Integrand size = 22, antiderivative size = 360 \[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}+\frac {p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{\sqrt {f} \sqrt {g}}-\frac {p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} (e+d x)}{\left (i d \sqrt {f}+e \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{\sqrt {f} \sqrt {g}}+\frac {i p \operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \operatorname {PolyLog}\left (2,\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 \sqrt {f} \sqrt {g}}+\frac {i p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} (e+d x)}{\left (i d \sqrt {f}+e \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 \sqrt {f} \sqrt {g}} \]
arctan(x*g^(1/2)/f^(1/2))*ln(c*(d+e/x)^p)/f^(1/2)/g^(1/2)+p*arctan(x*g^(1/ 2)/f^(1/2))*ln(2*f^(1/2)/(f^(1/2)-I*x*g^(1/2)))/f^(1/2)/g^(1/2)-p*arctan(x *g^(1/2)/f^(1/2))*ln(2*(d*x+e)*f^(1/2)*g^(1/2)/(I*d*f^(1/2)+e*g^(1/2))/(f^ (1/2)-I*x*g^(1/2)))/f^(1/2)/g^(1/2)+1/2*I*p*polylog(2,-I*x*g^(1/2)/f^(1/2) )/f^(1/2)/g^(1/2)-1/2*I*p*polylog(2,I*x*g^(1/2)/f^(1/2))/f^(1/2)/g^(1/2)-1 /2*I*p*polylog(2,1-2*f^(1/2)/(f^(1/2)-I*x*g^(1/2)))/f^(1/2)/g^(1/2)+1/2*I* p*polylog(2,1-2*(d*x+e)*f^(1/2)*g^(1/2)/(I*d*f^(1/2)+e*g^(1/2))/(f^(1/2)-I *x*g^(1/2)))/f^(1/2)/g^(1/2)
Time = 0.16 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )+p \log \left (\frac {\sqrt {g} x}{\sqrt {-f}}\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )-p \log \left (\frac {\sqrt {g} (e+d x)}{d \sqrt {-f}+e \sqrt {g}}\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )-\log \left (c \left (d+\frac {e}{x}\right )^p\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )-p \log \left (\frac {f \sqrt {g} x}{(-f)^{3/2}}\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )+p \log \left (-\frac {\sqrt {g} (e+d x)}{d \sqrt {-f}-e \sqrt {g}}\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )-p \operatorname {PolyLog}\left (2,\frac {d \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {-f}+e \sqrt {g}}\right )+p \operatorname {PolyLog}\left (2,\frac {d \left (\sqrt {-f}+\sqrt {g} x\right )}{d \sqrt {-f}-e \sqrt {g}}\right )-p \operatorname {PolyLog}\left (2,1+\frac {\sqrt {g} x}{\sqrt {-f}}\right )+p \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {g} x}{(-f)^{3/2}}\right )}{2 \sqrt {-f} \sqrt {g}} \]
(Log[c*(d + e/x)^p]*Log[Sqrt[-f] - Sqrt[g]*x] + p*Log[(Sqrt[g]*x)/Sqrt[-f] ]*Log[Sqrt[-f] - Sqrt[g]*x] - p*Log[(Sqrt[g]*(e + d*x))/(d*Sqrt[-f] + e*Sq rt[g])]*Log[Sqrt[-f] - Sqrt[g]*x] - Log[c*(d + e/x)^p]*Log[Sqrt[-f] + Sqrt [g]*x] - p*Log[(f*Sqrt[g]*x)/(-f)^(3/2)]*Log[Sqrt[-f] + Sqrt[g]*x] + p*Log [-((Sqrt[g]*(e + d*x))/(d*Sqrt[-f] - e*Sqrt[g]))]*Log[Sqrt[-f] + Sqrt[g]*x ] - p*PolyLog[2, (d*(Sqrt[-f] - Sqrt[g]*x))/(d*Sqrt[-f] + e*Sqrt[g])] + p* PolyLog[2, (d*(Sqrt[-f] + Sqrt[g]*x))/(d*Sqrt[-f] - e*Sqrt[g])] - p*PolyLo g[2, 1 + (Sqrt[g]*x)/Sqrt[-f]] + p*PolyLog[2, 1 + (f*Sqrt[g]*x)/(-f)^(3/2) ])/(2*Sqrt[-f]*Sqrt[g])
Time = 0.55 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2920, 27, 2005, 5411, 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.
| \(\displaystyle \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx\) |
| \(\Big \downarrow \) 2920 |
| \(\displaystyle e p \int \frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f} \sqrt {g} \left (d+\frac {e}{x}\right ) x^2}dx+\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e p \int \frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\left (d+\frac {e}{x}\right ) x^2}dx}{\sqrt {f} \sqrt {g}}+\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \frac {e p \int \frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{x (e+d x)}dx}{\sqrt {f} \sqrt {g}}+\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \frac {e p \int \left (\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{e x}-\frac {d \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{e (e+d x)}\right )dx}{\sqrt {f} \sqrt {g}}+\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}+\frac {e p \left (-\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} (d x+e)}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (e \sqrt {g}+i d \sqrt {f}\right )}\right )}{e}+\frac {\arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{e}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} (e+d x)}{\left (i \sqrt {f} d+e \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 e}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 e}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 e}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 e}\right )}{\sqrt {f} \sqrt {g}}\) |
(ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[c*(d + e/x)^p])/(Sqrt[f]*Sqrt[g]) + (e*p* ((ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/e - (ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f]*Sqrt[g]*(e + d*x))/((I*d*Sqr t[f] + e*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/e + ((I/2)*PolyLog[2, ((-I)*S qrt[g]*x)/Sqrt[f]])/e - ((I/2)*PolyLog[2, (I*Sqrt[g]*x)/Sqrt[f]])/e - ((I/ 2)*PolyLog[2, 1 - (2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/e + ((I/2)*PolyLog [2, 1 - (2*Sqrt[f]*Sqrt[g]*(e + d*x))/((I*d*Sqrt[f] + e*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/e))/(Sqrt[f]*Sqrt[g])
3.3.64.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_))^(p_.)]*(b_.))/((f_) + (g_.) *(x_)^2), x_Symbol] :> With[{u = IntHide[1/(f + g*x^2), x]}, Simp[u*(a + b* Log[c*(d + e*x^n)^p]), x] - Simp[b*e*n*p Int[u*(x^(n - 1)/(d + e*x^n)), x ], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
\[\int \frac {\ln \left (c \left (d +\frac {e}{x}\right )^{p}\right )}{g \,x^{2}+f}d x\]
\[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\int { \frac {\log \left (c {\left (d + \frac {e}{x}\right )}^{p}\right )}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\frac {e p {\left (\frac {4 \, \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (d + \frac {e}{x}\right )}{e} - \frac {{\left (\pi - 2 \, \arctan \left (\frac {{\left (d^{2} x + d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f + e^{2} g}, \frac {d e g x + e^{2} g}{d^{2} f + e^{2} g}\right )\right )} \log \left (g x^{2} + f\right ) - 4 \, \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) + 2 \, \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {d^{2} g x^{2} + 2 \, d e g x + e^{2} g}{d^{2} f + e^{2} g}\right ) + 2 i \, {\rm Li}_2\left (\frac {i \, \sqrt {g} x + \sqrt {f}}{\sqrt {f}}\right ) - 2 i \, {\rm Li}_2\left (-\frac {i \, \sqrt {g} x - \sqrt {f}}{\sqrt {f}}\right ) + 2 i \, {\rm Li}_2\left (\frac {d e g x + d^{2} f - {\left (i \, d^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f + 2 i \, d e \sqrt {f} \sqrt {g} - e^{2} g}\right ) - 2 i \, {\rm Li}_2\left (\frac {d e g x + d^{2} f + {\left (i \, d^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f - 2 i \, d e \sqrt {f} \sqrt {g} - e^{2} g}\right )}{e}\right )}}{4 \, \sqrt {f g}} - \frac {p \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (d + \frac {e}{x}\right )}{\sqrt {f g}} + \frac {\arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (c {\left (d + \frac {e}{x}\right )}^{p}\right )}{\sqrt {f g}} \]
1/4*e*p*(4*arctan(g*x/sqrt(f*g))*log(d + e/x)/e - ((pi - 2*arctan2((d^2*x + d*e)*sqrt(f)*sqrt(g)/(d^2*f + e^2*g), (d*e*g*x + e^2*g)/(d^2*f + e^2*g)) )*log(g*x^2 + f) - 4*arctan(sqrt(g)*x/sqrt(f))*log(sqrt(g)*x/sqrt(f)) + 2* arctan(sqrt(g)*x/sqrt(f))*log((d^2*g*x^2 + 2*d*e*g*x + e^2*g)/(d^2*f + e^2 *g)) + 2*I*dilog((I*sqrt(g)*x + sqrt(f))/sqrt(f)) - 2*I*dilog(-(I*sqrt(g)* x - sqrt(f))/sqrt(f)) + 2*I*dilog((d*e*g*x + d^2*f - (I*d^2*x - I*d*e)*sqr t(f)*sqrt(g))/(d^2*f + 2*I*d*e*sqrt(f)*sqrt(g) - e^2*g)) - 2*I*dilog((d*e* g*x + d^2*f + (I*d^2*x - I*d*e)*sqrt(f)*sqrt(g))/(d^2*f - 2*I*d*e*sqrt(f)* sqrt(g) - e^2*g)))/e)/sqrt(f*g) - p*arctan(g*x/sqrt(f*g))*log(d + e/x)/sqr t(f*g) + arctan(g*x/sqrt(f*g))*log(c*(d + e/x)^p)/sqrt(f*g)
\[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\int { \frac {\log \left (c {\left (d + \frac {e}{x}\right )}^{p}\right )}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right )}{f+g x^2} \, dx=\int \frac {\ln \left (c\,{\left (d+\frac {e}{x}\right )}^p\right )}{g\,x^2+f} \,d x \]