\(\int \frac {x^5 (a+b \log (c (d f+d g x^2)^p))}{\sqrt {f+g x^2}} \, dx\) [690]

Optimal result

Integrand size = 34, antiderivative size = 175 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=-\frac {2 b f^2 p \sqrt {f+g x^2}}{g^3}+\frac {4 b f p \left (f+g x^2\right )^{3/2}}{9 g^3}-\frac {2 b p \left (f+g x^2\right )^{5/2}}{25 g^3}+\frac {f^2 \sqrt {f+g x^2} \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{g^3}-\frac {2 f \left (f+g x^2\right )^{3/2} \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{3 g^3}+\frac {\left (f+g x^2\right )^{5/2} \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{5 g^3} \] 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*l 
n(c*(d*g*x^2+d*f)^p))/g^3
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {\sqrt {f+g x^2} \left (15 a \left (8 f^2-4 f g x^2+3 g^2 x^4\right )-2 b p \left (184 f^2-32 f g x^2+9 g^2 x^4\right )+15 b \left (8 f^2-4 f g x^2+3 g^2 x^4\right ) \log \left (c \left (d \left (f+g x^2\right )\right )^p\right )\right )}{225 g^3} \] 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 = 132, normalized size of antiderivative = 0.75 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=-\frac {{\left (9 \, {\left (2 \, b g^{2} p - 5 \, a g^{2}\right )} x^{4} + 368 \, b f^{2} p - 120 \, a f^{2} - 4 \, {\left (16 \, b f g p - 15 \, a f g\right )} x^{2} - 15 \, {\left (3 \, b g^{2} p x^{4} - 4 \, b f g p x^{2} + 8 \, b f^{2} p\right )} \log \left (d g x^{2} + d f\right ) - 15 \, {\left (3 \, b g^{2} x^{4} - 4 \, b f g x^{2} + 8 \, b f^{2}\right )} \log \left (c\right )\right )} \sqrt {g x^{2} + f}}{225 \, g^{3}} \] Input:

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

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 [A] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.53 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\begin {cases} \frac {8 a f^{2} \sqrt {f + g x^{2}}}{15 g^{3}} - \frac {4 a f x^{2} \sqrt {f + g x^{2}}}{15 g^{2}} + \frac {a x^{4} \sqrt {f + g x^{2}}}{5 g} - \frac {368 b f^{2} p \sqrt {f + g x^{2}}}{225 g^{3}} + \frac {8 b f^{2} \sqrt {f + g x^{2}} \log {\left (c \left (d f + d g x^{2}\right )^{p} \right )}}{15 g^{3}} + \frac {64 b f p x^{2} \sqrt {f + g x^{2}}}{225 g^{2}} - \frac {4 b f x^{2} \sqrt {f + g x^{2}} \log {\left (c \left (d f + d g x^{2}\right )^{p} \right )}}{15 g^{2}} - \frac {2 b p x^{4} \sqrt {f + g x^{2}}}{25 g} + \frac {b x^{4} \sqrt {f + g x^{2}} \log {\left (c \left (d f + d g x^{2}\right )^{p} \right )}}{5 g} & \text {for}\: g \neq 0 \\\frac {x^{6} \left (a + b \log {\left (c \left (d f\right )^{p} \right )}\right )}{6 \sqrt {f}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((8*a*f**2*sqrt(f + g*x**2)/(15*g**3) - 4*a*f*x**2*sqrt(f + g*x** 
2)/(15*g**2) + a*x**4*sqrt(f + g*x**2)/(5*g) - 368*b*f**2*p*sqrt(f + g*x** 
2)/(225*g**3) + 8*b*f**2*sqrt(f + g*x**2)*log(c*(d*f + d*g*x**2)**p)/(15*g 
**3) + 64*b*f*p*x**2*sqrt(f + g*x**2)/(225*g**2) - 4*b*f*x**2*sqrt(f + g*x 
**2)*log(c*(d*f + d*g*x**2)**p)/(15*g**2) - 2*b*p*x**4*sqrt(f + g*x**2)/(2 
5*g) + b*x**4*sqrt(f + g*x**2)*log(c*(d*f + d*g*x**2)**p)/(5*g), Ne(g, 0)) 
, (x**6*(a + b*log(c*(d*f)**p))/(6*sqrt(f)), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 173, 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="m 
axima")
 

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 - 2/225*(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 = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.09 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=-\frac {\frac {{\left (18 \, {\left (g x^{2} + f\right )}^{\frac {5}{2}} - 100 \, {\left (g x^{2} + f\right )}^{\frac {3}{2}} f + 450 \, \sqrt {g x^{2} + f} f^{2} - 15 \, {\left (3 \, {\left (g x^{2} + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x^{2} + f} f^{2}\right )} \log \left (d g x^{2} + d f\right )\right )} b p}{g^{2}} - \frac {15 \, {\left (3 \, {\left (g x^{2} + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x^{2} + f} f^{2}\right )} b \log \left (c\right )}{g^{2}} - \frac {15 \, {\left (3 \, {\left (g x^{2} + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x^{2} + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x^{2} + f} f^{2}\right )} a}{g^{2}}}{225 \, g} \] Input:

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

Output:

-1/225*((18*(g*x^2 + f)^(5/2) - 100*(g*x^2 + f)^(3/2)*f + 450*sqrt(g*x^2 + 
 f)*f^2 - 15*(3*(g*x^2 + f)^(5/2) - 10*(g*x^2 + f)^(3/2)*f + 15*sqrt(g*x^2 
 + f)*f^2)*log(d*g*x^2 + d*f))*b*p/g^2 - 15*(3*(g*x^2 + f)^(5/2) - 10*(g*x 
^2 + f)^(3/2)*f + 15*sqrt(g*x^2 + f)*f^2)*b*log(c)/g^2 - 15*(3*(g*x^2 + f) 
^(5/2) - 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. \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d\,g\,x^2+d\,f\right )}^p\right )\right )}{\sqrt {g\,x^2+f}} \,d x \] 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 = 0.16 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.45 \[ \int \frac {x^5 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {\sqrt {g \,x^{2}+f}\, \left (120 \,\mathrm {log}\left (\frac {f^{p} d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+f}\, x +f +g \,x^{2}\right )^{2 p} c}{\left (\sqrt {f}\, \sqrt {g \,x^{2}+f}+\sqrt {g}\, \sqrt {f}\, x \right )^{2 p}}\right ) b \,f^{2}-60 \,\mathrm {log}\left (\frac {f^{p} d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+f}\, x +f +g \,x^{2}\right )^{2 p} c}{\left (\sqrt {f}\, \sqrt {g \,x^{2}+f}+\sqrt {g}\, \sqrt {f}\, x \right )^{2 p}}\right ) b f g \,x^{2}+45 \,\mathrm {log}\left (\frac {f^{p} d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+f}\, x +f +g \,x^{2}\right )^{2 p} c}{\left (\sqrt {f}\, \sqrt {g \,x^{2}+f}+\sqrt {g}\, \sqrt {f}\, x \right )^{2 p}}\right ) b \,g^{2} x^{4}+120 a \,f^{2}-60 a f g \,x^{2}+45 a \,g^{2} x^{4}-368 b \,f^{2} p +64 b f g p \,x^{2}-18 b \,g^{2} p \,x^{4}\right )}{225 g^{3}} \] 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((f**p*d**p*(sqrt(g)*sqrt(f + g*x**2)*x + f + g* 
x**2)**(2*p)*c)/(sqrt(f)*sqrt(f + g*x**2) + sqrt(g)*sqrt(f)*x)**(2*p))*b*f 
**2 - 60*log((f**p*d**p*(sqrt(g)*sqrt(f + g*x**2)*x + f + g*x**2)**(2*p)*c 
)/(sqrt(f)*sqrt(f + g*x**2) + sqrt(g)*sqrt(f)*x)**(2*p))*b*f*g*x**2 + 45*l 
og((f**p*d**p*(sqrt(g)*sqrt(f + g*x**2)*x + f + g*x**2)**(2*p)*c)/(sqrt(f) 
*sqrt(f + g*x**2) + sqrt(g)*sqrt(f)*x)**(2*p))*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)