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

3.3.51.1 Optimal result
3.3.51.2 Mathematica [A] (verified)
3.3.51.3 Rubi [A] (verified)
3.3.51.4 Maple [A] (verified)
3.3.51.5 Fricas [F]
3.3.51.6 Sympy [F(-1)]
3.3.51.7 Maxima [F]
3.3.51.8 Giac [F]
3.3.51.9 Mupad [F(-1)]

3.3.51.1 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*x+d)*ln(-e*(x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2 
)-e*b^(1/2)))/d+p*ln(e*x+d)*ln(e*(-x*(-a)^(1/2)+b^(1/2))/(d*(-a)^(1/2)+e*b 
^(1/2)))/d-1/2*p*polylog(2,1+b/a/x^2)/d-2*p*polylog(2,1+e*x/d)/d+p*polylog 
(2,(e*x+d)*(-a)^(1/2)/(d*(-a)^(1/2)-e*b^(1/2)))/d+p*polylog(2,(e*x+d)*(-a) 
^(1/2)/(d*(-a)^(1/2)+e*b^(1/2)))/d
 
3.3.51.2 Mathematica [A] (verified)

Time = 0.06 (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
 
3.3.51.3 Rubi [A] (verified)

Time = 0.65 (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
 

3.3.51.3.1 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]
 
3.3.51.4 Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.26

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

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/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))+di 
log((a*x+(-a*b)^(1/2))/(-a*b)^(1/2)))/a))-1/d*(-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)+ln(e*x+d)*ln(-e*x/d))))
 
3.3.51.5 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)
 
3.3.51.6 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
 
3.3.51.7 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)
 
3.3.51.8 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)
 
3.3.51.9 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)