\(\int \frac {x^5 (a+b \log (c (d f-d g x^2)^p))}{\sqrt {f-g x^2}} \, dx\) [703]

Optimal result

Integrand size = 36, antiderivative size = 185 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {2 b f^2 p \sqrt {f-g x^2}}{g^3}-\frac {4 b f p \left (f-g x^2\right )^{3/2}}{9 g^3}+\frac {2 b p \left (f-g x^2\right )^{5/2}}{25 g^3}-\frac {f^2 \sqrt {f-g x^2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{g^3}+\frac {2 f \left (f-g x^2\right )^{3/2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{3 g^3}-\frac {\left (f-g x^2\right )^{5/2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{5 g^3} \] Output:

2*b*f^2*p*(-g*x^2+f)^(1/2)/g^3-4/9*b*f*p*(-g*x^2+f)^(3/2)/g^3+2/25*b*p*(-g 
*x^2+f)^(5/2)/g^3-f^2*(-g*x^2+f)^(1/2)*(a+b*ln(c*(-d*g*x^2+d*f)^p))/g^3+2/ 
3*f*(-g*x^2+f)^(3/2)*(a+b*ln(c*(-d*g*x^2+d*f)^p))/g^3-1/5*(-g*x^2+f)^(5/2) 
*(a+b*ln(c*(-d*g*x^2+d*f)^p))/g^3
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.58 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {\sqrt {f-g x^2} \left (-15 a \left (8 f^2+4 f g x^2+3 g^2 x^4\right )+2 b p \left (184 f^2+32 f g x^2+9 g^2 x^4\right )-15 b \left (8 f^2+4 f g x^2+3 g^2 x^4\right ) \log \left (c \left (d \left (f-g x^2\right )\right )^p\right )\right )}{225 g^3} \] Input:

Integrate[(x^5*(a + b*Log[c*(d*f - d*g*x^2)^p]))/Sqrt[f - g*x^2],x]
 

Output:

(Sqrt[f - g*x^2]*(-15*a*(8*f^2 + 4*f*g*x^2 + 3*g^2*x^4) + 2*b*p*(184*f^2 + 
 32*f*g*x^2 + 9*g^2*x^4) - 15*b*(8*f^2 + 4*f*g*x^2 + 3*g^2*x^4)*Log[c*(d*( 
f - g*x^2))^p]))/(225*g^3)
 

Rubi [F]

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.

\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx\)

\(\Big \downarrow \) 2929

\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}}dx\)

Input:

Int[(x^5*(a + b*Log[c*(d*f - d*g*x^2)^p]))/Sqrt[f - g*x^2],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2929

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((h_.)* 
(x_))^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Unintegrable[(h*x) 
^m*(f + g*x^s)^r*(a + b*Log[c*(d + e*x^n)^p])^q, x] /; FreeQ[{a, b, c, d, e 
, f, g, h, m, n, p, q, r, s}, x]
 

Maple [F]

\[\int \frac {x^{5} \left (a +b \ln \left (c \left (-d g \,x^{2}+d f \right )^{p}\right )\right )}{\sqrt {-g \,x^{2}+f}}d x\]

Input:

int(x^5*(a+b*ln(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Output:

int(x^5*(a+b*ln(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.72 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {{\left (9 \, {\left (2 \, b g^{2} p - 5 \, a g^{2}\right )} x^{4} + 368 \, b f^{2} p - 120 \, a f^{2} + 4 \, {\left (16 \, b f g p - 15 \, a f g\right )} x^{2} - 15 \, {\left (3 \, b g^{2} p x^{4} + 4 \, b f g p x^{2} + 8 \, b f^{2} p\right )} \log \left (-d g x^{2} + d f\right ) - 15 \, {\left (3 \, b g^{2} x^{4} + 4 \, b f g x^{2} + 8 \, b f^{2}\right )} \log \left (c\right )\right )} \sqrt {-g x^{2} + f}}{225 \, g^{3}} \] Input:

integrate(x^5*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"fricas")
 

Output:

1/225*(9*(2*b*g^2*p - 5*a*g^2)*x^4 + 368*b*f^2*p - 120*a*f^2 + 4*(16*b*f*g 
*p - 15*a*f*g)*x^2 - 15*(3*b*g^2*p*x^4 + 4*b*f*g*p*x^2 + 8*b*f^2*p)*log(-d 
*g*x^2 + d*f) - 15*(3*b*g^2*x^4 + 4*b*f*g*x^2 + 8*b*f^2)*log(c))*sqrt(-g*x 
^2 + f)/g^3
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\text {Timed out} \] Input:

integrate(x**5*(a+b*ln(c*(-d*g*x**2+d*f)**p))/(-g*x**2+f)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=-\frac {1}{15} \, {\left (\frac {3 \, \sqrt {-g x^{2} + f} x^{4}}{g} + \frac {4 \, \sqrt {-g x^{2} + f} f x^{2}}{g^{2}} + \frac {8 \, \sqrt {-g x^{2} + f} f^{2}}{g^{3}}\right )} b \log \left ({\left (-d g x^{2} + d f\right )}^{p} c\right ) - \frac {1}{15} \, {\left (\frac {3 \, \sqrt {-g x^{2} + f} x^{4}}{g} + \frac {4 \, \sqrt {-g x^{2} + f} f x^{2}}{g^{2}} + \frac {8 \, \sqrt {-g x^{2} + f} f^{2}}{g^{3}}\right )} a + \frac {2 \, {\left (9 \, {\left (-g x^{2} + f\right )}^{\frac {5}{2}} - 50 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f + 225 \, \sqrt {-g x^{2} + f} f^{2}\right )} b p}{225 \, g^{3}} \] Input:

integrate(x^5*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"maxima")
 

Output:

-1/15*(3*sqrt(-g*x^2 + f)*x^4/g + 4*sqrt(-g*x^2 + f)*f*x^2/g^2 + 8*sqrt(-g 
*x^2 + f)*f^2/g^3)*b*log((-d*g*x^2 + d*f)^p*c) - 1/15*(3*sqrt(-g*x^2 + f)* 
x^4/g + 4*sqrt(-g*x^2 + f)*f*x^2/g^2 + 8*sqrt(-g*x^2 + f)*f^2/g^3)*a + 2/2 
25*(9*(-g*x^2 + f)^(5/2) - 50*(-g*x^2 + f)^(3/2)*f + 225*sqrt(-g*x^2 + f)* 
f^2)*b*p/g^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.37 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=-\frac {b p {\left (\frac {15 \, {\left (3 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} - 10 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {-g x^{2} + f} f^{2}\right )} \log \left (-d g x^{2} + d f\right )}{g^{2}} - \frac {2 \, {\left (9 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} - 50 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f + 225 \, \sqrt {-g x^{2} + f} f^{2}\right )}}{g^{2}}\right )} + \frac {15 \, {\left (3 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} - 10 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {-g x^{2} + f} f^{2}\right )} b \log \left (c\right )}{g^{2}} + \frac {15 \, {\left (3 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} - 10 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {-g x^{2} + f} f^{2}\right )} a}{g^{2}}}{225 \, g} \] Input:

integrate(x^5*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"giac")
 

Output:

-1/225*(b*p*(15*(3*(g*x^2 - f)^2*sqrt(-g*x^2 + f) - 10*(-g*x^2 + f)^(3/2)* 
f + 15*sqrt(-g*x^2 + f)*f^2)*log(-d*g*x^2 + d*f)/g^2 - 2*(9*(g*x^2 - f)^2* 
sqrt(-g*x^2 + f) - 50*(-g*x^2 + f)^(3/2)*f + 225*sqrt(-g*x^2 + f)*f^2)/g^2 
) + 15*(3*(g*x^2 - f)^2*sqrt(-g*x^2 + f) - 10*(-g*x^2 + f)^(3/2)*f + 15*sq 
rt(-g*x^2 + f)*f^2)*b*log(c)/g^2 + 15*(3*(g*x^2 - f)^2*sqrt(-g*x^2 + f) - 
10*(-g*x^2 + f)^(3/2)*f + 15*sqrt(-g*x^2 + f)*f^2)*a/g^2)/g
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d\,f-d\,g\,x^2\right )}^p\right )\right )}{\sqrt {f-g\,x^2}} \,d x \] Input:

int((x^5*(a + b*log(c*(d*f - d*g*x^2)^p)))/(f - g*x^2)^(1/2),x)
 

Output:

int((x^5*(a + b*log(c*(d*f - d*g*x^2)^p)))/(f - g*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.73 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {\sqrt {-g \,x^{2}+f}\, \left (-120 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b \,f^{2}-60 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b f g \,x^{2}-45 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b \,g^{2} x^{4}-120 a \,f^{2}-60 a f g \,x^{2}-45 a \,g^{2} x^{4}+368 b \,f^{2} p +64 b f g p \,x^{2}+18 b \,g^{2} p \,x^{4}\right )}{225 g^{3}} \] Input:

int(x^5*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Output:

(sqrt(f - g*x**2)*( - 120*log(d**p*(f - g*x**2)**p*c)*b*f**2 - 60*log(d**p 
*(f - g*x**2)**p*c)*b*f*g*x**2 - 45*log(d**p*(f - g*x**2)**p*c)*b*g**2*x** 
4 - 120*a*f**2 - 60*a*f*g*x**2 - 45*a*g**2*x**4 + 368*b*f**2*p + 64*b*f*g* 
p*x**2 + 18*b*g**2*p*x**4))/(225*g**3)