Integrand size = 20, antiderivative size = 229 \[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\frac {\log \left (c (d+e x)^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}}-\frac {\log \left (c (d+e x)^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}}-\frac {p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}} \]
1/2*ln(c*(e*x+d)^p)*ln(e*((-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2)+d*g^(1/2)))/ (-f)^(1/2)/g^(1/2)-1/2*ln(c*(e*x+d)^p)*ln(e*((-f)^(1/2)+x*g^(1/2))/(e*(-f) ^(1/2)-d*g^(1/2)))/(-f)^(1/2)/g^(1/2)-1/2*p*polylog(2,-(e*x+d)*g^(1/2)/(e* (-f)^(1/2)-d*g^(1/2)))/(-f)^(1/2)/g^(1/2)+1/2*p*polylog(2,(e*x+d)*g^(1/2)/ (e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(1/2)/g^(1/2)
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.78 \[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\frac {\log \left (c (d+e x)^p\right ) \left (\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )-\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )\right )-p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}} \]
(Log[c*(d + e*x)^p]*(Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g ])] - Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])]) - p*PolyLo g[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + p*PolyLog[2, (Sqrt [g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*Sqrt[-f]*Sqrt[g])
Time = 0.43 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2856, 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.
| \(\displaystyle \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx\) |
| \(\Big \downarrow \) 2856 |
| \(\displaystyle \int \left (\frac {\sqrt {-f} \log \left (c (d+e x)^p\right )}{2 f \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {\sqrt {-f} \log \left (c (d+e x)^p\right )}{2 f \left (\sqrt {-f}+\sqrt {g} x\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log \left (c (d+e x)^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 \sqrt {-f} \sqrt {g}}-\frac {\log \left (c (d+e x)^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}}-\frac {p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 \sqrt {-f} \sqrt {g}}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 \sqrt {-f} \sqrt {g}}\) |
(Log[c*(d + e*x)^p]*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g] )])/(2*Sqrt[-f]*Sqrt[g]) - (Log[c*(d + e*x)^p]*Log[(e*(Sqrt[-f] + Sqrt[g]* x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*Sqrt[-f]*Sqrt[g]) - (p*PolyLog[2, -((Sqr t[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*Sqrt[-f]*Sqrt[g]) + (p*Poly Log[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*Sqrt[-f]*Sqrt[g])
3.3.63.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.92 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.71
| method | result | size |
| risch | \(-\frac {\arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) p \ln \left (e x +d \right )}{\sqrt {f g}}+\frac {\arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) \ln \left (\left (e x +d \right )^{p}\right )}{\sqrt {f g}}+\frac {p \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 \sqrt {-f g}}-\frac {p \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 \sqrt {-f g}}+\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 \sqrt {-f g}}-\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 \sqrt {-f g}}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{\sqrt {f g}}\) | \(392\) |
-1/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*p*ln(e*x+d)+1 /(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*ln((e*x+d)^p)+1 /2*p*ln(e*x+d)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1 /2)+d*g))-1/2*p*ln(e*x+d)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/( e*(-f*g)^(1/2)-d*g))+1/2*p/(-f*g)^(1/2)*dilog((e*(-f*g)^(1/2)-g*(e*x+d)+d* g)/(e*(-f*g)^(1/2)+d*g))-1/2*p/(-f*g)^(1/2)*dilog((e*(-f*g)^(1/2)+g*(e*x+d )-d*g)/(e*(-f*g)^(1/2)-d*g))+(1/2*I*Pi*csgn(I*(e*x+d)^p)*csgn(I*c*(e*x+d)^ p)^2-1/2*I*Pi*csgn(I*(e*x+d)^p)*csgn(I*c*(e*x+d)^p)*csgn(I*c)-1/2*I*Pi*csg n(I*c*(e*x+d)^p)^3+1/2*I*Pi*csgn(I*c*(e*x+d)^p)^2*csgn(I*c)+ln(c))/(f*g)^( 1/2)*arctan(g*x/(f*g)^(1/2))
\[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\int { \frac {\log \left ({\left (e x + d\right )}^{p} c\right )}{g x^{2} + f} \,d x } \]
\[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\int \frac {\log {\left (c \left (d + e x\right )^{p} \right )}}{f + g x^{2}}\, dx \]
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.35 \[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\frac {e p {\left (\frac {2 \, \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (e x + d\right )}{e} + \frac {\arctan \left (\frac {{\left (e^{2} x + d e\right )} \sqrt {f} \sqrt {g}}{e^{2} f + d^{2} g}, \frac {d e g x + d^{2} g}{e^{2} f + d^{2} g}\right ) \log \left (g x^{2} + f\right ) - \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {e^{2} g x^{2} + 2 \, d e g x + d^{2} g}{e^{2} f + d^{2} g}\right ) - i \, {\rm Li}_2\left (\frac {d e g x + e^{2} f - {\left (i \, e^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{e^{2} f + 2 i \, d e \sqrt {f} \sqrt {g} - d^{2} g}\right ) + i \, {\rm Li}_2\left (\frac {d e g x + e^{2} f + {\left (i \, e^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{e^{2} f - 2 i \, d e \sqrt {f} \sqrt {g} - d^{2} g}\right )}{e}\right )}}{2 \, \sqrt {f g}} - \frac {p \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (e x + d\right )}{\sqrt {f g}} + \frac {\arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left ({\left (e x + d\right )}^{p} c\right )}{\sqrt {f g}} \]
1/2*e*p*(2*arctan(g*x/sqrt(f*g))*log(e*x + d)/e + (arctan2((e^2*x + d*e)*s qrt(f)*sqrt(g)/(e^2*f + d^2*g), (d*e*g*x + d^2*g)/(e^2*f + d^2*g))*log(g*x ^2 + f) - arctan(sqrt(g)*x/sqrt(f))*log((e^2*g*x^2 + 2*d*e*g*x + d^2*g)/(e ^2*f + d^2*g)) - I*dilog((d*e*g*x + e^2*f - (I*e^2*x - I*d*e)*sqrt(f)*sqrt (g))/(e^2*f + 2*I*d*e*sqrt(f)*sqrt(g) - d^2*g)) + I*dilog((d*e*g*x + e^2*f + (I*e^2*x - I*d*e)*sqrt(f)*sqrt(g))/(e^2*f - 2*I*d*e*sqrt(f)*sqrt(g) - d ^2*g)))/e)/sqrt(f*g) - p*arctan(g*x/sqrt(f*g))*log(e*x + d)/sqrt(f*g) + ar ctan(g*x/sqrt(f*g))*log((e*x + d)^p*c)/sqrt(f*g)
\[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\int { \frac {\log \left ({\left (e x + d\right )}^{p} c\right )}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {\log \left (c (d+e x)^p\right )}{f+g x^2} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x\right )}^p\right )}{g\,x^2+f} \,d x \]