\(\int \frac {\log (c (a+\frac {b}{x^2})^p)}{x (d+e x)} \, dx\) [251]

Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )}{2 d}-\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{d}-\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}+\frac {p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{d}-\frac {p \operatorname {PolyLog}\left (2,1+\frac {b}{a x^2}\right )}{2 d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{d}-\frac {2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d} \] Output:

-1/2*ln(c*(a+b/x^2)^p)*ln(-b/a/x^2)/d-ln(c*(a+b/x^2)^p)*ln(e*x+d)/d-2*p*ln 
(-e*x/d)*ln(e*x+d)/d+p*ln(e*(b^(1/2)-(-a)^(1/2)*x)/((-a)^(1/2)*d+b^(1/2)*e 
))*ln(e*x+d)/d+p*ln(-e*(b^(1/2)+(-a)^(1/2)*x)/((-a)^(1/2)*d-b^(1/2)*e))*ln 
(e*x+d)/d-1/2*p*polylog(2,1+b/a/x^2)/d+p*polylog(2,(-a)^(1/2)*(e*x+d)/((-a 
)^(1/2)*d-b^(1/2)*e))/d+p*polylog(2,(-a)^(1/2)*(e*x+d)/((-a)^(1/2)*d+b^(1/ 
2)*e))/d-2*p*polylog(2,1+e*x/d)/d
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )}{2 d}-\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{d}-\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}+\frac {p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{d}-\frac {p \operatorname {PolyLog}\left (2,\frac {a+\frac {b}{x^2}}{a}\right )}{2 d}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{d} \] Input:

Integrate[Log[c*(a + b/x^2)^p]/(x*(d + e*x)),x]
 

Output:

-1/2*(Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))])/d - (Log[c*(a + b/x^2)^p]*Lo 
g[d + e*x])/d - (2*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*(Sq 
rt[b] + Sqrt[-a]*x))/(Sqrt[-a]*d - Sqrt[b]*e))]*Log[d + e*x])/d - (p*PolyL 
og[2, (a + b/x^2)/a])/(2*d) - (2*p*PolyLog[2, (d + e*x)/d])/d + (p*PolyLog 
[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt[b]*e)])/d + (p*PolyLog[2, (Sqr 
t[-a]*(d + e*x))/(Sqrt[-a]*d + Sqrt[b]*e)])/d
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 287, 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.

\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx\)

\(\Big \downarrow \) 2916

\(\displaystyle \int \left (\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d x}-\frac {e \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d (d+e x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d}-\frac {\log \left (-\frac {b}{a x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{d}+\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{d}-\frac {p \operatorname {PolyLog}\left (2,\frac {b}{a x^2}+1\right )}{2 d}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d}-\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}\)

Input:

Int[Log[c*(a + b/x^2)^p]/(x*(d + e*x)),x]
 

Output:

-1/2*(Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))])/d - (Log[c*(a + b/x^2)^p]*Lo 
g[d + e*x])/d - (2*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*(Sq 
rt[b] + Sqrt[-a]*x))/(Sqrt[-a]*d - Sqrt[b]*e))]*Log[d + e*x])/d - (p*PolyL 
og[2, 1 + b/(a*x^2)])/(2*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 + Sqr 
t[b]*e)])/d - (2*p*PolyLog[2, 1 + (e*x)/d])/d
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2916

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]
 

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.26

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (x \right )}{d}+2 p b \left (\frac {\frac {\ln \left (x \right )^{2}}{2 b}-\frac {\left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-a x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {a x +\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {-a x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {a x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 a}\right ) a}{b}}{d}-\frac {\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b}-\frac {\left (\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\ln \left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )}{2 a}\right ) a}{b}}{d}\right )\)	\(362\)

Input:

int(ln(c*(a+b/x^2)^p)/x/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

-ln(c*(a+b/x^2)^p)*ln(e*x+d)/d+ln(c*(a+b/x^2)^p)/d*ln(x)+2*p*b*(1/d*(1/2*l 
n(x)^2/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)/b*a)-1/d*((dilog(-e*x/d)+ln(e*x+d)*ln 
(-e*x/d))/b-(1/2*ln(e*x+d)*(ln((e*(-a*b)^(1/2)+d*a-a*(e*x+d))/(e*(-a*b)^(1 
/2)+d*a))+ln((e*(-a*b)^(1/2)-d*a+a*(e*x+d))/(e*(-a*b)^(1/2)-d*a)))/a+1/2*( 
dilog((e*(-a*b)^(1/2)+d*a-a*(e*x+d))/(e*(-a*b)^(1/2)+d*a))+dilog((e*(-a*b) 
^(1/2)-d*a+a*(e*x+d))/(e*(-a*b)^(1/2)-d*a)))/a)/b*a))
 

Fricas [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:

integrate(log(c*(a+b/x^2)^p)/x/(e*x+d),x, algorithm="fricas")
 

Output:

integral(log(c*((a*x^2 + b)/x^2)^p)/(e*x^2 + d*x), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\text {Timed out} \] Input:

integrate(ln(c*(a+b/x**2)**p)/x/(e*x+d),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:

integrate(log(c*(a+b/x^2)^p)/x/(e*x+d),x, algorithm="maxima")
 

Output:

integrate(log((a + b/x^2)^p*c)/((e*x + d)*x), x)
 

Giac [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:

integrate(log(c*(a+b/x^2)^p)/x/(e*x+d),x, algorithm="giac")
 

Output:

integrate(log((a + b/x^2)^p*c)/((e*x + d)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{x\,\left (d+e\,x\right )} \,d x \] Input:

int(log(c*(a + b/x^2)^p)/(x*(d + e*x)),x)
 

Output:

int(log(c*(a + b/x^2)^p)/(x*(d + e*x)), x)
 

Reduce [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x (d+e x)} \, dx=\int \frac {\mathrm {log}\left (\frac {\left (a \,x^{2}+b \right )^{p} c}{x^{2 p}}\right )}{e \,x^{2}+d x}d x \] Input:

int(log(c*(a+b/x^2)^p)/x/(e*x+d),x)
 

Output:

int(log(((a*x**2 + b)**p*c)/x**(2*p))/(d*x + e*x**2),x)