\(\int \frac {x \log (c (a+\frac {b}{x^2})^p)}{d+e x} \, dx\) [249]

Optimal result

Integrand size = 21, antiderivative size = 291 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {2 \sqrt {b} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} e}+\frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {d \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{e^2}-\frac {2 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^2}-\frac {2 d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.93 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {-\frac {2 \sqrt {b} e p \arctan \left (\frac {\sqrt {b}}{\sqrt {a} x}\right )}{\sqrt {a}}+e x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-d \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)-2 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)+d p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+d p \log \left (\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{-\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )-2 d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \] Input:

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

Output:

((-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
 

Rubi [A] (verified)

Time = 1.05 (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

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

Input:

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

Output:

(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
 

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 = 1.91 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.02

method	result	size
parts	\(\frac {x \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{e}-\frac {d \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}+2 p b \,e^{2} \left (\frac {\arctan \left (\frac {-2 d a +2 a \left (e x +d \right )}{2 e \sqrt {a b}}\right )}{e^{3} \sqrt {a b}}+\frac {d \left (-\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{2}}-\frac {\left (-\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\ln \left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )\right )}{2 a}-\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )}{2 a}\right ) a}{b \,e^{2}}\right )}{e^{2}}\right )\)	\(296\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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