\(\int \frac {x^7 (a+b \log (c (d f-d g x^2)^p))}{\sqrt {f-g x^2}} \, dx\) [702]

Optimal result

Integrand size = 36, antiderivative size = 248 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {2 b f^3 p \sqrt {f-g x^2}}{g^4}-\frac {2 b f^2 p \left (f-g x^2\right )^{3/2}}{3 g^4}+\frac {6 b f p \left (f-g x^2\right )^{5/2}}{25 g^4}-\frac {2 b p \left (f-g x^2\right )^{7/2}}{49 g^4}-\frac {f^3 \sqrt {f-g x^2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{g^4}+\frac {f^2 \left (f-g x^2\right )^{3/2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{g^4}-\frac {3 f \left (f-g x^2\right )^{5/2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{5 g^4}+\frac {\left (f-g x^2\right )^{7/2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{7 g^4} \] Output:

2*b*f^3*p*(-g*x^2+f)^(1/2)/g^4-2/3*b*f^2*p*(-g*x^2+f)^(3/2)/g^4+6/25*b*f*p 
*(-g*x^2+f)^(5/2)/g^4-2/49*b*p*(-g*x^2+f)^(7/2)/g^4-f^3*(-g*x^2+f)^(1/2)*( 
a+b*ln(c*(-d*g*x^2+d*f)^p))/g^4+f^2*(-g*x^2+f)^(3/2)*(a+b*ln(c*(-d*g*x^2+d 
*f)^p))/g^4-3/5*f*(-g*x^2+f)^(5/2)*(a+b*ln(c*(-d*g*x^2+d*f)^p))/g^4+1/7*(- 
g*x^2+f)^(7/2)*(a+b*ln(c*(-d*g*x^2+d*f)^p))/g^4
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.57 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {\sqrt {f-g x^2} \left (-105 a \left (16 f^3+8 f^2 g x^2+6 f g^2 x^4+5 g^3 x^6\right )+2 b p \left (2816 f^3+568 f^2 g x^2+216 f g^2 x^4+75 g^3 x^6\right )-105 b \left (16 f^3+8 f^2 g x^2+6 f g^2 x^4+5 g^3 x^6\right ) \log \left (c \left (d \left (f-g x^2\right )\right )^p\right )\right )}{3675 g^4} \] Input:

Integrate[(x^7*(a + b*Log[c*(d*f - d*g*x^2)^p]))/Sqrt[f - g*x^2],x]
 

Output:

(Sqrt[f - g*x^2]*(-105*a*(16*f^3 + 8*f^2*g*x^2 + 6*f*g^2*x^4 + 5*g^3*x^6) 
+ 2*b*p*(2816*f^3 + 568*f^2*g*x^2 + 216*f*g^2*x^4 + 75*g^3*x^6) - 105*b*(1 
6*f^3 + 8*f^2*g*x^2 + 6*f*g^2*x^4 + 5*g^3*x^6)*Log[c*(d*(f - g*x^2))^p]))/ 
(3675*g^4)
 

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.

\(\displaystyle \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx\)

\(\Big \downarrow \) 2929

\(\displaystyle \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}}dx\)

Input:

Int[(x^7*(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]

\[\int \frac {x^{7} \left (a +b \ln \left (c \left (-d g \,x^{2}+d f \right )^{p}\right )\right )}{\sqrt {-g \,x^{2}+f}}d x\]

Input:

int(x^7*(a+b*ln(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Output:

int(x^7*(a+b*ln(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.74 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {{\left (75 \, {\left (2 \, b g^{3} p - 7 \, a g^{3}\right )} x^{6} + 5632 \, b f^{3} p + 18 \, {\left (24 \, b f g^{2} p - 35 \, a f g^{2}\right )} x^{4} - 1680 \, a f^{3} + 8 \, {\left (142 \, b f^{2} g p - 105 \, a f^{2} g\right )} x^{2} - 105 \, {\left (5 \, b g^{3} p x^{6} + 6 \, b f g^{2} p x^{4} + 8 \, b f^{2} g p x^{2} + 16 \, b f^{3} p\right )} \log \left (-d g x^{2} + d f\right ) - 105 \, {\left (5 \, b g^{3} x^{6} + 6 \, b f g^{2} x^{4} + 8 \, b f^{2} g x^{2} + 16 \, b f^{3}\right )} \log \left (c\right )\right )} \sqrt {-g x^{2} + f}}{3675 \, g^{4}} \] Input:

integrate(x^7*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"fricas")
 

Output:

1/3675*(75*(2*b*g^3*p - 7*a*g^3)*x^6 + 5632*b*f^3*p + 18*(24*b*f*g^2*p - 3 
5*a*f*g^2)*x^4 - 1680*a*f^3 + 8*(142*b*f^2*g*p - 105*a*f^2*g)*x^2 - 105*(5 
*b*g^3*p*x^6 + 6*b*f*g^2*p*x^4 + 8*b*f^2*g*p*x^2 + 16*b*f^3*p)*log(-d*g*x^ 
2 + d*f) - 105*(5*b*g^3*x^6 + 6*b*f*g^2*x^4 + 8*b*f^2*g*x^2 + 16*b*f^3)*lo 
g(c))*sqrt(-g*x^2 + f)/g^4
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\text {Timed out} \] Input:

integrate(x**7*(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 = 240, normalized size of antiderivative = 0.97 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=-\frac {1}{35} \, {\left (\frac {5 \, \sqrt {-g x^{2} + f} x^{6}}{g} + \frac {6 \, \sqrt {-g x^{2} + f} f x^{4}}{g^{2}} + \frac {8 \, \sqrt {-g x^{2} + f} f^{2} x^{2}}{g^{3}} + \frac {16 \, \sqrt {-g x^{2} + f} f^{3}}{g^{4}}\right )} b \log \left ({\left (-d g x^{2} + d f\right )}^{p} c\right ) - \frac {1}{35} \, {\left (\frac {5 \, \sqrt {-g x^{2} + f} x^{6}}{g} + \frac {6 \, \sqrt {-g x^{2} + f} f x^{4}}{g^{2}} + \frac {8 \, \sqrt {-g x^{2} + f} f^{2} x^{2}}{g^{3}} + \frac {16 \, \sqrt {-g x^{2} + f} f^{3}}{g^{4}}\right )} a - \frac {2 \, {\left (75 \, {\left (-g x^{2} + f\right )}^{\frac {7}{2}} - 441 \, {\left (-g x^{2} + f\right )}^{\frac {5}{2}} f + 1225 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f^{2} - 3675 \, \sqrt {-g x^{2} + f} f^{3}\right )} b p}{3675 \, g^{4}} \] Input:

integrate(x^7*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"maxima")
 

Output:

-1/35*(5*sqrt(-g*x^2 + f)*x^6/g + 6*sqrt(-g*x^2 + f)*f*x^4/g^2 + 8*sqrt(-g 
*x^2 + f)*f^2*x^2/g^3 + 16*sqrt(-g*x^2 + f)*f^3/g^4)*b*log((-d*g*x^2 + d*f 
)^p*c) - 1/35*(5*sqrt(-g*x^2 + f)*x^6/g + 6*sqrt(-g*x^2 + f)*f*x^4/g^2 + 8 
*sqrt(-g*x^2 + f)*f^2*x^2/g^3 + 16*sqrt(-g*x^2 + f)*f^3/g^4)*a - 2/3675*(7 
5*(-g*x^2 + f)^(7/2) - 441*(-g*x^2 + f)^(5/2)*f + 1225*(-g*x^2 + f)^(3/2)* 
f^2 - 3675*sqrt(-g*x^2 + f)*f^3)*b*p/g^4
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.44 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=-\frac {b p {\left (\frac {105 \, {\left (5 \, {\left (g x^{2} - f\right )}^{3} \sqrt {-g x^{2} + f} + 21 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} f - 35 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f^{2} + 35 \, \sqrt {-g x^{2} + f} f^{3}\right )} \log \left (-d g x^{2} + d f\right )}{g^{3}} - \frac {2 \, {\left (75 \, {\left (g x^{2} - f\right )}^{3} \sqrt {-g x^{2} + f} + 441 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} f - 1225 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f^{2} + 3675 \, \sqrt {-g x^{2} + f} f^{3}\right )}}{g^{3}}\right )} + \frac {105 \, {\left (5 \, {\left (g x^{2} - f\right )}^{3} \sqrt {-g x^{2} + f} + 21 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} f - 35 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f^{2} + 35 \, \sqrt {-g x^{2} + f} f^{3}\right )} b \log \left (c\right )}{g^{3}} + \frac {105 \, {\left (5 \, {\left (g x^{2} - f\right )}^{3} \sqrt {-g x^{2} + f} + 21 \, {\left (g x^{2} - f\right )}^{2} \sqrt {-g x^{2} + f} f - 35 \, {\left (-g x^{2} + f\right )}^{\frac {3}{2}} f^{2} + 35 \, \sqrt {-g x^{2} + f} f^{3}\right )} a}{g^{3}}}{3675 \, g} \] Input:

integrate(x^7*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x, algorithm= 
"giac")
 

Output:

-1/3675*(b*p*(105*(5*(g*x^2 - f)^3*sqrt(-g*x^2 + f) + 21*(g*x^2 - f)^2*sqr 
t(-g*x^2 + f)*f - 35*(-g*x^2 + f)^(3/2)*f^2 + 35*sqrt(-g*x^2 + f)*f^3)*log 
(-d*g*x^2 + d*f)/g^3 - 2*(75*(g*x^2 - f)^3*sqrt(-g*x^2 + f) + 441*(g*x^2 - 
 f)^2*sqrt(-g*x^2 + f)*f - 1225*(-g*x^2 + f)^(3/2)*f^2 + 3675*sqrt(-g*x^2 
+ f)*f^3)/g^3) + 105*(5*(g*x^2 - f)^3*sqrt(-g*x^2 + f) + 21*(g*x^2 - f)^2* 
sqrt(-g*x^2 + f)*f - 35*(-g*x^2 + f)^(3/2)*f^2 + 35*sqrt(-g*x^2 + f)*f^3)* 
b*log(c)/g^3 + 105*(5*(g*x^2 - f)^3*sqrt(-g*x^2 + f) + 21*(g*x^2 - f)^2*sq 
rt(-g*x^2 + f)*f - 35*(-g*x^2 + f)^(3/2)*f^2 + 35*sqrt(-g*x^2 + f)*f^3)*a/ 
g^3)/g
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\int \frac {x^7\,\left (a+b\,\ln \left (c\,{\left (d\,f-d\,g\,x^2\right )}^p\right )\right )}{\sqrt {f-g\,x^2}} \,d x \] Input:

int((x^7*(a + b*log(c*(d*f - d*g*x^2)^p)))/(f - g*x^2)^(1/2),x)
 

Output:

int((x^7*(a + b*log(c*(d*f - d*g*x^2)^p)))/(f - g*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.76 \[ \int \frac {x^7 \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{\sqrt {f-g x^2}} \, dx=\frac {\sqrt {-g \,x^{2}+f}\, \left (-1680 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b \,f^{3}-840 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b \,f^{2} g \,x^{2}-630 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b f \,g^{2} x^{4}-525 \,\mathrm {log}\left (d^{p} \left (-g \,x^{2}+f \right )^{p} c \right ) b \,g^{3} x^{6}-1680 a \,f^{3}-840 a \,f^{2} g \,x^{2}-630 a f \,g^{2} x^{4}-525 a \,g^{3} x^{6}+5632 b \,f^{3} p +1136 b \,f^{2} g p \,x^{2}+432 b f \,g^{2} p \,x^{4}+150 b \,g^{3} p \,x^{6}\right )}{3675 g^{4}} \] Input:

int(x^7*(a+b*log(c*(-d*g*x^2+d*f)^p))/(-g*x^2+f)^(1/2),x)
 

Output:

(sqrt(f - g*x**2)*( - 1680*log(d**p*(f - g*x**2)**p*c)*b*f**3 - 840*log(d* 
*p*(f - g*x**2)**p*c)*b*f**2*g*x**2 - 630*log(d**p*(f - g*x**2)**p*c)*b*f* 
g**2*x**4 - 525*log(d**p*(f - g*x**2)**p*c)*b*g**3*x**6 - 1680*a*f**3 - 84 
0*a*f**2*g*x**2 - 630*a*f*g**2*x**4 - 525*a*g**3*x**6 + 5632*b*f**3*p + 11 
36*b*f**2*g*p*x**2 + 432*b*f*g**2*p*x**4 + 150*b*g**3*p*x**6))/(3675*g**4)