\(\int \frac {a+b \log (c (d f-d g x^2)^p)}{x^6 \sqrt {f-g x^2}} \, dx\) [713]

Optimal result

Integrand size = 36, antiderivative size = 217 \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\frac {2 b g p \sqrt {f-g x^2}}{15 f^2 x^3}+\frac {4 b g^2 p \sqrt {f-g x^2}}{5 f^3 x}-\frac {16 b g^{5/2} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f-g x^2}}\right )}{15 f^3}-\frac {\sqrt {f-g x^2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{5 f x^5}-\frac {4 g \sqrt {f-g x^2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{15 f^2 x^3}-\frac {8 g^2 \sqrt {f-g x^2} \left (a+b \log \left (c \left (d f-d g x^2\right )^p\right )\right )}{15 f^3 x} \] Output:

2/15*b*g*p*(-g*x^2+f)^(1/2)/f^2/x^3+4/5*b*g^2*p*(-g*x^2+f)^(1/2)/f^3/x-16/ 
15*b*g^(5/2)*p*arctan(g^(1/2)*x/(-g*x^2+f)^(1/2))/f^3-1/5*(-g*x^2+f)^(1/2) 
*(a+b*ln(c*(-d*g*x^2+d*f)^p))/f/x^5-4/15*g*(-g*x^2+f)^(1/2)*(a+b*ln(c*(-d* 
g*x^2+d*f)^p))/f^2/x^3-8/15*g^2*(-g*x^2+f)^(1/2)*(a+b*ln(c*(-d*g*x^2+d*f)^ 
p))/f^3/x
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.62 \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=-\frac {16 b g^{5/2} p x^5 \arctan \left (\frac {\sqrt {g} x}{\sqrt {f-g x^2}}\right )+\sqrt {f-g x^2} \left (-2 b g p x^2 \left (f+6 g x^2\right )+a \left (3 f^2+4 f g x^2+8 g^2 x^4\right )+b \left (3 f^2+4 f g x^2+8 g^2 x^4\right ) \log \left (c \left (d \left (f-g x^2\right )\right )^p\right )\right )}{15 f^3 x^5} \] Input:

Integrate[(a + b*Log[c*(d*f - d*g*x^2)^p])/(x^6*Sqrt[f - g*x^2]),x]
 

Output:

-1/15*(16*b*g^(5/2)*p*x^5*ArcTan[(Sqrt[g]*x)/Sqrt[f - g*x^2]] + Sqrt[f - g 
*x^2]*(-2*b*g*p*x^2*(f + 6*g*x^2) + a*(3*f^2 + 4*f*g*x^2 + 8*g^2*x^4) + b* 
(3*f^2 + 4*f*g*x^2 + 8*g^2*x^4)*Log[c*(d*(f - g*x^2))^p]))/(f^3*x^5)
 

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.

\(\displaystyle \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx\)

\(\Big \downarrow \) 2929

\(\displaystyle \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}}dx\)

Input:

Int[(a + b*Log[c*(d*f - d*g*x^2)^p])/(x^6*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]

\[\int \frac {a +b \ln \left (c \left (-d g \,x^{2}+d f \right )^{p}\right )}{x^{6} \sqrt {-g \,x^{2}+f}}d x\]

Input:

int((a+b*ln(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x)
 

Output:

int((a+b*ln(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.59 \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\left [\frac {8 \, b \sqrt {-g} g^{2} p x^{5} \log \left (2 \, g x^{2} - 2 \, \sqrt {-g x^{2} + f} \sqrt {-g} x - f\right ) + {\left (4 \, {\left (3 \, b g^{2} p - 2 \, a g^{2}\right )} x^{4} - 3 \, a f^{2} + 2 \, {\left (b f g p - 2 \, a f g\right )} x^{2} - {\left (8 \, b g^{2} p x^{4} + 4 \, b f g p x^{2} + 3 \, b f^{2} p\right )} \log \left (-d g x^{2} + d f\right ) - {\left (8 \, b g^{2} x^{4} + 4 \, b f g x^{2} + 3 \, b f^{2}\right )} \log \left (c\right )\right )} \sqrt {-g x^{2} + f}}{15 \, f^{3} x^{5}}, \frac {16 \, b g^{\frac {5}{2}} p x^{5} \arctan \left (\frac {\sqrt {-g x^{2} + f} \sqrt {g} x}{g x^{2} - f}\right ) + {\left (4 \, {\left (3 \, b g^{2} p - 2 \, a g^{2}\right )} x^{4} - 3 \, a f^{2} + 2 \, {\left (b f g p - 2 \, a f g\right )} x^{2} - {\left (8 \, b g^{2} p x^{4} + 4 \, b f g p x^{2} + 3 \, b f^{2} p\right )} \log \left (-d g x^{2} + d f\right ) - {\left (8 \, b g^{2} x^{4} + 4 \, b f g x^{2} + 3 \, b f^{2}\right )} \log \left (c\right )\right )} \sqrt {-g x^{2} + f}}{15 \, f^{3} x^{5}}\right ] \] Input:

integrate((a+b*log(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x, algorithm= 
"fricas")
 

Output:

[1/15*(8*b*sqrt(-g)*g^2*p*x^5*log(2*g*x^2 - 2*sqrt(-g*x^2 + f)*sqrt(-g)*x 
- f) + (4*(3*b*g^2*p - 2*a*g^2)*x^4 - 3*a*f^2 + 2*(b*f*g*p - 2*a*f*g)*x^2 
- (8*b*g^2*p*x^4 + 4*b*f*g*p*x^2 + 3*b*f^2*p)*log(-d*g*x^2 + d*f) - (8*b*g 
^2*x^4 + 4*b*f*g*x^2 + 3*b*f^2)*log(c))*sqrt(-g*x^2 + f))/(f^3*x^5), 1/15* 
(16*b*g^(5/2)*p*x^5*arctan(sqrt(-g*x^2 + f)*sqrt(g)*x/(g*x^2 - f)) + (4*(3 
*b*g^2*p - 2*a*g^2)*x^4 - 3*a*f^2 + 2*(b*f*g*p - 2*a*f*g)*x^2 - (8*b*g^2*p 
*x^4 + 4*b*f*g*p*x^2 + 3*b*f^2*p)*log(-d*g*x^2 + d*f) - (8*b*g^2*x^4 + 4*b 
*f*g*x^2 + 3*b*f^2)*log(c))*sqrt(-g*x^2 + f))/(f^3*x^5)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*(-d*g*x**2+d*f)**p))/x**6/(-g*x**2+f)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\int { \frac {b \log \left ({\left (-d g x^{2} + d f\right )}^{p} c\right ) + a}{\sqrt {-g x^{2} + f} x^{6}} \,d x } \] Input:

integrate((a+b*log(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x, algorithm= 
"maxima")
 

Output:

-1/15*a*(8*sqrt(-g*x^2 + f)*g^2/(f^3*x) + 4*sqrt(-g*x^2 + f)*g/(f^2*x^3) + 
 3*sqrt(-g*x^2 + f)/(f*x^5)) - 1/15*b*(integrate((16*g^3*p*x^6 + 8*f*g^2*p 
*x^4 + 6*f^2*g*p*x^2 - 15*f^3*p*log(d) - 15*f^3*log(c))/(sqrt(-g*x^2 + f)* 
x^6), x)/f^3 - (8*g^3*x^6 - 4*f*g^2*x^4 - f^2*g*x^2 - 3*f^3)*log((-g*x^2 + 
 f)^p)/(sqrt(-g*x^2 + f)*f^3*x^5))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (189) = 378\).

Time = 1.03 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.14 \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=-\frac {8}{15} \, b p {\left (\frac {\sqrt {-g} g^{2} \log \left ({\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2}\right )}{f^{3}} - \frac {\sqrt {-g} g^{2} \log \left ({\left | g x^{2} - f \right |}\right )}{f^{3}} - \frac {2 \, {\left (10 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{4} - 5 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} f + f^{2}\right )} \sqrt {-g} g^{2} \log \left (-d g x^{2} + d f\right )}{{\left ({\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} - f\right )}^{5}} + \frac {2 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{4} \sqrt {-g} g^{2} - 7 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} f \sqrt {-g} g^{2} + 3 \, f^{2} \sqrt {-g} g^{2}}{{\left ({\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} - f\right )}^{3} f^{2}}\right )} + \frac {16 \, {\left (10 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{4} - 5 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} f + f^{2}\right )} b \sqrt {-g} g^{2} \log \left (c\right )}{15 \, {\left ({\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} - f\right )}^{5}} + \frac {16 \, {\left (10 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{4} - 5 \, {\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} f + f^{2}\right )} a \sqrt {-g} g^{2}}{15 \, {\left ({\left (\sqrt {-g} x - \sqrt {-g x^{2} + f}\right )}^{2} - f\right )}^{5}} \] Input:

integrate((a+b*log(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x, algorithm= 
"giac")
 

Output:

-8/15*b*p*(sqrt(-g)*g^2*log((sqrt(-g)*x - sqrt(-g*x^2 + f))^2)/f^3 - sqrt( 
-g)*g^2*log(abs(g*x^2 - f))/f^3 - 2*(10*(sqrt(-g)*x - sqrt(-g*x^2 + f))^4 
- 5*(sqrt(-g)*x - sqrt(-g*x^2 + f))^2*f + f^2)*sqrt(-g)*g^2*log(-d*g*x^2 + 
 d*f)/((sqrt(-g)*x - sqrt(-g*x^2 + f))^2 - f)^5 + (2*(sqrt(-g)*x - sqrt(-g 
*x^2 + f))^4*sqrt(-g)*g^2 - 7*(sqrt(-g)*x - sqrt(-g*x^2 + f))^2*f*sqrt(-g) 
*g^2 + 3*f^2*sqrt(-g)*g^2)/(((sqrt(-g)*x - sqrt(-g*x^2 + f))^2 - f)^3*f^2) 
) + 16/15*(10*(sqrt(-g)*x - sqrt(-g*x^2 + f))^4 - 5*(sqrt(-g)*x - sqrt(-g* 
x^2 + f))^2*f + f^2)*b*sqrt(-g)*g^2*log(c)/((sqrt(-g)*x - sqrt(-g*x^2 + f) 
)^2 - f)^5 + 16/15*(10*(sqrt(-g)*x - sqrt(-g*x^2 + f))^4 - 5*(sqrt(-g)*x - 
 sqrt(-g*x^2 + f))^2*f + f^2)*a*sqrt(-g)*g^2/((sqrt(-g)*x - sqrt(-g*x^2 + 
f))^2 - f)^5
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d\,f-d\,g\,x^2\right )}^p\right )}{x^6\,\sqrt {f-g\,x^2}} \,d x \] Input:

int((a + b*log(c*(d*f - d*g*x^2)^p))/(x^6*(f - g*x^2)^(1/2)),x)
                                                                                    
                                                                                    
 

Output:

int((a + b*log(c*(d*f - d*g*x^2)^p))/(x^6*(f - g*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.50 \[ \int \frac {a+b \log \left (c \left (d f-d g x^2\right )^p\right )}{x^6 \sqrt {f-g x^2}} \, dx=\frac {-16 \sqrt {g}\, \mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right ) b \,g^{2} p \,x^{5}-3 \sqrt {-g \,x^{2}+f}\, \mathrm {log}\left (\frac {f^{p} d^{p} \left (-\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p} c}{\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p}}\right ) b \,f^{2}-4 \sqrt {-g \,x^{2}+f}\, \mathrm {log}\left (\frac {f^{p} d^{p} \left (-\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p} c}{\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p}}\right ) b f g \,x^{2}-8 \sqrt {-g \,x^{2}+f}\, \mathrm {log}\left (\frac {f^{p} d^{p} \left (-\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p} c}{\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {g}\, x}{\sqrt {f}}\right )}{2}\right )^{2}+1\right )^{2 p}}\right ) b \,g^{2} x^{4}-3 \sqrt {-g \,x^{2}+f}\, a \,f^{2}-4 \sqrt {-g \,x^{2}+f}\, a f g \,x^{2}-8 \sqrt {-g \,x^{2}+f}\, a \,g^{2} x^{4}+2 \sqrt {-g \,x^{2}+f}\, b f g p \,x^{2}+12 \sqrt {-g \,x^{2}+f}\, b \,g^{2} p \,x^{4}}{15 f^{3} x^{5}} \] Input:

int((a+b*log(c*(-d*g*x^2+d*f)^p))/x^6/(-g*x^2+f)^(1/2),x)
 

Output:

( - 16*sqrt(g)*asin((sqrt(g)*x)/sqrt(f))*b*g**2*p*x**5 - 3*sqrt(f - g*x**2 
)*log((f**p*d**p*( - tan(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p)*c)/(t 
an(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p))*b*f**2 - 4*sqrt(f - g*x**2 
)*log((f**p*d**p*( - tan(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p)*c)/(t 
an(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p))*b*f*g*x**2 - 8*sqrt(f - g* 
x**2)*log((f**p*d**p*( - tan(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p)*c 
)/(tan(asin((sqrt(g)*x)/sqrt(f))/2)**2 + 1)**(2*p))*b*g**2*x**4 - 3*sqrt(f 
 - g*x**2)*a*f**2 - 4*sqrt(f - g*x**2)*a*f*g*x**2 - 8*sqrt(f - g*x**2)*a*g 
**2*x**4 + 2*sqrt(f - g*x**2)*b*f*g*p*x**2 + 12*sqrt(f - g*x**2)*b*g**2*p* 
x**4)/(15*f**3*x**5)