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \pi b d g -i \pi b e f \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) \pi b d g -i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) b d g +i \pi b e f \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 \ln \left (e x +d \right ) b e g n x -2 \ln \left (-g x -f \right ) b e g n x +2 \ln \left (e x +d \right ) b e f n -2 \ln \left (-g x -f \right ) b e f n +2 \ln \left (c \right ) b d g -2 \ln \left (c \right ) b e f +2 a d g -2 a e f}{2 \left (g x +f \right ) g \left (d g -e f \right )}\) | \(354\) |
-(ln(e*x+d)*x*b*e^2*g*n-ln(g*x+f)*x*b*e^2*g*n+ln(e*x+d)*b*e^2*f*n-ln(g*x+f )*b*e^2*f*n+ln(c*(e*x+d)^n)*b*d*e*g-ln(c*(e*x+d)^n)*b*e^2*f+a*d*e*g-a*e^2* f)/(d*g-e*f)/(g*x+f)/e/g
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=-\frac {a e f - a d g - {\left (b e g n x + b d g n\right )} \log \left (e x + d\right ) + {\left (b e g n x + b e f n\right )} \log \left (g x + f\right ) + {\left (b e f - b d g\right )} \log \left (c\right )}{e f^{2} g - d f g^{2} + {\left (e f g^{2} - d g^{3}\right )} x} \]
-(a*e*f - a*d*g - (b*e*g*n*x + b*d*g*n)*log(e*x + d) + (b*e*g*n*x + b*e*f* n)*log(g*x + f) + (b*e*f - b*d*g)*log(c))/(e*f^2*g - d*f*g^2 + (e*f*g^2 - d*g^3)*x)
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (61) = 122\).
Time = 3.12 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.50 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=\begin {cases} \frac {a x + \frac {b d \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - b n x + b x \log {\left (c \left (d + e x\right )^{n} \right )}}{f^{2}} & \text {for}\: g = 0 \\- \frac {a}{f g + g^{2} x} - \frac {b n}{f g + g^{2} x} - \frac {b \log {\left (c \left (\frac {e f}{g} + e x\right )^{n} \right )}}{f g + g^{2} x} & \text {for}\: d = \frac {e f}{g} \\- \frac {a d g}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} + \frac {a e f}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} - \frac {b d g \log {\left (c \left (d + e x\right )^{n} \right )}}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} + \frac {b e f n \log {\left (\frac {f}{g} + x \right )}}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} + \frac {b e g n x \log {\left (\frac {f}{g} + x \right )}}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} - \frac {b e g x \log {\left (c \left (d + e x\right )^{n} \right )}}{d f g^{2} + d g^{3} x - e f^{2} g - e f g^{2} x} & \text {otherwise} \end {cases} \]
Piecewise(((a*x + b*d*log(c*(d + e*x)**n)/e - b*n*x + b*x*log(c*(d + e*x)* *n))/f**2, Eq(g, 0)), (-a/(f*g + g**2*x) - b*n/(f*g + g**2*x) - b*log(c*(e *f/g + e*x)**n)/(f*g + g**2*x), Eq(d, e*f/g)), (-a*d*g/(d*f*g**2 + d*g**3* x - e*f**2*g - e*f*g**2*x) + a*e*f/(d*f*g**2 + d*g**3*x - e*f**2*g - e*f*g **2*x) - b*d*g*log(c*(d + e*x)**n)/(d*f*g**2 + d*g**3*x - e*f**2*g - e*f*g **2*x) + b*e*f*n*log(f/g + x)/(d*f*g**2 + d*g**3*x - e*f**2*g - e*f*g**2*x ) + b*e*g*n*x*log(f/g + x)/(d*f*g**2 + d*g**3*x - e*f**2*g - e*f*g**2*x) - b*e*g*x*log(c*(d + e*x)**n)/(d*f*g**2 + d*g**3*x - e*f**2*g - e*f*g**2*x) , True))
Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=b e n {\left (\frac {\log \left (e x + d\right )}{e f g - d g^{2}} - \frac {\log \left (g x + f\right )}{e f g - d g^{2}}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{g^{2} x + f g} - \frac {a}{g^{2} x + f g} \]
b*e*n*(log(e*x + d)/(e*f*g - d*g^2) - log(g*x + f)/(e*f*g - d*g^2)) - b*lo g((e*x + d)^n*c)/(g^2*x + f*g) - a/(g^2*x + f*g)
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=\frac {b e n \log \left (e x + d\right )}{e f g - d g^{2}} - \frac {b e n \log \left (g x + f\right )}{e f g - d g^{2}} - \frac {b n \log \left (e x + d\right )}{g^{2} x + f g} - \frac {b \log \left (c\right ) + a}{g^{2} x + f g} \]
b*e*n*log(e*x + d)/(e*f*g - d*g^2) - b*e*n*log(g*x + f)/(e*f*g - d*g^2) - b*n*log(e*x + d)/(g^2*x + f*g) - (b*log(c) + a)/(g^2*x + f*g)
Time = 1.57 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx=-\frac {a}{x\,g^2+f\,g}-\frac {b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{g\,\left (f+g\,x\right )}+\frac {b\,e\,n\,\mathrm {atan}\left (\frac {e\,f\,2{}\mathrm {i}+e\,g\,x\,2{}\mathrm {i}}{d\,g-e\,f}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{g\,\left (d\,g-e\,f\right )} \]