Integrand size = 22, antiderivative size = 22 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=-48 f p^3 x+\frac {3 d g p^3 x^2}{e}-\frac {3 g p^3 \left (d+e x^2\right )^2}{16 e^2}+\frac {48 \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {24 i \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}-\frac {48 \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}+24 f p^2 x \log \left (c \left (d+e x^2\right )^p\right )-\frac {3 d g p^2 \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e^2}+\frac {3 g p^2 \left (d+e x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{8 e^2}-\frac {24 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{\sqrt {e}}-6 f p x \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {3 d g p \left (d+e x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right )}{2 e^2}-\frac {3 g p \left (d+e x^2\right )^2 \log ^2\left (c \left (d+e x^2\right )^p\right )}{8 e^2}+f x \log ^3\left (c \left (d+e x^2\right )^p\right )-\frac {d g \left (d+e x^2\right ) \log ^3\left (c \left (d+e x^2\right )^p\right )}{2 e^2}+\frac {g \left (d+e x^2\right )^2 \log ^3\left (c \left (d+e x^2\right )^p\right )}{4 e^2}-\frac {24 i \sqrt {d} f p^3 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}+6 d f p \text {Int}\left (\frac {\log ^2\left (c \left (d+e x^2\right )^p\right )}{d+e x^2},x\right ) \]
-48*f*p^3*x+3*d*g*p^3*x^2/e-3/16*g*p^3*(e*x^2+d)^2/e^2+24*f*p^2*x*ln(c*(e* x^2+d)^p)-3*d*g*p^2*(e*x^2+d)*ln(c*(e*x^2+d)^p)/e^2+3/8*g*p^2*(e*x^2+d)^2* ln(c*(e*x^2+d)^p)/e^2-6*f*p*x*ln(c*(e*x^2+d)^p)^2+3/2*d*g*p*(e*x^2+d)*ln(c *(e*x^2+d)^p)^2/e^2-3/8*g*p*(e*x^2+d)^2*ln(c*(e*x^2+d)^p)^2/e^2+f*x*ln(c*( e*x^2+d)^p)^3-1/2*d*g*(e*x^2+d)*ln(c*(e*x^2+d)^p)^3/e^2+1/4*g*(e*x^2+d)^2* ln(c*(e*x^2+d)^p)^3/e^2+48*f*p^3*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1/2) -24*I*f*p^3*polylog(2,1-2*d^(1/2)/(d^(1/2)+I*x*e^(1/2)))*d^(1/2)/e^(1/2)-2 4*f*p^2*arctan(x*e^(1/2)/d^(1/2))*ln(c*(e*x^2+d)^p)*d^(1/2)/e^(1/2)-48*f*p ^3*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)-24*I*f*p^3*arctan(x*e^(1/2)/d^(1/2))^2*d^(1/2)/e^(1/2)+6*d*f*p*Unin tegrable(ln(c*(e*x^2+d)^p)^2/(e*x^2+d),x)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1051\) vs. \(2(518)=1036\).
Time = 1.30 (sec) , antiderivative size = 1051, normalized size of antiderivative = 47.77 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {1}{4} g x^4 \log ^3\left (c \left (d+e x^2\right )^p\right )+\frac {6 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-p \log \left (d+e x^2\right )+\log \left (c \left (d+e x^2\right )^p\right )\right )^2}{\sqrt {e}}+3 f p x \log \left (d+e x^2\right ) \left (-p \log \left (d+e x^2\right )+\log \left (c \left (d+e x^2\right )^p\right )\right )^2+f x \left (-p \log \left (d+e x^2\right )+\log \left (c \left (d+e x^2\right )^p\right )\right )^2 \left (-6 p-p \log \left (d+e x^2\right )+\log \left (c \left (d+e x^2\right )^p\right )\right )-\frac {3}{4} g p \left (-\frac {7 d p^2 x^2}{2 e}+\frac {p^2 x^4}{4}+\frac {d^2 p^2 \log \left (d+e x^2\right )}{2 e^2}+\frac {3 d^2 p \log \left (c \left (d+e x^2\right )^p\right )}{e^2}+\frac {3 d p x^2 \log \left (c \left (d+e x^2\right )^p\right )}{e}-\frac {1}{2} p x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac {3 d^2 \log ^2\left (c \left (d+e x^2\right )^p\right )}{2 e^2}-\frac {d x^2 \log ^2\left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} x^4 \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {d^2 \log ^3\left (c \left (d+e x^2\right )^p\right )}{3 e^2 p}\right )+3 f p^2 \left (-p \log \left (d+e x^2\right )+\log \left (c \left (d+e x^2\right )^p\right )\right ) \left (x \log ^2\left (d+e x^2\right )-\frac {4 \left (-i \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2+\sqrt {e} x \left (-2+\log \left (d+e x^2\right )\right )-\sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-2+2 \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )+\log \left (d+e x^2\right )\right )-i \sqrt {d} \operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} x}{-i \sqrt {d}+\sqrt {e} x}\right )\right )}{\sqrt {e}}\right )+\frac {f p^3 \left (-48 \sqrt {-d^2} \sqrt {d+e x^2} \sqrt {1-\frac {d}{d+e x^2}} \arcsin \left (\frac {\sqrt {d}}{\sqrt {d+e x^2}}\right )-6 \sqrt {-d^2} \sqrt {1-\frac {d}{d+e x^2}} \left (8 \sqrt {d} \, _4F_3\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2},\frac {3}{2};\frac {d}{d+e x^2}\right )+4 \sqrt {d} \, _3F_2\left (\frac {1}{2},\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {d}{d+e x^2}\right ) \log \left (d+e x^2\right )+\sqrt {d+e x^2} \arcsin \left (\frac {\sqrt {d}}{\sqrt {d+e x^2}}\right ) \log ^2\left (d+e x^2\right )\right )+\sqrt {-d} e x^2 \left (-48+24 \log \left (d+e x^2\right )-6 \log ^2\left (d+e x^2\right )+\log ^3\left (d+e x^2\right )\right )+24 d \sqrt {e x^2} \text {arctanh}\left (\frac {\sqrt {e x^2}}{\sqrt {-d}}\right ) \left (\log \left (d+e x^2\right )-\log \left (\frac {d+e x^2}{d}\right )\right )+6 (-d)^{3/2} \sqrt {1-\frac {d+e x^2}{d}} \left (\log ^2\left (\frac {d+e x^2}{d}\right )-4 \log \left (\frac {d+e x^2}{d}\right ) \log \left (\frac {1}{2} \left (1+\sqrt {1-\frac {d+e x^2}{d}}\right )\right )+2 \log ^2\left (\frac {1}{2} \left (1+\sqrt {1-\frac {d+e x^2}{d}}\right )\right )-4 \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {1}{2} \sqrt {1-\frac {d+e x^2}{d}}\right )\right )\right )}{\sqrt {-d} e x} \]
(g*x^4*Log[c*(d + e*x^2)^p]^3)/4 + (6*Sqrt[d]*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[ d]]*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p])^2)/Sqrt[e] + 3*f*p*x*Log[ d + e*x^2]*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p])^2 + f*x*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p])^2*(-6*p - p*Log[d + e*x^2] + Log[c*(d + e*x^2)^p]) - (3*g*p*((-7*d*p^2*x^2)/(2*e) + (p^2*x^4)/4 + (d^2*p^2*Log[d + e*x^2])/(2*e^2) + (3*d^2*p*Log[c*(d + e*x^2)^p])/e^2 + (3*d*p*x^2*Log[c* (d + e*x^2)^p])/e - (p*x^4*Log[c*(d + e*x^2)^p])/2 - (3*d^2*Log[c*(d + e*x ^2)^p]^2)/(2*e^2) - (d*x^2*Log[c*(d + e*x^2)^p]^2)/e + (x^4*Log[c*(d + e*x ^2)^p]^2)/2 + (d^2*Log[c*(d + e*x^2)^p]^3)/(3*e^2*p)))/4 + 3*f*p^2*(-(p*Lo g[d + e*x^2]) + Log[c*(d + e*x^2)^p])*(x*Log[d + e*x^2]^2 - (4*((-I)*Sqrt[ d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]^2 + Sqrt[e]*x*(-2 + Log[d + e*x^2]) - Sqrt[ d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(-2 + 2*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e ]*x)] + Log[d + e*x^2]) - I*Sqrt[d]*PolyLog[2, (I*Sqrt[d] + Sqrt[e]*x)/((- I)*Sqrt[d] + Sqrt[e]*x)]))/Sqrt[e]) + (f*p^3*(-48*Sqrt[-d^2]*Sqrt[d + e*x^ 2]*Sqrt[1 - d/(d + e*x^2)]*ArcSin[Sqrt[d]/Sqrt[d + e*x^2]] - 6*Sqrt[-d^2]* Sqrt[1 - d/(d + e*x^2)]*(8*Sqrt[d]*HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/2}, d/(d + e*x^2)] + 4*Sqrt[d]*HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, d/(d + e*x^2)]*Log[d + e*x^2] + Sqrt[d + e*x^2]*ArcSin[S qrt[d]/Sqrt[d + e*x^2]]*Log[d + e*x^2]^2) + Sqrt[-d]*e*x^2*(-48 + 24*Log[d + e*x^2] - 6*Log[d + e*x^2]^2 + Log[d + e*x^2]^3) + 24*d*Sqrt[e*x^2]*A...
Not integrable
Time = 0.92 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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.
\(\displaystyle \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2921 |
\(\displaystyle \int \left (f \log ^3\left (c \left (d+e x^2\right )^p\right )+g x^3 \log ^3\left (c \left (d+e x^2\right )^p\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 6 d f p \int \frac {\log ^2\left (c \left (e x^2+d\right )^p\right )}{e x^2+d}dx-\frac {24 \sqrt {d} f p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{\sqrt {e}}-\frac {24 i \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}+\frac {48 \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {48 \sqrt {d} f p^3 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}+\frac {3 g p^2 \left (d+e x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{8 e^2}-\frac {3 d g p^2 \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e^2}+\frac {g \left (d+e x^2\right )^2 \log ^3\left (c \left (d+e x^2\right )^p\right )}{4 e^2}-\frac {d g \left (d+e x^2\right ) \log ^3\left (c \left (d+e x^2\right )^p\right )}{2 e^2}-\frac {3 g p \left (d+e x^2\right )^2 \log ^2\left (c \left (d+e x^2\right )^p\right )}{8 e^2}+\frac {3 d g p \left (d+e x^2\right ) \log ^2\left (c \left (d+e x^2\right )^p\right )}{2 e^2}+24 f p^2 x \log \left (c \left (d+e x^2\right )^p\right )+f x \log ^3\left (c \left (d+e x^2\right )^p\right )-6 f p x \log ^2\left (c \left (d+e x^2\right )^p\right )-\frac {3 g p^3 \left (d+e x^2\right )^2}{16 e^2}-\frac {24 i \sqrt {d} f p^3 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{\sqrt {e}}+\frac {3 d g p^3 x^2}{e}-48 f p^3 x\) |
3.3.99.3.1 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]))
Not integrable
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
\[\int \left (g \,x^{3}+f \right ) {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}^{3}d x\]
Not integrable
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\int { {\left (g x^{3} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{3} \,d x } \]
Not integrable
Time = 10.50 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\int \left (f + g x^{3}\right ) \log {\left (c \left (d + e x^{2}\right )^{p} \right )}^{3}\, dx \]
Exception generated. \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]
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
Not integrable
Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\int { {\left (g x^{3} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )^{3} \,d x } \]
Not integrable
Time = 1.58 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (f+g x^3\right ) \log ^3\left (c \left (d+e x^2\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}^3\,\left (g\,x^3+f\right ) \,d x \]