Integrand size = 20, antiderivative size = 241 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{e}+\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}+\frac {2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e} \]
ln(c*(a+b/x^2)^p)*ln(e*x+d)/e+2*p*ln(-e*x/d)*ln(e*x+d)/e-p*ln(e*x+d)*ln(-e *(x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)-e*b^(1/2)))/e-p*ln(e*x+d)*ln(e*(-x*( -a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)+e*b^(1/2)))/e+2*p*polylog(2,1+e*x/d)/e-p* polylog(2,(e*x+d)*(-a)^(1/2)/(d*(-a)^(1/2)-e*b^(1/2)))/e-p*polylog(2,(e*x+ d)*(-a)^(1/2)/(d*(-a)^(1/2)+e*b^(1/2)))/e
Time = 0.04 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{e}+\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{e}+\frac {2 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e} \]
(Log[c*(a + b/x^2)^p]*Log[d + e*x])/e + (2*p*Log[-((e*x)/d)]*Log[d + e*x]) /e - (p*Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e *x])/e - (p*Log[-((e*(Sqrt[b] + Sqrt[-a]*x))/(Sqrt[-a]*d - Sqrt[b]*e))]*Lo g[d + e*x])/e + (2*p*PolyLog[2, (d + e*x)/d])/e - (p*PolyLog[2, (Sqrt[-a]* (d + e*x))/(Sqrt[-a]*d - Sqrt[b]*e)])/e - (p*PolyLog[2, (Sqrt[-a]*(d + e*x ))/(Sqrt[-a]*d + Sqrt[b]*e)])/e
Time = 0.59 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2912, 2005, 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.
| \(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx\) |
| \(\Big \downarrow \) 2912 |
| \(\displaystyle \frac {2 b p \int \frac {\log (d+e x)}{\left (a+\frac {b}{x^2}\right ) x^3}dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \frac {2 b p \int \frac {\log (d+e x)}{x \left (a x^2+b\right )}dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \frac {2 b p \int \left (\frac {\log (d+e x)}{b x}-\frac {a x \log (d+e x)}{b \left (a x^2+b\right )}\right )dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}+\frac {2 b p \left (-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{2 b}-\frac {\log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{2 b}-\frac {\log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{2 b}+\frac {\operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{b}+\frac {\log \left (-\frac {e x}{d}\right ) \log (d+e x)}{b}\right )}{e}\) |
(Log[c*(a + b/x^2)^p]*Log[d + e*x])/e + (2*b*p*((Log[-((e*x)/d)]*Log[d + e *x])/b - (Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x])/(2*b) - (Log[-((e*(Sqrt[b] + Sqrt[-a]*x))/(Sqrt[-a]*d - Sqrt[b]*e)) ]*Log[d + e*x])/(2*b) - PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt [b]*e)]/(2*b) - PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d + Sqrt[b]*e)]/ (2*b) + PolyLog[2, 1 + (e*x)/d]/b))/e
3.3.50.3.1 Defintions of rubi rules used
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_.)]*(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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_. )*(x_)), x_Symbol] :> Simp[Log[f + g*x]*((a + b*Log[c*(d + e*x^n)^p])/g), x ] - Simp[b*e*n*(p/g) Int[x^(n - 1)*(Log[f + g*x]/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]
Time = 1.30 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.97
| method | result | size |
| parts | \(\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{e}+2 p b e \left (\frac {a \left (-\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\ln \left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )\right )}{2 a}-\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+a d -a \left (e x +d \right )}{e \sqrt {-a b}+a d}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-a d +a \left (e x +d \right )}{e \sqrt {-a b}-a d}\right )}{2 a}\right )}{b \,e^{2}}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{2}}\right )\) | \(234\) |
ln(c*(a+b/x^2)^p)*ln(e*x+d)/e+2*p*b*e*(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)+1/b/e^2*(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d)))
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int \frac {\log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{d + e x}\, dx \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{d+e\,x} \,d x \]