Integrand size = 21, antiderivative size = 291 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {2 \sqrt {b} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} e}+\frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {d \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{e^2}-\frac {2 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^2}-\frac {2 d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \]
x*ln(c*(a+b/x^2)^p)/e-d*ln(c*(a+b/x^2)^p)*ln(e*x+d)/e^2-2*d*p*ln(-e*x/d)*l n(e*x+d)/e^2+d*p*ln(e*x+d)*ln(-e*(x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)-e*b^ (1/2)))/e^2+d*p*ln(e*x+d)*ln(e*(-x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)+e*b^( 1/2)))/e^2-2*d*p*polylog(2,1+e*x/d)/e^2+d*p*polylog(2,(e*x+d)*(-a)^(1/2)/( d*(-a)^(1/2)-e*b^(1/2)))/e^2+d*p*polylog(2,(e*x+d)*(-a)^(1/2)/(d*(-a)^(1/2 )+e*b^(1/2)))/e^2+2*p*arctan(x*a^(1/2)/b^(1/2))*b^(1/2)/e/a^(1/2)
Time = 0.11 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.93 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {-\frac {2 \sqrt {b} e p \arctan \left (\frac {\sqrt {b}}{\sqrt {a} x}\right )}{\sqrt {a}}+e x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-d \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)-2 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)+d p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+d p \log \left (\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{-\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )-2 d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \]
((-2*Sqrt[b]*e*p*ArcTan[Sqrt[b]/(Sqrt[a]*x)])/Sqrt[a] + e*x*Log[c*(a + b/x ^2)^p] - d*Log[c*(a + b/x^2)^p]*Log[d + e*x] - 2*d*p*Log[-((e*x)/d)]*Log[d + e*x] + d*p*Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log [d + e*x] + d*p*Log[(e*(Sqrt[b] + Sqrt[-a]*x))/(-(Sqrt[-a]*d) + Sqrt[b]*e) ]*Log[d + e*x] + d*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt[b] *e)] + d*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d + Sqrt[b]*e)] - 2*d *p*PolyLog[2, 1 + (e*x)/d])/e^2
Time = 0.61 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2916, 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.
| \(\displaystyle \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx\) |
| \(\Big \downarrow \) 2916 |
| \(\displaystyle \int \left (\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {d \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {b} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} e}-\frac {d \log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^2}+\frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^2}+\frac {d p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^2}+\frac {d p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^2}-\frac {2 d p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e^2}-\frac {2 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}\) |
(2*Sqrt[b]*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(Sqrt[a]*e) + (x*Log[c*(a + b/x^ 2)^p])/e - (d*Log[c*(a + b/x^2)^p]*Log[d + e*x])/e^2 - (2*d*p*Log[-((e*x)/ d)]*Log[d + e*x])/e^2 + (d*p*Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x])/e^2 + (d*p*Log[-((e*(Sqrt[b] + Sqrt[-a]*x))/(Sqr t[-a]*d - Sqrt[b]*e))]*Log[d + e*x])/e^2 + (d*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt[b]*e)])/e^2 + (d*p*PolyLog[2, (Sqrt[-a]*(d + e*x) )/(Sqrt[-a]*d + Sqrt[b]*e)])/e^2 - (2*d*p*PolyLog[2, 1 + (e*x)/d])/e^2
3.3.49.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log [c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g , n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
Time = 1.36 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.02
| method | result | size |
| parts | \(\frac {x \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{e}-\frac {d \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}+2 p b \,e^{2} \left (\frac {\arctan \left (\frac {-2 a d +2 a \left (e x +d \right )}{2 e \sqrt {a b}}\right )}{e^{3} \sqrt {a b}}+\frac {d \left (-\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{2}}-\frac {a \left (-\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\ln \left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )\right )}{2 a}-\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )}{2 a}\right )}{b \,e^{2}}\right )}{e^{2}}\right )\) | \(296\) |
x*ln(c*(a+b/x^2)^p)/e-d*ln(c*(a+b/x^2)^p)*ln(e*x+d)/e^2+2*p*b*e^2*(1/e^3/( a*b)^(1/2)*arctan(1/2*(-2*a*d+2*a*(e*x+d))/e/(a*b)^(1/2))+1/e^2*d*(-1/b/e^ 2*(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))-a/b/e^2*(-1/2*ln(e*x+d)*(ln((e*(-a* b)^(1/2)+a*d-a*(e*x+d))/(e*(-a*b)^(1/2)+a*d))+ln((e*(-a*b)^(1/2)-a*d+a*(e* x+d))/(e*(-a*b)^(1/2)-a*d)))/a-1/2*(dilog((e*(-a*b)^(1/2)+a*d-a*(e*x+d))/( e*(-a*b)^(1/2)+a*d))+dilog((e*(-a*b)^(1/2)-a*d+a*(e*x+d))/(e*(-a*b)^(1/2)- a*d)))/a)))
\[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\text {Timed out} \]
\[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
\[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int \frac {x\,\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{d+e\,x} \,d x \]