Integrand size = 18, antiderivative size = 336 \[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {184 a^2 p^2 x}{75 b^2}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125}-\frac {184 a^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {4 i a^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}+\frac {8 a^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {4 i a^{5/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}} \]
184/75*a^2*p^2*x/b^2-64/225*a*p^2*x^3/b+8/125*p^2*x^5-184/75*a^(5/2)*p^2*a rctan(x*b^(1/2)/a^(1/2))/b^(5/2)+4/5*I*a^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2 ))^2/b^(5/2)-4/5*a^2*p*x*ln(c*(b*x^2+a)^p)/b^2+4/15*a*p*x^3*ln(c*(b*x^2+a) ^p)/b-4/25*p*x^5*ln(c*(b*x^2+a)^p)+4/5*a^(5/2)*p*arctan(x*b^(1/2)/a^(1/2)) *ln(c*(b*x^2+a)^p)/b^(5/2)+1/5*x^5*ln(c*(b*x^2+a)^p)^2+8/5*a^(5/2)*p^2*arc tan(x*b^(1/2)/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(5/2)+4/5*I*a ^(5/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.74 \[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {900 i a^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2+60 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-46 p+30 p \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )+15 \log \left (c \left (a+b x^2\right )^p\right )\right )+\sqrt {b} x \left (8 p^2 \left (345 a^2-40 a b x^2+9 b^2 x^4\right )-60 p \left (15 a^2-5 a b x^2+3 b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )+225 b^2 x^4 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )+900 i a^{5/2} p^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{1125 b^{5/2}} \]
((900*I)*a^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 + 60*a^(5/2)*p*ArcTan[( Sqrt[b]*x)/Sqrt[a]]*(-46*p + 30*p*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)] + 15*Log[c*(a + b*x^2)^p]) + Sqrt[b]*x*(8*p^2*(345*a^2 - 40*a*b*x^2 + 9*b ^2*x^4) - 60*p*(15*a^2 - 5*a*b*x^2 + 3*b^2*x^4)*Log[c*(a + b*x^2)^p] + 225 *b^2*x^4*Log[c*(a + b*x^2)^p]^2) + (900*I)*a^(5/2)*p^2*PolyLog[2, (I*Sqrt[ a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(1125*b^(5/2))
Time = 0.62 (sec) , antiderivative size = 320, normalized size of antiderivative = 0.95, 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.
| \(\displaystyle \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle \frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{5} b p \int \frac {x^6 \log \left (c \left (b x^2+a\right )^p\right )}{b x^2+a}dx\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle \frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{5} b p \int \left (\frac {\log \left (c \left (b x^2+a\right )^p\right ) x^4}{b}-\frac {a \log \left (c \left (b x^2+a\right )^p\right ) x^2}{b^2}-\frac {a^3 \log \left (c \left (b x^2+a\right )^p\right )}{b^3 \left (b x^2+a\right )}+\frac {a^2 \log \left (c \left (b x^2+a\right )^p\right )}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{5} b p \left (-\frac {a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^{7/2}}-\frac {i a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{b^{7/2}}+\frac {46 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 b^{7/2}}-\frac {2 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{b^{7/2}}-\frac {i a^{5/2} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 x \log \left (c \left (a+b x^2\right )^p\right )}{b^3}-\frac {46 a^2 p x}{15 b^3}-\frac {a x^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 b^2}+\frac {16 a p x^3}{45 b^2}+\frac {x^5 \log \left (c \left (a+b x^2\right )^p\right )}{5 b}-\frac {2 p x^5}{25 b}\right )\) |
(x^5*Log[c*(a + b*x^2)^p]^2)/5 - (4*b*p*((-46*a^2*p*x)/(15*b^3) + (16*a*p* x^3)/(45*b^2) - (2*p*x^5)/(25*b) + (46*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a ]])/(15*b^(7/2)) - (I*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/b^(7/2) - ( 2*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[ b]*x)])/b^(7/2) + (a^2*x*Log[c*(a + b*x^2)^p])/b^3 - (a*x^3*Log[c*(a + b*x ^2)^p])/(3*b^2) + (x^5*Log[c*(a + b*x^2)^p])/(5*b) - (a^(5/2)*ArcTan[(Sqrt [b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/b^(7/2) - (I*a^(5/2)*p*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/b^(7/2)))/5
3.1.84.3.1 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 = 0.86 (sec) , antiderivative size = 612, normalized size of antiderivative = 1.82
| method | result | size |
| risch | \(\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2} x^{5}}{5}-\frac {4 p \,x^{5} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{25}+\frac {4 p a \,x^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{15 b}-\frac {4 p \,a^{2} x \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 b^{2}}-\frac {4 p^{2} a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{5 b^{2} \sqrt {a b}}+\frac {4 p \,a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 b^{2} \sqrt {a b}}+\frac {8 p^{2} x^{5}}{125}-\frac {64 a \,p^{2} x^{3}}{225 b}+\frac {184 a^{2} p^{2} x}{75 b^{2}}-\frac {184 p^{2} a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{75 b^{2} \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 }\left (-\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^{3}}{2 b^{4} \underline {\hspace {1.25 ex}}\alpha }\right )\right )}{5}+\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^{5} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5}-\frac {2 p b \left (\frac {\frac {1}{5} x^{5} b^{2}-\frac {1}{3} a b \,x^{3}+a^{2} x}{b^{3}}-\frac {a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\right )}{5}\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^{5}}{20}\) | \(612\) |
1/5*ln((b*x^2+a)^p)^2*x^5-4/25*p*x^5*ln((b*x^2+a)^p)+4/15*p/b*a*x^3*ln((b* x^2+a)^p)-4/5*p/b^2*a^2*x*ln((b*x^2+a)^p)-4/5*p^2/b^2*a^3/(a*b)^(1/2)*arct an(b*x/(a*b)^(1/2))*ln(b*x^2+a)+4/5*p/b^2*a^3/(a*b)^(1/2)*arctan(b*x/(a*b) ^(1/2))*ln((b*x^2+a)^p)+8/125*p^2*x^5-64/225*a*p^2*x^3/b+184/75*a^2*p^2*x/ b^2-184/75*p^2/b^2*a^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))-4/5*p^2*b*Sum(- 1/2*(ln(x-_alpha)*ln(b*x^2+a)-2*b*(1/4/_alpha/b*ln(x-_alpha)^2+1/2*_alpha/ a*ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+1/2*_alpha/a*dilog(1/2*(x+_alpha) /_alpha)))*a^3/b^4/_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*csgn(I *c)+2*ln(c))*(1/5*x^5*ln((b*x^2+a)^p)-2/5*p*b*(1/b^3*(1/5*x^5*b^2-1/3*a*b* x^3+a^2*x)-a^3/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))+1/20*(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*csgn(I*c)+2*ln(c))^2*x^5
\[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
\[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}\, dx \]
\[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
1/5*p^2*x^5*log(b*x^2 + a)^2 + integrate(1/5*(5*b*x^6*log(c)^2 + 5*a*x^4*l og(c)^2 - 2*((2*p^2 - 5*p*log(c))*b*x^6 - 5*a*p*x^4*log(c))*log(b*x^2 + a) )/(b*x^2 + a), x)
\[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int { x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} \,d x } \]
Timed out. \[ \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\int x^4\,{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2 \,d x \]