Integrand size = 23, antiderivative size = 414 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {p}{2 d x^2}-\frac {2 e p}{d^2 x}-\frac {2 \sqrt {a} e p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {b} d^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 b d}+\frac {e \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^2 x}-\frac {e^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )}{2 d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{d^3}-\frac {2 e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x^2}\right )}{2 d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{d^3}-\frac {2 e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^3} \]
1/2*p/d/x^2-2*e*p/d^2/x-1/2*(a+b/x^2)*ln(c*(a+b/x^2)^p)/b/d+e*ln(c*(a+b/x^ 2)^p)/d^2/x-1/2*e^2*ln(c*(a+b/x^2)^p)*ln(-b/a/x^2)/d^3-e^2*ln(c*(a+b/x^2)^ p)*ln(e*x+d)/d^3-2*e^2*p*ln(-e*x/d)*ln(e*x+d)/d^3+e^2*p*ln(e*x+d)*ln(-e*(x *(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)-e*b^(1/2)))/d^3+e^2*p*ln(e*x+d)*ln(e*(- x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)+e*b^(1/2)))/d^3-1/2*e^2*p*polylog(2,1+ b/a/x^2)/d^3-2*e^2*p*polylog(2,1+e*x/d)/d^3+e^2*p*polylog(2,(e*x+d)*(-a)^( 1/2)/(d*(-a)^(1/2)-e*b^(1/2)))/d^3+e^2*p*polylog(2,(e*x+d)*(-a)^(1/2)/(d*( -a)^(1/2)+e*b^(1/2)))/d^3-2*e*p*arctan(x*a^(1/2)/b^(1/2))*a^(1/2)/d^2/b^(1 /2)
Time = 0.17 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.93 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {-\frac {4 d e p}{x}+\frac {4 \sqrt {a} d e p \arctan \left (\frac {\sqrt {b}}{\sqrt {a} x}\right )}{\sqrt {b}}+\frac {2 d e \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x}+d^2 \left (\frac {p}{x^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{b}\right )-e^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )-2 e^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)-4 e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)+2 e^2 p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+2 e^2 p \log \left (\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{-\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)-e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x^2}\right )+2 e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )+2 e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )-4 e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 d^3} \]
((-4*d*e*p)/x + (4*Sqrt[a]*d*e*p*ArcTan[Sqrt[b]/(Sqrt[a]*x)])/Sqrt[b] + (2 *d*e*Log[c*(a + b/x^2)^p])/x + d^2*(p/x^2 - ((a + b/x^2)*Log[c*(a + b/x^2) ^p])/b) - e^2*Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))] - 2*e^2*Log[c*(a + b/ x^2)^p]*Log[d + e*x] - 4*e^2*p*Log[-((e*x)/d)]*Log[d + e*x] + 2*e^2*p*Log[ (e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x] + 2*e^2* p*Log[(e*(Sqrt[b] + Sqrt[-a]*x))/(-(Sqrt[-a]*d) + Sqrt[b]*e)]*Log[d + e*x] - e^2*p*PolyLog[2, 1 + b/(a*x^2)] + 2*e^2*p*PolyLog[2, (Sqrt[-a]*(d + e*x ))/(Sqrt[-a]*d - Sqrt[b]*e)] + 2*e^2*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sq rt[-a]*d + Sqrt[b]*e)] - 4*e^2*p*PolyLog[2, 1 + (e*x)/d])/(2*d^3)
Time = 0.76 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
| \(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx\) |
| \(\Big \downarrow \) 2916 |
| \(\displaystyle \int \left (-\frac {e^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^3 (d+e x)}+\frac {e^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^3 x}-\frac {e \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^2 x^2}+\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {a} e p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {b} d^2}-\frac {e^2 \log \left (-\frac {b}{a x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^3}+\frac {e \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d^2 x}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 b d}-\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {b}{a x^2}+1\right )}{2 d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{d^3}-\frac {2 e^2 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^3}-\frac {2 e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}-\frac {2 e p}{d^2 x}+\frac {p}{2 d x^2}\) |
p/(2*d*x^2) - (2*e*p)/(d^2*x) - (2*Sqrt[a]*e*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]] )/(Sqrt[b]*d^2) - ((a + b/x^2)*Log[c*(a + b/x^2)^p])/(2*b*d) + (e*Log[c*(a + b/x^2)^p])/(d^2*x) - (e^2*Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))])/(2*d^ 3) - (e^2*Log[c*(a + b/x^2)^p]*Log[d + e*x])/d^3 - (2*e^2*p*Log[-((e*x)/d) ]*Log[d + e*x])/d^3 + (e^2*p*Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x])/d^3 + (e^2*p*Log[-((e*(Sqrt[b] + Sqrt[-a]*x))/(S qrt[-a]*d - Sqrt[b]*e))]*Log[d + e*x])/d^3 - (e^2*p*PolyLog[2, 1 + b/(a*x^ 2)])/(2*d^3) + (e^2*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt[b ]*e)])/d^3 + (e^2*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d + Sqrt[b]* e)])/d^3 - (2*e^2*p*PolyLog[2, 1 + (e*x)/d])/d^3
3.3.53.3.1 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 = 1.71 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.18
| method | result | size |
| parts | \(-\frac {e^{2} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {e \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{d^{2} x}+p b \left (\frac {1}{2 d b \,x^{2}}-\frac {2 e}{d^{2} b x}+\frac {a \ln \left (x \right )}{d \,b^{2}}-\frac {a \ln \left (x^{2} a +b \right )}{2 d \,b^{2}}-\frac {2 a e \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{d^{2} b \sqrt {a b}}+\frac {2 e^{2} \left (\frac {\ln \left (x \right )^{2}}{2 b}-\frac {a \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-a x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {a x +\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {-a x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {a x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 a}\right )}{b}\right )}{d^{3}}-\frac {2 e^{2} \left (-\frac {a \left (\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\ln \left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )}{2 a}\right )}{b}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b}\right )}{d^{3}}\right )\) | \(488\) |
-e^2*ln(c*(a+b/x^2)^p)*ln(e*x+d)/d^3-1/2*ln(c*(a+b/x^2)^p)/d/x^2+ln(c*(a+b /x^2)^p)*e^2/d^3*ln(x)+e*ln(c*(a+b/x^2)^p)/d^2/x+p*b*(1/2/d/b/x^2-2/d^2/b* e/x+1/d*a/b^2*ln(x)-1/2/d/b^2*a*ln(a*x^2+b)-2/d^2/b*a*e/(a*b)^(1/2)*arctan (a*x/(a*b)^(1/2))+2*e^2/d^3*(1/2/b*ln(x)^2-a/b*(1/2*ln(x)*(ln((-a*x+(-a*b) ^(1/2))/(-a*b)^(1/2))+ln((a*x+(-a*b)^(1/2))/(-a*b)^(1/2)))/a+1/2*(dilog((- a*x+(-a*b)^(1/2))/(-a*b)^(1/2))+dilog((a*x+(-a*b)^(1/2))/(-a*b)^(1/2)))/a) )-2*e^2/d^3*(-a/b*(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)+1/b*(dilog(-e*x/d)+l n(e*x+d)*ln(-e*x/d))))
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \]