Integrand size = 24, antiderivative size = 172 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=-\frac {a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (1+2 p)}+\frac {a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (1+p)}-\frac {a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (3+2 p)}+\frac {\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (2+p)} \]
-1/3*a^3*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(1+2*p)+1/2*a^2*(b*x^3+a) ^2*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(p+1)-a*(b*x^3+a)^3*(b^2*x^6+2*a*b*x^3+a^ 2)^p/b^4/(3+2*p)+1/6*(b*x^3+a)^4*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(2+p)
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.64 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (-3 a^3+3 a^2 b (1+2 p) x^3-3 a b^2 \left (1+3 p+2 p^2\right ) x^6+b^3 \left (3+11 p+12 p^2+4 p^3\right ) x^9\right )}{6 b^4 (1+p) (2+p) (1+2 p) (3+2 p)} \]
((a + b*x^3)*((a + b*x^3)^2)^p*(-3*a^3 + 3*a^2*b*(1 + 2*p)*x^3 - 3*a*b^2*( 1 + 3*p + 2*p^2)*x^6 + b^3*(3 + 11*p + 12*p^2 + 4*p^3)*x^9))/(6*b^4*(1 + p )*(2 + p)*(1 + 2*p)*(3 + 2*p))
Time = 0.31 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1385, 798, 53, 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.
| \(\displaystyle \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx\) |
| \(\Big \downarrow \) 1385 |
| \(\displaystyle \left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \int x^{11} \left (\frac {b x^3}{a}+1\right )^{2 p}dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{3} \left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \int x^9 \left (\frac {b x^3}{a}+1\right )^{2 p}dx^3\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {1}{3} \left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \int \left (-\frac {a^3 \left (\frac {b x^3}{a}+1\right )^{2 p}}{b^3}+\frac {3 a^3 \left (\frac {b x^3}{a}+1\right )^{2 p+1}}{b^3}-\frac {3 a^3 \left (\frac {b x^3}{a}+1\right )^{2 p+2}}{b^3}+\frac {a^3 \left (\frac {b x^3}{a}+1\right )^{2 p+3}}{b^3}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \left (\frac {3 a^4 \left (\frac {b x^3}{a}+1\right )^{2 (p+1)}}{2 b^4 (p+1)}+\frac {a^4 \left (\frac {b x^3}{a}+1\right )^{2 (p+2)}}{2 b^4 (p+2)}-\frac {a^4 \left (\frac {b x^3}{a}+1\right )^{2 p+1}}{b^4 (2 p+1)}-\frac {3 a^4 \left (\frac {b x^3}{a}+1\right )^{2 p+3}}{b^4 (2 p+3)}\right )\) |
((a^2 + 2*a*b*x^3 + b^2*x^6)^p*((3*a^4*(1 + (b*x^3)/a)^(2*(1 + p)))/(2*b^4 *(1 + p)) + (a^4*(1 + (b*x^3)/a)^(2*(2 + p)))/(2*b^4*(2 + p)) - (a^4*(1 + (b*x^3)/a)^(1 + 2*p))/(b^4*(1 + 2*p)) - (3*a^4*(1 + (b*x^3)/a)^(3 + 2*p))/ (b^4*(3 + 2*p))))/(3*(1 + (b*x^3)/a)^(2*p))
3.2.25.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* FracPart[p])) Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[2*p] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)]
Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.87
| method | result | size |
| gosper | \(-\frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} \left (-4 b^{3} p^{3} x^{9}-12 b^{3} p^{2} x^{9}-11 b^{3} p \,x^{9}-3 b^{3} x^{9}+6 a \,b^{2} p^{2} x^{6}+9 a \,b^{2} p \,x^{6}+3 b^{2} x^{6} a -6 a^{2} b p \,x^{3}-3 a^{2} b \,x^{3}+3 a^{3}\right ) \left (b \,x^{3}+a \right )}{6 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\) | \(150\) |
| risch | \(-\frac {\left (-4 b^{4} p^{3} x^{12}-12 b^{4} p^{2} x^{12}-11 b^{4} p \,x^{12}-4 a \,b^{3} p^{3} x^{9}-3 b^{4} x^{12}-6 a \,b^{3} p^{2} x^{9}-2 a p \,x^{9} b^{3}+6 a^{2} b^{2} p^{2} x^{6}+3 a^{2} p \,x^{6} b^{2}-6 a^{3} p \,x^{3} b +3 a^{4}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{6 \left (3+2 p \right ) \left (2+p \right ) \left (1+p \right ) \left (1+2 p \right ) b^{4}}\) | \(156\) |
| parallelrisch | \(\frac {4 x^{12} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{4} p^{3}+12 x^{12} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{4} p^{2}+11 x^{12} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{4} p +3 x^{12} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{4}+4 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p^{3}+6 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p^{2}+2 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p -6 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p^{2}-3 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p +6 x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{3} b p -3 \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{4}}{6 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\) | \(363\) |
-1/6*(b^2*x^6+2*a*b*x^3+a^2)^p*(-4*b^3*p^3*x^9-12*b^3*p^2*x^9-11*b^3*p*x^9 -3*b^3*x^9+6*a*b^2*p^2*x^6+9*a*b^2*p*x^6+3*a*b^2*x^6-6*a^2*b*p*x^3-3*a^2*b *x^3+3*a^3)*(b*x^3+a)/b^4/(4*p^4+20*p^3+35*p^2+25*p+6)
Time = 0.27 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{12} + 2 \, {\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{9} + 6 \, a^{3} b p x^{3} - 3 \, {\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{6} - 3 \, a^{4}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
1/6*((4*b^4*p^3 + 12*b^4*p^2 + 11*b^4*p + 3*b^4)*x^12 + 2*(2*a*b^3*p^3 + 3 *a*b^3*p^2 + a*b^3*p)*x^9 + 6*a^3*b*p*x^3 - 3*(2*a^2*b^2*p^2 + a^2*b^2*p)* x^6 - 3*a^4)*(b^2*x^6 + 2*a*b*x^3 + a^2)^p/(4*b^4*p^4 + 20*b^4*p^3 + 35*b^ 4*p^2 + 25*b^4*p + 6*b^4)
\[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\text {Too large to display} \]
Piecewise((x**12*(a**2)**p/12, Eq(b, 0)), (6*a**3*log(x - (-a/b)**(1/3))/( 18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*a**3 *log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54*a**2 *b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 12*a**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 11*a**3/(18*a**3*b **4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3* log(x - (-a/b)**(1/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b) **(2/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9 ) - 36*a**2*b*x**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x* *6 + 18*b**7*x**9) + 27*a**2*b*x**3/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54 *a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6*log(x - (-a/b)**(1/3))/(18*a **3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2* x**6*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54* a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 36*a*b**2*x**6*log(2)/(1 8*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b* *2*x**6/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*b**3*x**9*log(x - (-a/b)**(1/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*b**3*x**9*log(4*x**2 + 4*x*(-a/b)**(1/3 ) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6...
Time = 0.22 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.67 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{12} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{9} - 3 \, {\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 3 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{6 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \]
1/6*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^12 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^ 9 - 3*(2*p^2 + p)*a^2*b^2*x^6 + 6*a^3*b*p*x^3 - 3*a^4)*(b*x^3 + a)^(2*p)/( (4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (166) = 332\).
Time = 0.31 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.18 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {4 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{3} x^{12} + 12 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{2} x^{12} + 11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p x^{12} + 4 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{3} x^{9} + 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} x^{12} + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{2} x^{9} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p x^{9} - 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{6} - 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p x^{6} + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{3} b p x^{3} - 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{4}}{6 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
1/6*(4*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^4*p^3*x^12 + 12*(b^2*x^6 + 2*a*b*x^ 3 + a^2)^p*b^4*p^2*x^12 + 11*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^4*p*x^12 + 4* (b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p^3*x^9 + 3*(b^2*x^6 + 2*a*b*x^3 + a^2 )^p*b^4*x^12 + 6*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p^2*x^9 + 2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p*x^9 - 6*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^2*b^2 *p^2*x^6 - 3*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^2*b^2*p*x^6 + 6*(b^2*x^6 + 2* a*b*x^3 + a^2)^p*a^3*b*p*x^3 - 3*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^4)/(4*b^4 *p^4 + 20*b^4*p^3 + 35*b^4*p^2 + 25*b^4*p + 6*b^4)
Time = 8.39 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.20 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx={\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^{12}\,\left (4\,p^3+12\,p^2+11\,p+3\right )}{6\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {a^4}{2\,b^4\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a^3\,p\,x^3}{b^3\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a\,p\,x^9\,\left (2\,p^2+3\,p+1\right )}{3\,b\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {a^2\,p\,x^6\,\left (2\,p+1\right )}{2\,b^2\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}\right ) \]
(a^2 + b^2*x^6 + 2*a*b*x^3)^p*((x^12*(11*p + 12*p^2 + 4*p^3 + 3))/(6*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) - a^4/(2*b^4*(25*p + 35*p^2 + 20*p^3 + 4* p^4 + 6)) + (a^3*p*x^3)/(b^3*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) + (a*p* x^9*(3*p + 2*p^2 + 1))/(3*b*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) - (a^2*p *x^6*(2*p + 1))/(2*b^2*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)))