\(\int \frac {x^5 (a+b \log (c (4 d+d g x^2)^p))}{\sqrt {4+g x^2}} \, dx\) [664]

Optimal result

Integrand size = 34, antiderivative size = 168 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=-\frac {32 b p \sqrt {4+g x^2}}{g^3}+\frac {16 b p \left (4+g x^2\right )^{3/2}}{9 g^3}-\frac {2 b p \left (4+g x^2\right )^{5/2}}{25 g^3}+\frac {16 \sqrt {4+g x^2} \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{g^3}-\frac {8 \left (4+g x^2\right )^{3/2} \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{3 g^3}+\frac {\left (4+g x^2\right )^{5/2} \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{5 g^3} \] Output:

-32*b*p*(g*x^2+4)^(1/2)/g^3+16/9*b*p*(g*x^2+4)^(3/2)/g^3-2/25*b*p*(g*x^2+4 
)^(5/2)/g^3+16*(g*x^2+4)^(1/2)*(a+b*ln(c*(d*g*x^2+4*d)^p))/g^3-8/3*(g*x^2+ 
4)^(3/2)*(a+b*ln(c*(d*g*x^2+4*d)^p))/g^3+1/5*(g*x^2+4)^(5/2)*(a+b*ln(c*(d* 
g*x^2+4*d)^p))/g^3
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.54 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {\sqrt {4+g x^2} \left (15 a \left (128-16 g x^2+3 g^2 x^4\right )-2 b p \left (2944-128 g x^2+9 g^2 x^4\right )+15 b \left (128-16 g x^2+3 g^2 x^4\right ) \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )\right )}{225 g^3} \] Input:

Integrate[(x^5*(a + b*Log[c*(4*d + d*g*x^2)^p]))/Sqrt[4 + g*x^2],x]
 

Output:

(Sqrt[4 + g*x^2]*(15*a*(128 - 16*g*x^2 + 3*g^2*x^4) - 2*b*p*(2944 - 128*g* 
x^2 + 9*g^2*x^4) + 15*b*(128 - 16*g*x^2 + 3*g^2*x^4)*Log[c*(d*(4 + 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.

\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c \left (d g x^2+4 d\right )^p\right )\right )}{\sqrt {g x^2+4}} \, dx\)

\(\Big \downarrow \) 2929

\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c \left (d g x^2+4 d\right )^p\right )\right )}{\sqrt {g x^2+4}}dx\)

Input:

Int[(x^5*(a + b*Log[c*(4*d + d*g*x^2)^p]))/Sqrt[4 + 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]

\[\int \frac {x^{5} \left (a +b \ln \left (c \left (d g \,x^{2}+4 d \right )^{p}\right )\right )}{\sqrt {g \,x^{2}+4}}d x\]

Input:

int(x^5*(a+b*ln(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
 

Output:

int(x^5*(a+b*ln(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.69 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=-\frac {{\left (9 \, {\left (2 \, b g^{2} p - 5 \, a g^{2}\right )} x^{4} - 16 \, {\left (16 \, b g p - 15 \, a g\right )} x^{2} + 5888 \, b p - 15 \, {\left (3 \, b g^{2} p x^{4} - 16 \, b g p x^{2} + 128 \, b p\right )} \log \left (d g x^{2} + 4 \, d\right ) - 15 \, {\left (3 \, b g^{2} x^{4} - 16 \, b g x^{2} + 128 \, b\right )} \log \left (c\right ) - 1920 \, a\right )} \sqrt {g x^{2} + 4}}{225 \, g^{3}} \] Input:

integrate(x^5*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="f 
ricas")
 

Output:

-1/225*(9*(2*b*g^2*p - 5*a*g^2)*x^4 - 16*(16*b*g*p - 15*a*g)*x^2 + 5888*b* 
p - 15*(3*b*g^2*p*x^4 - 16*b*g*p*x^2 + 128*b*p)*log(d*g*x^2 + 4*d) - 15*(3 
*b*g^2*x^4 - 16*b*g*x^2 + 128*b)*log(c) - 1920*a)*sqrt(g*x^2 + 4)/g^3
 

Sympy [A] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.49 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\begin {cases} \frac {a x^{4} \sqrt {g x^{2} + 4}}{5 g} - \frac {16 a x^{2} \sqrt {g x^{2} + 4}}{15 g^{2}} + \frac {128 a \sqrt {g x^{2} + 4}}{15 g^{3}} - \frac {2 b p x^{4} \sqrt {g x^{2} + 4}}{25 g} + \frac {b x^{4} \sqrt {g x^{2} + 4} \log {\left (c \left (d g x^{2} + 4 d\right )^{p} \right )}}{5 g} + \frac {256 b p x^{2} \sqrt {g x^{2} + 4}}{225 g^{2}} - \frac {16 b x^{2} \sqrt {g x^{2} + 4} \log {\left (c \left (d g x^{2} + 4 d\right )^{p} \right )}}{15 g^{2}} - \frac {5888 b p \sqrt {g x^{2} + 4}}{225 g^{3}} + \frac {128 b \sqrt {g x^{2} + 4} \log {\left (c \left (d g x^{2} + 4 d\right )^{p} \right )}}{15 g^{3}} & \text {for}\: g \neq 0 \\\frac {x^{6} \left (\frac {a}{2} + \frac {b \log {\left (c \left (4 d\right )^{p} \right )}}{2}\right )}{6} & \text {otherwise} \end {cases} \] Input:

integrate(x**5*(a+b*ln(c*(d*g*x**2+4*d)**p))/(g*x**2+4)**(1/2),x)
 

Output:

Piecewise((a*x**4*sqrt(g*x**2 + 4)/(5*g) - 16*a*x**2*sqrt(g*x**2 + 4)/(15* 
g**2) + 128*a*sqrt(g*x**2 + 4)/(15*g**3) - 2*b*p*x**4*sqrt(g*x**2 + 4)/(25 
*g) + b*x**4*sqrt(g*x**2 + 4)*log(c*(d*g*x**2 + 4*d)**p)/(5*g) + 256*b*p*x 
**2*sqrt(g*x**2 + 4)/(225*g**2) - 16*b*x**2*sqrt(g*x**2 + 4)*log(c*(d*g*x* 
*2 + 4*d)**p)/(15*g**2) - 5888*b*p*sqrt(g*x**2 + 4)/(225*g**3) + 128*b*sqr 
t(g*x**2 + 4)*log(c*(d*g*x**2 + 4*d)**p)/(15*g**3), Ne(g, 0)), (x**6*(a/2 
+ b*log(c*(4*d)**p)/2)/6, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {1}{15} \, {\left (\frac {3 \, \sqrt {g x^{2} + 4} x^{4}}{g} - \frac {16 \, \sqrt {g x^{2} + 4} x^{2}}{g^{2}} + \frac {128 \, \sqrt {g x^{2} + 4}}{g^{3}}\right )} b \log \left ({\left (d g x^{2} + 4 \, d\right )}^{p} c\right ) + \frac {1}{15} \, {\left (\frac {3 \, \sqrt {g x^{2} + 4} x^{4}}{g} - \frac {16 \, \sqrt {g x^{2} + 4} x^{2}}{g^{2}} + \frac {128 \, \sqrt {g x^{2} + 4}}{g^{3}}\right )} a - \frac {2 \, {\left (9 \, {\left (g x^{2} + 4\right )}^{\frac {5}{2}} - 200 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 3600 \, \sqrt {g x^{2} + 4}\right )} b p}{225 \, g^{3}} \] Input:

integrate(x^5*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="m 
axima")
 

Output:

1/15*(3*sqrt(g*x^2 + 4)*x^4/g - 16*sqrt(g*x^2 + 4)*x^2/g^2 + 128*sqrt(g*x^ 
2 + 4)/g^3)*b*log((d*g*x^2 + 4*d)^p*c) + 1/15*(3*sqrt(g*x^2 + 4)*x^4/g - 1 
6*sqrt(g*x^2 + 4)*x^2/g^2 + 128*sqrt(g*x^2 + 4)/g^3)*a - 2/225*(9*(g*x^2 + 
 4)^(5/2) - 200*(g*x^2 + 4)^(3/2) + 3600*sqrt(g*x^2 + 4))*b*p/g^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=-\frac {\frac {{\left (18 \, {\left (g x^{2} + 4\right )}^{\frac {5}{2}} - 15 \, {\left (3 \, {\left (g x^{2} + 4\right )}^{\frac {5}{2}} - 40 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 240 \, \sqrt {g x^{2} + 4}\right )} \log \left (d g x^{2} + 4 \, d\right ) - 400 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 7200 \, \sqrt {g x^{2} + 4}\right )} b p}{g^{2}} - \frac {15 \, {\left (3 \, {\left (g x^{2} + 4\right )}^{\frac {5}{2}} - 40 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 240 \, \sqrt {g x^{2} + 4}\right )} b \log \left (c\right )}{g^{2}} - \frac {15 \, {\left (3 \, {\left (g x^{2} + 4\right )}^{\frac {5}{2}} - 40 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 240 \, \sqrt {g x^{2} + 4}\right )} a}{g^{2}}}{225 \, g} \] Input:

integrate(x^5*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="g 
iac")
 

Output:

-1/225*((18*(g*x^2 + 4)^(5/2) - 15*(3*(g*x^2 + 4)^(5/2) - 40*(g*x^2 + 4)^( 
3/2) + 240*sqrt(g*x^2 + 4))*log(d*g*x^2 + 4*d) - 400*(g*x^2 + 4)^(3/2) + 7 
200*sqrt(g*x^2 + 4))*b*p/g^2 - 15*(3*(g*x^2 + 4)^(5/2) - 40*(g*x^2 + 4)^(3 
/2) + 240*sqrt(g*x^2 + 4))*b*log(c)/g^2 - 15*(3*(g*x^2 + 4)^(5/2) - 40*(g* 
x^2 + 4)^(3/2) + 240*sqrt(g*x^2 + 4))*a/g^2)/g
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d\,g\,x^2+4\,d\right )}^p\right )\right )}{\sqrt {g\,x^2+4}} \,d x \] Input:

int((x^5*(a + b*log(c*(4*d + d*g*x^2)^p)))/(g*x^2 + 4)^(1/2),x)
 

Output:

int((x^5*(a + b*log(c*(4*d + d*g*x^2)^p)))/(g*x^2 + 4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.40 \[ \int \frac {x^5 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {\sqrt {g \,x^{2}+4}\, \left (45 \,\mathrm {log}\left (\frac {d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+4}\, x +g \,x^{2}+4\right )^{2 p} 4^{p} c}{\left (2 \sqrt {g \,x^{2}+4}+2 \sqrt {g}\, x \right )^{2 p}}\right ) b \,g^{2} x^{4}-240 \,\mathrm {log}\left (\frac {d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+4}\, x +g \,x^{2}+4\right )^{2 p} 4^{p} c}{\left (2 \sqrt {g \,x^{2}+4}+2 \sqrt {g}\, x \right )^{2 p}}\right ) b g \,x^{2}+1920 \,\mathrm {log}\left (\frac {d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+4}\, x +g \,x^{2}+4\right )^{2 p} 4^{p} c}{\left (2 \sqrt {g \,x^{2}+4}+2 \sqrt {g}\, x \right )^{2 p}}\right ) b +45 a \,g^{2} x^{4}-240 a g \,x^{2}+1920 a -18 b \,g^{2} p \,x^{4}+256 b g p \,x^{2}-5888 b p \right )}{225 g^{3}} \] Input:

int(x^5*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
 

Output:

(sqrt(g*x**2 + 4)*(45*log((d**p*(sqrt(g)*sqrt(g*x**2 + 4)*x + g*x**2 + 4)* 
*(2*p)*4**p*c)/(2*sqrt(g*x**2 + 4) + 2*sqrt(g)*x)**(2*p))*b*g**2*x**4 - 24 
0*log((d**p*(sqrt(g)*sqrt(g*x**2 + 4)*x + g*x**2 + 4)**(2*p)*4**p*c)/(2*sq 
rt(g*x**2 + 4) + 2*sqrt(g)*x)**(2*p))*b*g*x**2 + 1920*log((d**p*(sqrt(g)*s 
qrt(g*x**2 + 4)*x + g*x**2 + 4)**(2*p)*4**p*c)/(2*sqrt(g*x**2 + 4) + 2*sqr 
t(g)*x)**(2*p))*b + 45*a*g**2*x**4 - 240*a*g*x**2 + 1920*a - 18*b*g**2*p*x 
**4 + 256*b*g*p*x**2 - 5888*b*p))/(225*g**3)