Integrand size = 27, antiderivative size = 290 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}+\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{3/2}}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{3/2}}-\frac {b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{3/2}}+\frac {b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{3/2}} \] Output:
b*e*n*ln(x)/d/f-b*e*n*ln(e*x+d)/d/f-(a+b*ln(c*(e*x+d)^n))/f/x+1/2*g^(1/2)* (a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)-g^(1/2)*x)/(e*(-f)^(1/2)+d*g^(1/2)) )/(-f)^(3/2)-1/2*g^(1/2)*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)+g^(1/2)*x) /(e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(3/2)-1/2*b*g^(1/2)*n*polylog(2,-g^(1/2)*( e*x+d)/(e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(3/2)+1/2*b*g^(1/2)*n*polylog(2,g^(1 /2)*(e*x+d)/(e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(3/2)
Time = 0.22 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\frac {f \left (2 b e (-f)^{3/2} n x (\log (x)-\log (d+e x))+2 d \sqrt {-f} f \left (a+b \log \left (c (d+e x)^n\right )\right )+d f \sqrt {g} x \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )-d f \sqrt {g} x \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )-b d f \sqrt {g} n x \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b d f \sqrt {g} n x \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )\right )}{2 d (-f)^{7/2} x} \] Input:
Integrate[(a + b*Log[c*(d + e*x)^n])/(x^2*(f + g*x^2)),x]
Output:
(f*(2*b*e*(-f)^(3/2)*n*x*(Log[x] - Log[d + e*x]) + 2*d*Sqrt[-f]*f*(a + b*L og[c*(d + e*x)^n]) + d*f*Sqrt[g]*x*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt [-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])] - d*f*Sqrt[g]*x*(a + b*Log[c* (d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])] - b *d*f*Sqrt[g]*n*x*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]) )] + b*d*f*Sqrt[g]*n*x*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt [g])]))/(2*d*(-f)^(7/2)*x)
Time = 0.98 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2863, 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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f x^2}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f \left (f+g x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}-\frac {b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{3/2}}+\frac {b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 (-f)^{3/2}}+\frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f}\) |
Input:
Int[(a + b*Log[c*(d + e*x)^n])/(x^2*(f + g*x^2)),x]
Output:
(b*e*n*Log[x])/(d*f) - (b*e*n*Log[d + e*x])/(d*f) - (a + b*Log[c*(d + e*x) ^n])/(f*x) + (Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g ]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*(-f)^(3/2)) - (Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*( -f)^(3/2)) - (b*Sqrt[g]*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d *Sqrt[g]))])/(2*(-f)^(3/2)) + (b*Sqrt[g]*n*PolyLog[2, (Sqrt[g]*(d + e*x))/ (e*Sqrt[-f] + d*Sqrt[g])])/(2*(-f)^(3/2))
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.29 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.70
method | result | size |
risch | \(\frac {b g \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {g f}}\right ) n \ln \left (e x +d \right )}{f \sqrt {g f}}-\frac {b g \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {g f}}\right ) \ln \left (\left (e x +d \right )^{n}\right )}{f \sqrt {g f}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f x}-\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f \sqrt {-g f}}+\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f \sqrt {-g f}}-\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f \sqrt {-g f}}+\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f \sqrt {-g f}}+\frac {b e n \ln \left (e x \right )}{f d}-\frac {b e n \ln \left (e x +d \right )}{d f}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{f x}-\frac {g \arctan \left (\frac {g x}{\sqrt {g f}}\right )}{f \sqrt {g f}}\right )\) | \(492\) |
Input:
int((a+b*ln(c*(e*x+d)^n))/x^2/(g*x^2+f),x,method=_RETURNVERBOSE)
Output:
b*g/f/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(g*f)^(1/2))*n*ln(e*x+d )-b*g/f/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(g*f)^(1/2))*ln((e*x+ d)^n)-b*ln((e*x+d)^n)/f/x-1/2*b*n*g/f*ln(e*x+d)/(-g*f)^(1/2)*ln((e*(-g*f)^ (1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))+1/2*b*n*g/f*ln(e*x+d)/(-g*f)^(1 /2)*ln((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))-1/2*b*n*g/f/(- g*f)^(1/2)*dilog((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))+1/2* b*n*g/f/(-g*f)^(1/2)*dilog((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)- d*g))+b*e*n/f/d*ln(e*x)-b*e*n*ln(e*x+d)/d/f+(1/2*I*b*Pi*csgn(I*(e*x+d)^n)* csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csg n(I*c)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^2*c sgn(I*c)+b*ln(c)+a)*(-1/f/x-g/f/(g*f)^(1/2)*arctan(g*x/(g*f)^(1/2)))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f),x, algorithm="fricas")
Output:
integral((b*log((e*x + d)^n*c) + a)/(g*x^4 + f*x^2), x)
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(e*x+d)**n))/x**2/(g*x**2+f),x)
Output:
Timed out
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f),x, algorithm="maxima")
Output:
-a*(g*arctan(g*x/sqrt(f*g))/(sqrt(f*g)*f) + 1/(f*x)) + b*integrate((log((e *x + d)^n) + log(c))/(g*x^4 + f*x^2), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f),x, algorithm="giac")
Output:
integrate((b*log((e*x + d)^n*c) + a)/((g*x^2 + f)*x^2), x)
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^2\,\left (g\,x^2+f\right )} \,d x \] Input:
int((a + b*log(c*(d + e*x)^n))/(x^2*(f + g*x^2)),x)
Output:
int((a + b*log(c*(d + e*x)^n))/(x^2*(f + g*x^2)), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )} \, dx=\frac {-\sqrt {g}\, \sqrt {f}\, \mathit {atan} \left (\frac {g x}{\sqrt {g}\, \sqrt {f}}\right ) a x +\left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g \,x^{4}+f \,x^{2}}d x \right ) b \,f^{2} x -a f}{f^{2} x} \] Input:
int((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f),x)
Output:
( - sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*a*x + int(log((d + e*x)* *n*c)/(f*x**2 + g*x**4),x)*b*f**2*x - a*f)/(f**2*x)