Integrand size = 34, antiderivative size = 113 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {2 b f p \sqrt {f+g x^2}}{g^2}-\frac {2 b p \left (f+g x^2\right )^{3/2}}{9 g^2}-\frac {f \sqrt {f+g x^2} \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{g^2}+\frac {\left (f+g x^2\right )^{3/2} \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{3 g^2} \] Output:
2*b*f*p*(g*x^2+f)^(1/2)/g^2-2/9*b*p*(g*x^2+f)^(3/2)/g^2-f*(g*x^2+f)^(1/2)* (a+b*ln(c*(d*g*x^2+d*f)^p))/g^2+1/3*(g*x^2+f)^(3/2)*(a+b*ln(c*(d*g*x^2+d*f )^p))/g^2
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {\sqrt {f+g x^2} \left (-6 a f+16 b f p+3 a g x^2-2 b g p x^2+b \left (-6 f+3 g x^2\right ) \log \left (c \left (d \left (f+g x^2\right )\right )^p\right )\right )}{9 g^2} \] Input:
Integrate[(x^3*(a + b*Log[c*(d*f + d*g*x^2)^p]))/Sqrt[f + g*x^2],x]
Output:
(Sqrt[f + g*x^2]*(-6*a*f + 16*b*f*p + 3*a*g*x^2 - 2*b*g*p*x^2 + b*(-6*f + 3*g*x^2)*Log[c*(d*(f + 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.
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx\) |
\(\Big \downarrow \) 2929 |
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}}dx\) |
Input:
Int[(x^3*(a + b*Log[c*(d*f + d*g*x^2)^p]))/Sqrt[f + 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]
\[\int \frac {x^{3} \left (a +b \ln \left (c \left (d g \,x^{2}+d f \right )^{p}\right )\right )}{\sqrt {g \,x^{2}+f}}d x\]
Input:
int(x^3*(a+b*ln(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x)
Output:
int(x^3*(a+b*ln(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x)
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {{\left (16 \, b f p - {\left (2 \, b g p - 3 \, a g\right )} x^{2} - 6 \, a f + 3 \, {\left (b g p x^{2} - 2 \, b f p\right )} \log \left (d g x^{2} + d f\right ) + 3 \, {\left (b g x^{2} - 2 \, b f\right )} \log \left (c\right )\right )} \sqrt {g x^{2} + f}}{9 \, g^{2}} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x, algorithm="f ricas")
Output:
1/9*(16*b*f*p - (2*b*g*p - 3*a*g)*x^2 - 6*a*f + 3*(b*g*p*x^2 - 2*b*f*p)*lo g(d*g*x^2 + d*f) + 3*(b*g*x^2 - 2*b*f)*log(c))*sqrt(g*x^2 + f)/g^2
Time = 1.69 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.53 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\begin {cases} - \frac {2 a f \sqrt {f + g x^{2}}}{3 g^{2}} + \frac {a x^{2} \sqrt {f + g x^{2}}}{3 g} + \frac {16 b f p \sqrt {f + g x^{2}}}{9 g^{2}} - \frac {2 b f \sqrt {f + g x^{2}} \log {\left (c \left (d f + d g x^{2}\right )^{p} \right )}}{3 g^{2}} - \frac {2 b p x^{2} \sqrt {f + g x^{2}}}{9 g} + \frac {b x^{2} \sqrt {f + g x^{2}} \log {\left (c \left (d f + d g x^{2}\right )^{p} \right )}}{3 g} & \text {for}\: g \neq 0 \\\frac {x^{4} \left (a + b \log {\left (c \left (d f\right )^{p} \right )}\right )}{4 \sqrt {f}} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(a+b*ln(c*(d*g*x**2+d*f)**p))/(g*x**2+f)**(1/2),x)
Output:
Piecewise((-2*a*f*sqrt(f + g*x**2)/(3*g**2) + a*x**2*sqrt(f + g*x**2)/(3*g ) + 16*b*f*p*sqrt(f + g*x**2)/(9*g**2) - 2*b*f*sqrt(f + g*x**2)*log(c*(d*f + d*g*x**2)**p)/(3*g**2) - 2*b*p*x**2*sqrt(f + g*x**2)/(9*g) + b*x**2*sqr t(f + g*x**2)*log(c*(d*f + d*g*x**2)**p)/(3*g), Ne(g, 0)), (x**4*(a + b*lo g(c*(d*f)**p))/(4*sqrt(f)), True))
Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {1}{3} \, {\left (\frac {\sqrt {g x^{2} + f} x^{2}}{g} - \frac {2 \, \sqrt {g x^{2} + f} f}{g^{2}}\right )} b \log \left ({\left (d g x^{2} + d f\right )}^{p} c\right ) + \frac {1}{3} \, {\left (\frac {\sqrt {g x^{2} + f} x^{2}}{g} - \frac {2 \, \sqrt {g x^{2} + f} f}{g^{2}}\right )} a - \frac {2 \, {\left ({\left (g x^{2} + f\right )}^{\frac {3}{2}} - 9 \, \sqrt {g x^{2} + f} f\right )} b p}{9 \, g^{2}} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x, algorithm="m axima")
Output:
1/3*(sqrt(g*x^2 + f)*x^2/g - 2*sqrt(g*x^2 + f)*f/g^2)*b*log((d*g*x^2 + d*f )^p*c) + 1/3*(sqrt(g*x^2 + f)*x^2/g - 2*sqrt(g*x^2 + f)*f/g^2)*a - 2/9*((g *x^2 + f)^(3/2) - 9*sqrt(g*x^2 + f)*f)*b*p/g^2
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {\frac {{\left (3 \, {\left ({\left (g x^{2} + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x^{2} + f} f\right )} \log \left (d g x^{2} + d f\right ) - 2 \, {\left (g x^{2} + f\right )}^{\frac {3}{2}} + 18 \, \sqrt {g x^{2} + f} f\right )} b p}{g} + \frac {3 \, {\left ({\left (g x^{2} + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x^{2} + f} f\right )} b \log \left (c\right )}{g} + \frac {3 \, {\left ({\left (g x^{2} + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x^{2} + f} f\right )} a}{g}}{9 \, g} \] Input:
integrate(x^3*(a+b*log(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x, algorithm="g iac")
Output:
1/9*((3*((g*x^2 + f)^(3/2) - 3*sqrt(g*x^2 + f)*f)*log(d*g*x^2 + d*f) - 2*( g*x^2 + f)^(3/2) + 18*sqrt(g*x^2 + f)*f)*b*p/g + 3*((g*x^2 + f)^(3/2) - 3* sqrt(g*x^2 + f)*f)*b*log(c)/g + 3*((g*x^2 + f)^(3/2) - 3*sqrt(g*x^2 + f)*f )*a/g)/g
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d\,g\,x^2+d\,f\right )}^p\right )\right )}{\sqrt {g\,x^2+f}} \,d x \] Input:
int((x^3*(a + b*log(c*(d*f + d*g*x^2)^p)))/(f + g*x^2)^(1/2),x)
Output:
int((x^3*(a + b*log(c*(d*f + d*g*x^2)^p)))/(f + g*x^2)^(1/2), x)
Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.42 \[ \int \frac {x^3 \left (a+b \log \left (c \left (d f+d g x^2\right )^p\right )\right )}{\sqrt {f+g x^2}} \, dx=\frac {\sqrt {g \,x^{2}+f}\, \left (-6 \,\mathrm {log}\left (\frac {f^{p} d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+f}\, x +f +g \,x^{2}\right )^{2 p} c}{\left (\sqrt {f}\, \sqrt {g \,x^{2}+f}+\sqrt {g}\, \sqrt {f}\, x \right )^{2 p}}\right ) b f +3 \,\mathrm {log}\left (\frac {f^{p} d^{p} \left (\sqrt {g}\, \sqrt {g \,x^{2}+f}\, x +f +g \,x^{2}\right )^{2 p} c}{\left (\sqrt {f}\, \sqrt {g \,x^{2}+f}+\sqrt {g}\, \sqrt {f}\, x \right )^{2 p}}\right ) b g \,x^{2}-6 a f +3 a g \,x^{2}+16 b f p -2 b g p \,x^{2}\right )}{9 g^{2}} \] Input:
int(x^3*(a+b*log(c*(d*g*x^2+d*f)^p))/(g*x^2+f)^(1/2),x)
Output:
(sqrt(f + g*x**2)*( - 6*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 + 3*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*g*x**2 - 6*a*f + 3 *a*g*x**2 + 16*b*f*p - 2*b*g*p*x**2))/(9*g**2)