\(\int \frac {x \log (c (a+b x^2)^p)}{d+e x} \, dx\) [228]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 256 \[ \int \frac {x \log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=-\frac {2 p x}{e}+\frac {2 \sqrt {a} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} e}+\frac {d p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{e^2}+\frac {x \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac {d \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.93 \[ \int \frac {x \log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=\frac {-2 e p x+\frac {2 \sqrt {a} e p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}+d p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)+d p \log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{-\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)+e x \log \left (c \left (a+b x^2\right )^p\right )-d \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+d p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e^2} \] Input:

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

Output:

(-2*e*p*x + (2*Sqrt[a]*e*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b] + d*p*Log[ 
(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d + e*x] + d*p*Lo 
g[(e*(Sqrt[-a] + Sqrt[b]*x))/(-(Sqrt[b]*d) + Sqrt[-a]*e)]*Log[d + e*x] + e 
*x*Log[c*(a + b*x^2)^p] - d*Log[d + e*x]*Log[c*(a + b*x^2)^p] + d*p*PolyLo 
g[2, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)] + d*p*PolyLog[2, (Sqrt[ 
b]*(d + e*x))/(Sqrt[b]*d + Sqrt[-a]*e)])/e^2
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 256, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*p*x)/e + (2*Sqrt[a]*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*e) + (d*p* 
Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d + e*x])/e^2 
 + (d*p*Log[-((e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*d - Sqrt[-a]*e))]*Log[d 
+ e*x])/e^2 + (x*Log[c*(a + b*x^2)^p])/e - (d*Log[d + e*x]*Log[c*(a + b*x^ 
2)^p])/e^2 + (d*p*PolyLog[2, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)] 
)/e^2 + (d*p*PolyLog[2, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d + Sqrt[-a]*e)])/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 = 1.69 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03

method result size
parts \(\frac {x \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{e}-\frac {d \ln \left (e x +d \right ) \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{e^{2}}-\frac {2 p b \left (\frac {e x +d}{b}-\frac {a e \arctan \left (\frac {2 \left (e x +d \right ) b -2 b d}{2 e \sqrt {a b}}\right )}{b \sqrt {a b}}+d \left (-\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )+\ln \left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )\right )}{2 b}-\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{2 b}\right )\right )}{e^{2}}\) \(263\)
risch \(\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) x}{e}-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {2 p x}{e}-\frac {2 p d}{e^{2}}+\frac {2 p a \arctan \left (\frac {2 \left (e x +d \right ) b -2 b d}{2 e \sqrt {a b}}\right )}{e \sqrt {a b}}+\frac {p d \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )}{e^{2}}+\frac {p d \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{e^{2}}+\frac {p d \operatorname {dilog}\left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )}{e^{2}}+\frac {p d \operatorname {dilog}\left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{e^{2}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\) \(407\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int \frac {x \log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=\frac {8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) e \,p^{2}-4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b e \,x^{3}+b d \,x^{2}+a e x +a d}d x \right ) a b d e p +4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) x}{b e \,x^{3}+b d \,x^{2}+a e x +a d}d x \right ) b^{2} d^{2} p -{\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} b d +4 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b e p x -8 b e \,p^{2} x}{4 b \,e^{2} p} \] Input:

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

Output:

(8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*e*p**2 - 4*int(log((a + b 
*x**2)**p*c)/(a*d + a*e*x + b*d*x**2 + b*e*x**3),x)*a*b*d*e*p + 4*int((log 
((a + b*x**2)**p*c)*x)/(a*d + a*e*x + b*d*x**2 + b*e*x**3),x)*b**2*d**2*p 
- log((a + b*x**2)**p*c)**2*b*d + 4*log((a + b*x**2)**p*c)*b*e*p*x - 8*b*e 
*p**2*x)/(4*b*e**2*p)