Integrand size = 18, antiderivative size = 215 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {a^2 p^2 x^2}{b^2}-\frac {a p^2 \left (a+b x^2\right )^2}{4 b^3}+\frac {p^2 \left (a+b x^2\right )^3}{27 b^3}-\frac {a^3 p^2 \log ^2\left (a+b x^2\right )}{6 b^3}-\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right ) \]
a^2*p^2*x^2/b^2-1/4*a*p^2*(b*x^2+a)^2/b^3+1/27*p^2*(b*x^2+a)^3/b^3-1/6*a^3 *p^2*ln(b*x^2+a)^2/b^3-a^2*p*(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^3+1/2*a*p*(b*x^ 2+a)^2*ln(c*(b*x^2+a)^p)/b^3-1/9*p*(b*x^2+a)^3*ln(c*(b*x^2+a)^p)/b^3+1/3*a ^3*p*ln(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^3+1/6*x^6*ln(c*(b*x^2+a)^p)^2
Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {b p^2 x^2 \left (66 a^2-15 a b x^2+4 b^2 x^4\right )-30 a^3 p^2 \log \left (a+b x^2\right )-6 p \left (6 a^3+6 a^2 b x^2-3 a b^2 x^4+2 b^3 x^6\right ) \log \left (c \left (a+b x^2\right )^p\right )+18 \left (a^3+b^3 x^6\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{108 b^3} \]
(b*p^2*x^2*(66*a^2 - 15*a*b*x^2 + 4*b^2*x^4) - 30*a^3*p^2*Log[a + b*x^2] - 6*p*(6*a^3 + 6*a^2*b*x^2 - 3*a*b^2*x^4 + 2*b^3*x^6)*Log[c*(a + b*x^2)^p] + 18*(a^3 + b^3*x^6)*Log[c*(a + b*x^2)^p]^2)/(108*b^3)
Time = 0.41 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2904, 2845, 2858, 25, 27, 2772, 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.
| \(\displaystyle \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle \frac {1}{2} \int x^4 \log ^2\left (c \left (b x^2+a\right )^p\right )dx^2\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {2}{3} b p \int \frac {x^6 \log \left (c \left (b x^2+a\right )^p\right )}{b x^2+a}dx^2\right )\) |
| \(\Big \downarrow \) 2858 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {2}{3} p \int x^4 \log \left (c \left (b x^2+a\right )^p\right )d\left (b x^2+a\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{3} p \int -x^4 \log \left (c \left (b x^2+a\right )^p\right )d\left (b x^2+a\right )+\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 p \int -b^3 x^4 \log \left (c \left (b x^2+a\right )^p\right )d\left (b x^2+a\right )}{3 b^3}+\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )\) |
| \(\Big \downarrow \) 2772 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 p \left (-p \int \left (-\frac {x^4}{3}-3 a^2+\frac {3}{2} a \left (b x^2+a\right )+\frac {a^3 \log \left (b x^2+a\right )}{x^2}\right )d\left (b x^2+a\right )+a^3 \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-3 a^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} x^6 \log \left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} a x^4 \log \left (c \left (a+b x^2\right )^p\right )\right )}{3 b^3}+\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 p \left (a^3 \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-3 a^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-p \left (\frac {1}{2} a^3 \log ^2\left (a+b x^2\right )-3 a^2 \left (a+b x^2\right )+\frac {3 a x^4}{4}-\frac {x^6}{9}\right )-\frac {1}{3} x^6 \log \left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} a x^4 \log \left (c \left (a+b x^2\right )^p\right )\right )}{3 b^3}+\frac {1}{3} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )\) |
((x^6*Log[c*(a + b*x^2)^p]^2)/3 + (2*p*(-(p*((3*a*x^4)/4 - x^6/9 - 3*a^2*( a + b*x^2) + (a^3*Log[a + b*x^2]^2)/2)) + (3*a*x^4*Log[c*(a + b*x^2)^p])/2 - (x^6*Log[c*(a + b*x^2)^p])/3 - 3*a^2*(a + b*x^2)*Log[c*(a + b*x^2)^p] + a^3*Log[a + b*x^2]*Log[c*(a + b*x^2)^p]))/(3*b^3))/2
3.1.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ .))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] && !(EqQ[q , 1] && EqQ[m, -1])
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^ n])^p/(g*(q + 1))), x] - Simp[b*e*n*(p/(g*(q + 1))) Int[(f + g*x)^(q + 1) *((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && In tegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ .)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e Subst[In t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
Time = 0.91 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(-\frac {-18 x^{6} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} b^{3}+12 x^{6} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{3} p -4 x^{6} b^{3} p^{2}-18 x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a \,b^{2} p +15 x^{4} a \,b^{2} p^{2}+36 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} b p -66 x^{2} a^{2} b \,p^{2}+102 \ln \left (b \,x^{2}+a \right ) a^{3} p^{2}-18 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{3}-36 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{3} p +66 a^{3} p^{2}}{108 b^{3}}\) | \(190\) |
| risch | \(\text {Expression too large to display}\) | \(1436\) |
-1/108*(-18*x^6*ln(c*(b*x^2+a)^p)^2*b^3+12*x^6*ln(c*(b*x^2+a)^p)*b^3*p-4*x ^6*b^3*p^2-18*x^4*ln(c*(b*x^2+a)^p)*a*b^2*p+15*x^4*a*b^2*p^2+36*x^2*ln(c*( b*x^2+a)^p)*a^2*b*p-66*x^2*a^2*b*p^2+102*ln(b*x^2+a)*a^3*p^2-18*ln(c*(b*x^ 2+a)^p)^2*a^3-36*ln(c*(b*x^2+a)^p)*a^3*p+66*a^3*p^2)/b^3
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {4 \, b^{3} p^{2} x^{6} + 18 \, b^{3} x^{6} \log \left (c\right )^{2} - 15 \, a b^{2} p^{2} x^{4} + 66 \, a^{2} b p^{2} x^{2} + 18 \, {\left (b^{3} p^{2} x^{6} + a^{3} p^{2}\right )} \log \left (b x^{2} + a\right )^{2} - 6 \, {\left (2 \, b^{3} p^{2} x^{6} - 3 \, a b^{2} p^{2} x^{4} + 6 \, a^{2} b p^{2} x^{2} + 11 \, a^{3} p^{2} - 6 \, {\left (b^{3} p x^{6} + a^{3} p\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (2 \, b^{3} p x^{6} - 3 \, a b^{2} p x^{4} + 6 \, a^{2} b p x^{2}\right )} \log \left (c\right )}{108 \, b^{3}} \]
1/108*(4*b^3*p^2*x^6 + 18*b^3*x^6*log(c)^2 - 15*a*b^2*p^2*x^4 + 66*a^2*b*p ^2*x^2 + 18*(b^3*p^2*x^6 + a^3*p^2)*log(b*x^2 + a)^2 - 6*(2*b^3*p^2*x^6 - 3*a*b^2*p^2*x^4 + 6*a^2*b*p^2*x^2 + 11*a^3*p^2 - 6*(b^3*p*x^6 + a^3*p)*log (c))*log(b*x^2 + a) - 6*(2*b^3*p*x^6 - 3*a*b^2*p*x^4 + 6*a^2*b*p*x^2)*log( c))/b^3
Time = 3.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.85 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} - \frac {11 a^{3} p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{18 b^{3}} + \frac {a^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{6 b^{3}} + \frac {11 a^{2} p^{2} x^{2}}{18 b^{2}} - \frac {a^{2} p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3 b^{2}} - \frac {5 a p^{2} x^{4}}{36 b} + \frac {a p x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 b} + \frac {p^{2} x^{6}}{27} - \frac {p x^{6} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{9} + \frac {x^{6} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{6} & \text {for}\: b \neq 0 \\\frac {x^{6} \log {\left (a^{p} c \right )}^{2}}{6} & \text {otherwise} \end {cases} \]
Piecewise((-11*a**3*p*log(c*(a + b*x**2)**p)/(18*b**3) + a**3*log(c*(a + b *x**2)**p)**2/(6*b**3) + 11*a**2*p**2*x**2/(18*b**2) - a**2*p*x**2*log(c*( a + b*x**2)**p)/(3*b**2) - 5*a*p**2*x**4/(36*b) + a*p*x**4*log(c*(a + b*x* *2)**p)/(6*b) + p**2*x**6/27 - p*x**6*log(c*(a + b*x**2)**p)/9 + x**6*log( c*(a + b*x**2)**p)**2/6, Ne(b, 0)), (x**6*log(a**p*c)**2/6, True))
Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.67 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{6} \, x^{6} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} + \frac {1}{18} \, b p {\left (\frac {6 \, a^{3} \log \left (b x^{2} + a\right )}{b^{4}} - \frac {2 \, b^{2} x^{6} - 3 \, a b x^{4} + 6 \, a^{2} x^{2}}{b^{3}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {{\left (4 \, b^{3} x^{6} - 15 \, a b^{2} x^{4} + 66 \, a^{2} b x^{2} - 18 \, a^{3} \log \left (b x^{2} + a\right )^{2} - 66 \, a^{3} \log \left (b x^{2} + a\right )\right )} p^{2}}{108 \, b^{3}} \]
1/6*x^6*log((b*x^2 + a)^p*c)^2 + 1/18*b*p*(6*a^3*log(b*x^2 + a)/b^4 - (2*b ^2*x^6 - 3*a*b*x^4 + 6*a^2*x^2)/b^3)*log((b*x^2 + a)^p*c) + 1/108*(4*b^3*x ^6 - 15*a*b^2*x^4 + 66*a^2*b*x^2 - 18*a^3*log(b*x^2 + a)^2 - 66*a^3*log(b* x^2 + a))*p^2/b^3
Time = 0.30 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.72 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{3} p^{2} \log \left (b x^{2} + a\right )^{2}}{6 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p^{2} \log \left (b x^{2} + a\right )^{2}}{2 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{3} p^{2} \log \left (b x^{2} + a\right )}{9 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{2} a p^{2} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} p \log \left (b x^{2} + a\right ) \log \left (c\right )}{3 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p \log \left (b x^{2} + a\right ) \log \left (c\right )}{b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} p^{2}}{27 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p^{2}}{4 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{3} p \log \left (c\right )}{9 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{2} a p \log \left (c\right )}{2 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} \log \left (c\right )^{2}}{6 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a \log \left (c\right )^{2}}{2 \, b^{3}} + \frac {{\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} a^{2} p^{2} - 2 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a^{2} p \log \left (c\right ) + {\left (b x^{2} + a\right )} a^{2} \log \left (c\right )^{2}}{2 \, b^{3}} \]
1/6*(b*x^2 + a)^3*p^2*log(b*x^2 + a)^2/b^3 - 1/2*(b*x^2 + a)^2*a*p^2*log(b *x^2 + a)^2/b^3 - 1/9*(b*x^2 + a)^3*p^2*log(b*x^2 + a)/b^3 + 1/2*(b*x^2 + a)^2*a*p^2*log(b*x^2 + a)/b^3 + 1/3*(b*x^2 + a)^3*p*log(b*x^2 + a)*log(c)/ b^3 - (b*x^2 + a)^2*a*p*log(b*x^2 + a)*log(c)/b^3 + 1/27*(b*x^2 + a)^3*p^2 /b^3 - 1/4*(b*x^2 + a)^2*a*p^2/b^3 - 1/9*(b*x^2 + a)^3*p*log(c)/b^3 + 1/2* (b*x^2 + a)^2*a*p*log(c)/b^3 + 1/6*(b*x^2 + a)^3*log(c)^2/b^3 - 1/2*(b*x^2 + a)^2*a*log(c)^2/b^3 + 1/2*((2*b*x^2 + (b*x^2 + a)*log(b*x^2 + a)^2 - 2* (b*x^2 + a)*log(b*x^2 + a) + 2*a)*a^2*p^2 - 2*(b*x^2 - (b*x^2 + a)*log(b*x ^2 + a) + a)*a^2*p*log(c) + (b*x^2 + a)*a^2*log(c)^2)/b^3
Time = 1.41 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.59 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {p^2\,x^6}{27}+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {x^6}{6}+\frac {a^3}{6\,b^3}\right )-\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {p\,x^6}{9}+\frac {a^2\,p\,x^2}{3\,b^2}-\frac {a\,p\,x^4}{6\,b}\right )-\frac {5\,a\,p^2\,x^4}{36\,b}-\frac {11\,a^3\,p^2\,\ln \left (b\,x^2+a\right )}{18\,b^3}+\frac {11\,a^2\,p^2\,x^2}{18\,b^2} \]