Integrand size = 22, antiderivative size = 548 \[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=8 f p^2 x-\frac {32 d g p^2 x}{9 e}+\frac {8}{27} g p^2 x^3-\frac {8 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {32 d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 e^{3/2}}+\frac {4 i \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}-\frac {4 i d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{3 e^{3/2}}+\frac {8 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}-\frac {8 d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{3 e^{3/2}}-4 f p x \log \left (c \left (d+e x^2\right )^p\right )+\frac {4 d g p x \log \left (c \left (d+e x^2\right )^p\right )}{3 e}-\frac {4}{9} g p x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {4 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{\sqrt {e}}-\frac {4 d^{3/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3 e^{3/2}}+f x \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {4 i \sqrt {d} f p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}-\frac {4 i d^{3/2} g p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{3 e^{3/2}} \]
8*f*p^2*x-32/9*d*g*p^2*x/e+8/27*g*p^2*x^3+32/9*d^(3/2)*g*p^2*arctan(x*e^(1 /2)/d^(1/2))/e^(3/2)-4/3*I*d^(3/2)*g*p^2*polylog(2,1-2*d^(1/2)/(d^(1/2)+I* x*e^(1/2)))/e^(3/2)-4*f*p*x*ln(c*(e*x^2+d)^p)+4/3*d*g*p*x*ln(c*(e*x^2+d)^p )/e-4/9*g*p*x^3*ln(c*(e*x^2+d)^p)-4/3*d^(3/2)*g*p*arctan(x*e^(1/2)/d^(1/2) )*ln(c*(e*x^2+d)^p)/e^(3/2)+f*x*ln(c*(e*x^2+d)^p)^2+1/3*g*x^3*ln(c*(e*x^2+ d)^p)^2-8/3*d^(3/2)*g*p^2*arctan(x*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+ I*x*e^(1/2)))/e^(3/2)+4*I*f*p^2*arctan(x*e^(1/2)/d^(1/2))^2*d^(1/2)/e^(1/2 )-8*f*p^2*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1/2)-4/3*I*d^(3/2)*g*p^2*ar ctan(x*e^(1/2)/d^(1/2))^2/e^(3/2)+4*f*p*arctan(x*e^(1/2)/d^(1/2))*ln(c*(e* x^2+d)^p)*d^(1/2)/e^(1/2)+8*f*p^2*arctan(x*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/( d^(1/2)+I*x*e^(1/2)))*d^(1/2)/e^(1/2)+4*I*f*p^2*polylog(2,1-2*d^(1/2)/(d^( 1/2)+I*x*e^(1/2)))*d^(1/2)/e^(1/2)
Time = 0.17 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.51 \[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-36 i \sqrt {d} (-3 e f+d g) p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2-12 \sqrt {d} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (2 (9 e f-4 d g) p+6 (-3 e f+d g) p \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )+(-9 e f+3 d g) \log \left (c \left (d+e x^2\right )^p\right )\right )+\sqrt {e} x \left (8 p^2 \left (27 e f-12 d g+e g x^2\right )-12 p \left (9 e f-3 d g+e g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )+9 e \left (3 f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right )\right )-36 i \sqrt {d} (-3 e f+d g) p^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} x}{-i \sqrt {d}+\sqrt {e} x}\right )}{27 e^{3/2}} \]
((-36*I)*Sqrt[d]*(-3*e*f + d*g)*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]^2 - 12*Sqr t[d]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(2*(9*e*f - 4*d*g)*p + 6*(-3*e*f + d*g) *p*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x)] + (-9*e*f + 3*d*g)*Log[c*(d + e*x^2)^p]) + Sqrt[e]*x*(8*p^2*(27*e*f - 12*d*g + e*g*x^2) - 12*p*(9*e*f - 3*d*g + e*g*x^2)*Log[c*(d + e*x^2)^p] + 9*e*(3*f + g*x^2)*Log[c*(d + e*x^2 )^p]^2) - (36*I)*Sqrt[d]*(-3*e*f + d*g)*p^2*PolyLog[2, (I*Sqrt[d] + Sqrt[e ]*x)/((-I)*Sqrt[d] + Sqrt[e]*x)])/(27*e^(3/2))
Time = 0.85 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2921, 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.
| \(\displaystyle \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2921 |
| \(\displaystyle \int \left (f \log ^2\left (c \left (d+e x^2\right )^p\right )+g x^2 \log ^2\left (c \left (d+e x^2\right )^p\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 d^{3/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3 e^{3/2}}+\frac {4 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{\sqrt {e}}-\frac {4 i d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{3 e^{3/2}}+\frac {32 d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 e^{3/2}}-\frac {8 d^{3/2} g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{3 e^{3/2}}+\frac {4 i \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}-\frac {8 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {8 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}+f x \log ^2\left (c \left (d+e x^2\right )^p\right )-4 f p x \log \left (c \left (d+e x^2\right )^p\right )+\frac {4 d g p x \log \left (c \left (d+e x^2\right )^p\right )}{3 e}+\frac {1}{3} g x^3 \log ^2\left (c \left (d+e x^2\right )^p\right )-\frac {4}{9} g p x^3 \log \left (c \left (d+e x^2\right )^p\right )-\frac {4 i d^{3/2} g p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 i \sqrt {d} f p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{\sqrt {e}}-\frac {32 d g p^2 x}{9 e}+8 f p^2 x+\frac {8}{27} g p^2 x^3\) |
8*f*p^2*x - (32*d*g*p^2*x)/(9*e) + (8*g*p^2*x^3)/27 - (8*Sqrt[d]*f*p^2*Arc Tan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] + (32*d^(3/2)*g*p^2*ArcTan[(Sqrt[e]*x)/S qrt[d]])/(9*e^(3/2)) + ((4*I)*Sqrt[d]*f*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]^2) /Sqrt[e] - (((4*I)/3)*d^(3/2)*g*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]^2)/e^(3/2) + (8*Sqrt[d]*f*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x)])/Sqrt[e] - (8*d^(3/2)*g*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log [(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x)])/(3*e^(3/2)) - 4*f*p*x*Log[c*(d + e* x^2)^p] + (4*d*g*p*x*Log[c*(d + e*x^2)^p])/(3*e) - (4*g*p*x^3*Log[c*(d + e *x^2)^p])/9 + (4*Sqrt[d]*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[c*(d + e*x^2) ^p])/Sqrt[e] - (4*d^(3/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[c*(d + e*x^2 )^p])/(3*e^(3/2)) + f*x*Log[c*(d + e*x^2)^p]^2 + (g*x^3*Log[c*(d + e*x^2)^ p]^2)/3 + ((4*I)*Sqrt[d]*f*p^2*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] + I*Sqr t[e]*x)])/Sqrt[e] - (((4*I)/3)*d^(3/2)*g*p^2*PolyLog[2, 1 - (2*Sqrt[d])/(S qrt[d] + I*Sqrt[e]*x)])/e^(3/2)
3.3.74.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> With[{t = ExpandIntegrand[(a + b*Log[ c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; FreeQ[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && Integ erQ[r] && IntegerQ[s] && (EqQ[q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.83 (sec) , antiderivative size = 729, normalized size of antiderivative = 1.33
1/3*ln((e*x^2+d)^p)^2*g*x^3+ln((e*x^2+d)^p)^2*x*f-4/9*p*g*x^3*ln((e*x^2+d) ^p)+4/3*p/e*g*d*x*ln((e*x^2+d)^p)-4*p*f*x*ln((e*x^2+d)^p)+4/3*p^2/e*g*d^2/ (d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*ln(e*x^2+d)-4/3*p/e*g*d^2/(d*e)^(1/2)* arctan(x*e/(d*e)^(1/2))*ln((e*x^2+d)^p)-4*p^2*d/(d*e)^(1/2)*arctan(x*e/(d* e)^(1/2))*f*ln(e*x^2+d)+4*p*d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f*ln((e* x^2+d)^p)+8/27*g*p^2*x^3-32/9*d*g*p^2*x/e+32/9*p^2/e*g*d^2/(d*e)^(1/2)*arc tan(x*e/(d*e)^(1/2))+8*f*p^2*x-8*p^2*d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)) *f-4/3*p^2*e*Sum(1/2*(ln(x-_alpha)*ln(e*x^2+d)-2*e*(1/4/_alpha/e*ln(x-_alp ha)^2+1/2*_alpha/d*ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+1/2*_alpha/d*dil og(1/2*(x+_alpha)/_alpha)))*d*(d*g-3*e*f)/e^3/_alpha,_alpha=RootOf(_Z^2*e+ d))+(I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-I*Pi*csgn(I*(e*x^2+d )^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-I*Pi*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*csg n(I*c*(e*x^2+d)^p)^2*csgn(I*c)+2*ln(c))*(1/3*ln((e*x^2+d)^p)*g*x^3+ln((e*x ^2+d)^p)*x*f-2/3*p*e*(1/e^2*(1/3*e*g*x^3-d*g*x+3*e*f*x)+d*(d*g-3*e*f)/e^2/ (d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))))+1/4*(I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I *c*(e*x^2+d)^p)^2-I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c) -I*Pi*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+2*ln( c))^2*(1/3*g*x^3+f*x)
\[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\int { {\left (g x^{2} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2} \,d x } \]
\[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\int \left (f + g x^{2}\right ) \log {\left (c \left (d + e x^{2}\right )^{p} \right )}^{2}\, dx \]
Exception generated. \[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\int { {\left (g x^{2} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{2} \,d x } \]
Timed out. \[ \int \left (f+g x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}^2\,\left (g\,x^2+f\right ) \,d x \]