Integrand size = 23, antiderivative size = 154 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {2 d f p x}{3 e}-\frac {2 d^2 g p x}{5 e^2}-\frac {2}{9} f p x^3+\frac {2 d g p x^3}{15 e}-\frac {2}{25} g p x^5-\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right ) \]
2/3*d*f*p*x/e-2/5*d^2*g*p*x/e^2-2/9*f*p*x^3+2/15*d*g*p*x^3/e-2/25*g*p*x^5- 2/3*d^(3/2)*f*p*arctan(x*e^(1/2)/d^(1/2))/e^(3/2)+2/5*d^(5/2)*g*p*arctan(x *e^(1/2)/d^(1/2))/e^(5/2)+1/3*f*x^3*ln(c*(e*x^2+d)^p)+1/5*g*x^5*ln(c*(e*x^ 2+d)^p)
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {30 d^{3/2} (-5 e f+3 d g) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {e} x \left (-2 p \left (45 d^2 g-15 d e \left (5 f+g x^2\right )+e^2 x^2 \left (25 f+9 g x^2\right )\right )+15 e^2 x^2 \left (5 f+3 g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{225 e^{5/2}} \]
(30*d^(3/2)*(-5*e*f + 3*d*g)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]] + Sqrt[e]*x*(-2 *p*(45*d^2*g - 15*d*e*(5*f + g*x^2) + e^2*x^2*(25*f + 9*g*x^2)) + 15*e^2*x ^2*(5*f + 3*g*x^2)*Log[c*(d + e*x^2)^p]))/(225*e^(5/2))
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2926, 2009}
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.
\(\displaystyle \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2926 |
\(\displaystyle \int \left (f x^2 \log \left (c \left (d+e x^2\right )^p\right )+g x^4 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^2 g p x}{5 e^2}+\frac {2 d f p x}{3 e}+\frac {2 d g p x^3}{15 e}-\frac {2}{9} f p x^3-\frac {2}{25} g p x^5\) |
(2*d*f*p*x)/(3*e) - (2*d^2*g*p*x)/(5*e^2) - (2*f*p*x^3)/9 + (2*d*g*p*x^3)/ (15*e) - (2*g*p*x^5)/25 - (2*d^(3/2)*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e ^(3/2)) + (2*d^(5/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*e^(5/2)) + (f*x^3 *Log[c*(d + e*x^2)^p])/3 + (g*x^5*Log[c*(d + e*x^2)^p])/5
3.4.18.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b *Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e , f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & & IntegerQ[s]
Time = 1.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77
method | result | size |
parts | \(\frac {g \,x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+\frac {f \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3}-\frac {2 p e \left (\frac {\frac {3}{5} e^{2} g \,x^{5}-d g \,x^{3} e +\frac {5}{3} f \,x^{3} e^{2}+3 d^{2} g x -5 d e f x}{e^{3}}-\frac {d^{2} \left (3 d g -5 e f \right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{3} \sqrt {d e}}\right )}{15}\) | \(118\) |
risch | \(\left (\frac {1}{5} g \,x^{5}+\frac {1}{3} f \,x^{3}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{10}+\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{6}-\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{6}+\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}-\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{10}+\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{6}+\frac {\ln \left (c \right ) g \,x^{5}}{5}-\frac {2 g p \,x^{5}}{25}+\frac {\ln \left (c \right ) f \,x^{3}}{3}+\frac {2 d g p \,x^{3}}{15 e}-\frac {2 f p \,x^{3}}{9}+\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (-\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}-\frac {\sqrt {-d e}\, p d \ln \left (-\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}+\frac {\sqrt {-d e}\, p d \ln \left (\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {2 d^{2} g p x}{5 e^{2}}+\frac {2 d f p x}{3 e}\) | \(453\) |
1/5*g*x^5*ln(c*(e*x^2+d)^p)+1/3*f*x^3*ln(c*(e*x^2+d)^p)-2/15*p*e*(1/e^3*(3 /5*e^2*g*x^5-d*g*x^3*e+5/3*f*x^3*e^2+3*d^2*g*x-5*d*e*f*x)-d^2*(3*d*g-5*e*f )/e^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))
Time = 0.34 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.95 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 15 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}, -\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}\right ] \]
[-1/225*(18*e^2*g*p*x^5 + 10*(5*e^2*f - 3*d*e*g)*p*x^3 + 15*(5*d*e*f - 3*d ^2*g)*p*sqrt(-d/e)*log((e*x^2 + 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) - 30*(5 *d*e*f - 3*d^2*g)*p*x - 15*(3*e^2*g*p*x^5 + 5*e^2*f*p*x^3)*log(e*x^2 + d) - 15*(3*e^2*g*x^5 + 5*e^2*f*x^3)*log(c))/e^2, -1/225*(18*e^2*g*p*x^5 + 10* (5*e^2*f - 3*d*e*g)*p*x^3 + 30*(5*d*e*f - 3*d^2*g)*p*sqrt(d/e)*arctan(e*x* sqrt(d/e)/d) - 30*(5*d*e*f - 3*d^2*g)*p*x - 15*(3*e^2*g*p*x^5 + 5*e^2*f*p* x^3)*log(e*x^2 + d) - 15*(3*e^2*g*x^5 + 5*e^2*f*x^3)*log(c))/e^2]
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (158) = 316\).
Time = 31.34 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.08 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} & \text {for}\: d = 0 \\\frac {2 d^{3} g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {d^{3} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} g p x}{5 e^{2}} + \frac {2 d f p x}{3 e} + \frac {2 d g p x^{3}}{15 e} - \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
Piecewise(((f*x**3/3 + g*x**5/5)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((f*x* *3/3 + g*x**5/5)*log(c*d**p), Eq(e, 0)), (-2*f*p*x**3/9 + f*x**3*log(c*(e* x**2)**p)/3 - 2*g*p*x**5/25 + g*x**5*log(c*(e*x**2)**p)/5, Eq(d, 0)), (2*d **3*g*p*log(x - sqrt(-d/e))/(5*e**3*sqrt(-d/e)) - d**3*g*log(c*(d + e*x**2 )**p)/(5*e**3*sqrt(-d/e)) - 2*d**2*f*p*log(x - sqrt(-d/e))/(3*e**2*sqrt(-d /e)) + d**2*f*log(c*(d + e*x**2)**p)/(3*e**2*sqrt(-d/e)) - 2*d**2*g*p*x/(5 *e**2) + 2*d*f*p*x/(3*e) + 2*d*g*p*x**3/(15*e) - 2*f*p*x**3/9 + f*x**3*log (c*(d + e*x**2)**p)/3 - 2*g*p*x**5/25 + g*x**5*log(c*(d + e*x**2)**p)/5, T rue))
Exception generated. \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{25} \, {\left (2 \, g p - 5 \, g \log \left (c\right )\right )} x^{5} - \frac {{\left (10 \, e f p - 6 \, d g p - 15 \, e f \log \left (c\right )\right )} x^{3}}{45 \, e} + \frac {1}{15} \, {\left (3 \, g p x^{5} + 5 \, f p x^{3}\right )} \log \left (e x^{2} + d\right ) + \frac {2 \, {\left (5 \, d e f p - 3 \, d^{2} g p\right )} x}{15 \, e^{2}} - \frac {2 \, {\left (5 \, d^{2} e f p - 3 \, d^{3} g p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{15 \, \sqrt {d e} e^{2}} \]
-1/25*(2*g*p - 5*g*log(c))*x^5 - 1/45*(10*e*f*p - 6*d*g*p - 15*e*f*log(c)) *x^3/e + 1/15*(3*g*p*x^5 + 5*f*p*x^3)*log(e*x^2 + d) + 2/15*(5*d*e*f*p - 3 *d^2*g*p)*x/e^2 - 2/15*(5*d^2*e*f*p - 3*d^3*g*p)*arctan(e*x/sqrt(d*e))/(sq rt(d*e)*e^2)
Time = 1.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^5}{5}+\frac {f\,x^3}{3}\right )-x^3\,\left (\frac {2\,f\,p}{9}-\frac {2\,d\,g\,p}{15\,e}\right )-\frac {2\,g\,p\,x^5}{25}+\frac {d\,x\,\left (\frac {2\,f\,p}{3}-\frac {2\,d\,g\,p}{5\,e}\right )}{e}+\frac {2\,d^{3/2}\,p\,\mathrm {atan}\left (\frac {d^{3/2}\,\sqrt {e}\,p\,x\,\left (3\,d\,g-5\,e\,f\right )}{3\,d^3\,g\,p-5\,d^2\,e\,f\,p}\right )\,\left (3\,d\,g-5\,e\,f\right )}{15\,e^{5/2}} \]