Integrand size = 34, antiderivative size = 111 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {8 b p \sqrt {4+g x^2}}{g^2}-\frac {2 b p \left (4+g x^2\right )^{3/2}}{9 g^2}-\frac {4 \sqrt {4+g x^2} \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{g^2}+\frac {\left (4+g x^2\right )^{3/2} \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{3 g^2} \] Output:
8*b*p*(g*x^2+4)^(1/2)/g^2-2/9*b*p*(g*x^2+4)^(3/2)/g^2-4*(g*x^2+4)^(1/2)*(a +b*ln(c*(d*g*x^2+4*d)^p))/g^2+1/3*(g*x^2+4)^(3/2)*(a+b*ln(c*(d*g*x^2+4*d)^ p))/g^2
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {\sqrt {4+g x^2} \left (-2 b p \left (-32+g x^2\right )+3 a \left (-8+g x^2\right )+3 b \left (-8+g x^2\right ) \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )\right )}{9 g^2} \] Input:
Integrate[(x^3*(a + b*Log[c*(4*d + d*g*x^2)^p]))/Sqrt[4 + g*x^2],x]
Output:
(Sqrt[4 + g*x^2]*(-2*b*p*(-32 + g*x^2) + 3*a*(-8 + g*x^2) + 3*b*(-8 + g*x^ 2)*Log[c*(d*(4 + g*x^2))^p]))/(9*g^2)
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.
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c \left (d g x^2+4 d\right )^p\right )\right )}{\sqrt {g x^2+4}} \, dx\) |
\(\Big \downarrow \) 2929 |
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c \left (d g x^2+4 d\right )^p\right )\right )}{\sqrt {g x^2+4}}dx\) |
Input:
Int[(x^3*(a + b*Log[c*(4*d + d*g*x^2)^p]))/Sqrt[4 + g*x^2],x]
Output:
$Aborted
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]
\[\int \frac {x^{3} \left (a +b \ln \left (c \left (d g \,x^{2}+4 d \right )^{p}\right )\right )}{\sqrt {g \,x^{2}+4}}d x\]
Input:
int(x^3*(a+b*ln(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
Output:
int(x^3*(a+b*ln(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=-\frac {{\left ({\left (2 \, b g p - 3 \, a g\right )} x^{2} - 64 \, b p - 3 \, {\left (b g p x^{2} - 8 \, b p\right )} \log \left (d g x^{2} + 4 \, d\right ) - 3 \, {\left (b g x^{2} - 8 \, b\right )} \log \left (c\right ) + 24 \, a\right )} \sqrt {g x^{2} + 4}}{9 \, g^{2}} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="f ricas")
Output:
-1/9*((2*b*g*p - 3*a*g)*x^2 - 64*b*p - 3*(b*g*p*x^2 - 8*b*p)*log(d*g*x^2 + 4*d) - 3*(b*g*x^2 - 8*b)*log(c) + 24*a)*sqrt(g*x^2 + 4)/g^2
Time = 1.60 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.50 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\begin {cases} \frac {a x^{2} \sqrt {g x^{2} + 4}}{3 g} - \frac {8 a \sqrt {g x^{2} + 4}}{3 g^{2}} - \frac {2 b p x^{2} \sqrt {g x^{2} + 4}}{9 g} + \frac {b x^{2} \sqrt {g x^{2} + 4} \log {\left (c \left (d g x^{2} + 4 d\right )^{p} \right )}}{3 g} + \frac {64 b p \sqrt {g x^{2} + 4}}{9 g^{2}} - \frac {8 b \sqrt {g x^{2} + 4} \log {\left (c \left (d g x^{2} + 4 d\right )^{p} \right )}}{3 g^{2}} & \text {for}\: g \neq 0 \\\frac {x^{4} \left (\frac {a}{2} + \frac {b \log {\left (c \left (4 d\right )^{p} \right )}}{2}\right )}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(a+b*ln(c*(d*g*x**2+4*d)**p))/(g*x**2+4)**(1/2),x)
Output:
Piecewise((a*x**2*sqrt(g*x**2 + 4)/(3*g) - 8*a*sqrt(g*x**2 + 4)/(3*g**2) - 2*b*p*x**2*sqrt(g*x**2 + 4)/(9*g) + b*x**2*sqrt(g*x**2 + 4)*log(c*(d*g*x* *2 + 4*d)**p)/(3*g) + 64*b*p*sqrt(g*x**2 + 4)/(9*g**2) - 8*b*sqrt(g*x**2 + 4)*log(c*(d*g*x**2 + 4*d)**p)/(3*g**2), Ne(g, 0)), (x**4*(a/2 + b*log(c*( 4*d)**p)/2)/4, True))
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {1}{3} \, {\left (\frac {\sqrt {g x^{2} + 4} x^{2}}{g} - \frac {8 \, \sqrt {g x^{2} + 4}}{g^{2}}\right )} b \log \left ({\left (d g x^{2} + 4 \, d\right )}^{p} c\right ) + \frac {1}{3} \, {\left (\frac {\sqrt {g x^{2} + 4} x^{2}}{g} - \frac {8 \, \sqrt {g x^{2} + 4}}{g^{2}}\right )} a - \frac {2 \, {\left ({\left (g x^{2} + 4\right )}^{\frac {3}{2}} - 36 \, \sqrt {g x^{2} + 4}\right )} b p}{9 \, g^{2}} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="m axima")
Output:
1/3*(sqrt(g*x^2 + 4)*x^2/g - 8*sqrt(g*x^2 + 4)/g^2)*b*log((d*g*x^2 + 4*d)^ p*c) + 1/3*(sqrt(g*x^2 + 4)*x^2/g - 8*sqrt(g*x^2 + 4)/g^2)*a - 2/9*((g*x^2 + 4)^(3/2) - 36*sqrt(g*x^2 + 4))*b*p/g^2
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {\frac {{\left (3 \, {\left ({\left (g x^{2} + 4\right )}^{\frac {3}{2}} - 12 \, \sqrt {g x^{2} + 4}\right )} \log \left (d g x^{2} + 4 \, d\right ) - 2 \, {\left (g x^{2} + 4\right )}^{\frac {3}{2}} + 72 \, \sqrt {g x^{2} + 4}\right )} b p}{g} + \frac {3 \, {\left ({\left (g x^{2} + 4\right )}^{\frac {3}{2}} - 12 \, \sqrt {g x^{2} + 4}\right )} b \log \left (c\right )}{g} + \frac {3 \, {\left ({\left (g x^{2} + 4\right )}^{\frac {3}{2}} - 12 \, \sqrt {g x^{2} + 4}\right )} a}{g}}{9 \, g} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x, algorithm="g iac")
Output:
1/9*((3*((g*x^2 + 4)^(3/2) - 12*sqrt(g*x^2 + 4))*log(d*g*x^2 + 4*d) - 2*(g *x^2 + 4)^(3/2) + 72*sqrt(g*x^2 + 4))*b*p/g + 3*((g*x^2 + 4)^(3/2) - 12*sq rt(g*x^2 + 4))*b*log(c)/g + 3*((g*x^2 + 4)^(3/2) - 12*sqrt(g*x^2 + 4))*a/g )/g
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d\,g\,x^2+4\,d\right )}^p\right )\right )}{\sqrt {g\,x^2+4}} \,d x \] Input:
int((x^3*(a + b*log(c*(4*d + d*g*x^2)^p)))/(g*x^2 + 4)^(1/2),x)
Output:
int((x^3*(a + b*log(c*(4*d + d*g*x^2)^p)))/(g*x^2 + 4)^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.39 \[ \int \frac {x^3 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {\sqrt {g \,x^{2}+4}\, \left (3 \,\mathrm {log}\left (\frac {d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+4}\, x +g \,x^{2}+4\right )^{2 p} 4^{p} c}{\left (2 \sqrt {g \,x^{2}+4}+2 \sqrt {g}\, x \right )^{2 p}}\right ) b g \,x^{2}-24 \,\mathrm {log}\left (\frac {d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+4}\, x +g \,x^{2}+4\right )^{2 p} 4^{p} c}{\left (2 \sqrt {g \,x^{2}+4}+2 \sqrt {g}\, x \right )^{2 p}}\right ) b +3 a g \,x^{2}-24 a -2 b g p \,x^{2}+64 b p \right )}{9 g^{2}} \] Input:
int(x^3*(a+b*log(c*(d*g*x^2+4*d)^p))/(g*x^2+4)^(1/2),x)
Output:
(sqrt(g*x**2 + 4)*(3*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*x**2 - 24*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 + 3*a*g*x**2 - 24*a - 2*b*g*p*x**2 + 64 *b*p))/(9*g**2)