Integrand size = 18, antiderivative size = 294 \[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {32 a^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {4 i a^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}-\frac {8 a^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4 i a^{3/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}} \]
-32/9*a*p^2*x/b+8/27*p^2*x^3+32/9*a^(3/2)*p^2*arctan(x*b^(1/2)/a^(1/2))/b^ (3/2)-4/3*I*a^(3/2)*p^2*arctan(x*b^(1/2)/a^(1/2))^2/b^(3/2)+4/3*a*p*x*ln(c *(b*x^2+a)^p)/b-4/9*p*x^3*ln(c*(b*x^2+a)^p)-4/3*a^(3/2)*p*arctan(x*b^(1/2) /a^(1/2))*ln(c*(b*x^2+a)^p)/b^(3/2)+1/3*x^3*ln(c*(b*x^2+a)^p)^2-8/3*a^(3/2 )*p^2*arctan(x*b^(1/2)/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(3/2 )-4/3*I*a^(3/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(3/2)
Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.76 \[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-36 i a^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2-12 a^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-8 p+6 p \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )+3 \log \left (c \left (a+b x^2\right )^p\right )\right )+\sqrt {b} x \left (8 p^2 \left (-12 a+b x^2\right )+12 p \left (3 a-b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )+9 b x^2 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )-36 i a^{3/2} p^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{27 b^{3/2}} \]
((-36*I)*a^(3/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 - 12*a^(3/2)*p*ArcTan[( Sqrt[b]*x)/Sqrt[a]]*(-8*p + 6*p*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)] + 3*Log[c*(a + b*x^2)^p]) + Sqrt[b]*x*(8*p^2*(-12*a + b*x^2) + 12*p*(3*a - b*x^2)*Log[c*(a + b*x^2)^p] + 9*b*x^2*Log[c*(a + b*x^2)^p]^2) - (36*I)*a^( 3/2)*p^2*PolyLog[2, (I*Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/( 27*b^(3/2))
Time = 0.54 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2907, 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.
| \(\displaystyle \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle \frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{3} b p \int \frac {x^4 \log \left (c \left (b x^2+a\right )^p\right )}{b x^2+a}dx\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle \frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{3} b p \int \left (\frac {\log \left (c \left (b x^2+a\right )^p\right ) a^2}{b^2 \left (b x^2+a\right )}-\frac {\log \left (c \left (b x^2+a\right )^p\right ) a}{b^2}+\frac {x^2 \log \left (c \left (b x^2+a\right )^p\right )}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{3} b p \left (\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^{5/2}}+\frac {i a^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{b^{5/2}}-\frac {8 a^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{5/2}}+\frac {2 a^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{b^{5/2}}+\frac {i a^{3/2} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{b^{5/2}}-\frac {a x \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {8 a p x}{3 b^2}+\frac {x^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {2 p x^3}{9 b}\right )\) |
(x^3*Log[c*(a + b*x^2)^p]^2)/3 - (4*b*p*((8*a*p*x)/(3*b^2) - (2*p*x^3)/(9* b) - (8*a^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(3*b^(5/2)) + (I*a^(3/2)*p* ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/b^(5/2) + (2*a^(3/2)*p*ArcTan[(Sqrt[b]*x)/S qrt[a]]*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/b^(5/2) - (a*x*Log[c*(a + b*x^2)^p])/b^2 + (x^3*Log[c*(a + b*x^2)^p])/(3*b) + (a^(3/2)*ArcTan[(Sqr t[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/b^(5/2) + (I*a^(3/2)*p*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/b^(5/2)))/3
3.1.85.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*((f_.)*( x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])^q /(f*(m + 1))), x] - Simp[b*e*n*p*(q/(f^n*(m + 1))) Int[(f*x)^(m + n)*((a + b*Log[c*(d + e*x^n)^p])^(q - 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d , e, f, m, p}, x] && IGtQ[q, 1] && IntegerQ[n] && NeQ[m, -1]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.59 (sec) , antiderivative size = 565, normalized size of antiderivative = 1.92
| method | result | size |
| risch | \(\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2} x^{3}}{3}-\frac {4 p \,x^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{9}+\frac {4 p a x \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 b}+\frac {4 p^{2} a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{3 b \sqrt {a b}}-\frac {4 p \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 b \sqrt {a b}}+\frac {8 p^{2} x^{3}}{27}-\frac {32 a \,p^{2} x}{9 b}+\frac {32 p^{2} a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{9 b \sqrt {a b}}-\frac {4 p^{2} b \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (b \,x^{2}+a \right )-2 b \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{4 \underline {\hspace {1.25 ex}}\alpha b}+\frac {\underline {\hspace {1.25 ex}}\alpha \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{2 a}+\frac {\underline {\hspace {1.25 ex}}\alpha \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{2 a}\right )\right ) a^{2}}{2 b^{3} \underline {\hspace {1.25 ex}}\alpha }\right )}{3}+\left (i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right ) \left (\frac {x^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3}-\frac {2 p b \left (\frac {\frac {1}{3} b \,x^{3}-a x}{b^{2}}+\frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\right )}{3}\right )+\frac {{\left (i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right )}^{2} x^{3}}{12}\) | \(565\) |
1/3*ln((b*x^2+a)^p)^2*x^3-4/9*p*x^3*ln((b*x^2+a)^p)+4/3*p/b*a*x*ln((b*x^2+ a)^p)+4/3*p^2/b*a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln(b*x^2+a)-4/3*p/ b*a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln((b*x^2+a)^p)+8/27*p^2*x^3-32/ 9*a*p^2*x/b+32/9*p^2/b*a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))-4/3*p^2*b*S um(1/2*(ln(x-_alpha)*ln(b*x^2+a)-2*b*(1/4/_alpha/b*ln(x-_alpha)^2+1/2*_alp ha/a*ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+1/2*_alpha/a*dilog(1/2*(x+_alp ha)/_alpha)))*a^2/b^3/_alpha,_alpha=RootOf(_Z^2*b+a))+(I*Pi*csgn(I*(b*x^2+ a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^ p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csg n(I*c)+2*ln(c))*(1/3*x^3*ln((b*x^2+a)^p)-2/3*p*b*(1/b^2*(1/3*b*x^3-a*x)+a^ 2/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))+1/12*(I*Pi*csgn(I*(b*x^2+a)^p) *csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*cs gn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c )+2*ln(c))^2*x^3
\[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
\[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}\, dx \]
\[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
1/3*p^2*x^3*log(b*x^2 + a)^2 + integrate(1/3*(3*b*x^4*log(c)^2 + 3*a*x^2*l og(c)^2 - 2*((2*p^2 - 3*p*log(c))*b*x^4 - 3*a*p*x^2*log(c))*log(b*x^2 + a) )/(b*x^2 + a), x)
\[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
Timed out. \[ \int x^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int x^2\,{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2 \,d x \]