Integrand size = 13, antiderivative size = 95 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=-\frac {a^3 \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}+\frac {a^2 \left (a+b x^3\right )^{2+p}}{b^4 (2+p)}-\frac {a \left (a+b x^3\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)} \] Output:
-1/3*a^3*(b*x^3+a)^(p+1)/b^4/(p+1)+a^2*(b*x^3+a)^(2+p)/b^4/(2+p)-a*(b*x^3+ a)^(3+p)/b^4/(3+p)+1/3*(b*x^3+a)^(4+p)/b^4/(4+p)
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=-\frac {a^3 \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}+\frac {a^2 \left (a+b x^3\right )^{2+p}}{b^4 (2+p)}-\frac {a \left (a+b x^3\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)} \] Input:
Integrate[x^11*(a + b*x^3)^p,x]
Output:
-1/3*(a^3*(a + b*x^3)^(1 + p))/(b^4*(1 + p)) + (a^2*(a + b*x^3)^(2 + p))/( b^4*(2 + p)) - (a*(a + b*x^3)^(3 + p))/(b^4*(3 + p)) + (a + b*x^3)^(4 + p) /(3*b^4*(4 + p))
Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {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.
\(\displaystyle \int x^{11} \left (a+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int x^9 \left (b x^3+a\right )^pdx^3\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {a^3 \left (b x^3+a\right )^p}{b^3}+\frac {3 a^2 \left (b x^3+a\right )^{p+1}}{b^3}-\frac {3 a \left (b x^3+a\right )^{p+2}}{b^3}+\frac {\left (b x^3+a\right )^{p+3}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {a^3 \left (a+b x^3\right )^{p+1}}{b^4 (p+1)}+\frac {3 a^2 \left (a+b x^3\right )^{p+2}}{b^4 (p+2)}-\frac {3 a \left (a+b x^3\right )^{p+3}}{b^4 (p+3)}+\frac {\left (a+b x^3\right )^{p+4}}{b^4 (p+4)}\right )\) |
Input:
Int[x^11*(a + b*x^3)^p,x]
Output:
(-((a^3*(a + b*x^3)^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*x^3)^(2 + p))/ (b^4*(2 + p)) - (3*a*(a + b*x^3)^(3 + p))/(b^4*(3 + p)) + (a + b*x^3)^(4 + p)/(b^4*(4 + p)))/3
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]]
Time = 0.87 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.39
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{p +1} \left (-b^{3} p^{3} x^{9}-6 b^{3} p^{2} x^{9}-11 b^{3} p \,x^{9}-6 b^{3} x^{9}+3 a \,b^{2} p^{2} x^{6}+9 a p \,x^{6} b^{2}+6 a \,b^{2} x^{6}-6 a^{2} p \,x^{3} b -6 a^{2} b \,x^{3}+6 a^{3}\right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(132\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{p} \left (-b^{3} p^{3} x^{9}-6 b^{3} p^{2} x^{9}-11 b^{3} p \,x^{9}-6 b^{3} x^{9}+3 a \,b^{2} p^{2} x^{6}+9 a p \,x^{6} b^{2}+6 a \,b^{2} x^{6}-6 a^{2} p \,x^{3} b -6 a^{2} b \,x^{3}+6 a^{3}\right ) \left (b \,x^{3}+a \right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(137\) |
risch | \(-\frac {\left (-b^{4} p^{3} x^{12}-6 b^{4} p^{2} x^{12}-11 b^{4} p \,x^{12}-a \,b^{3} p^{3} x^{9}-6 b^{4} x^{12}-3 a \,b^{3} p^{2} x^{9}-2 a p \,x^{9} b^{3}+3 a^{2} b^{2} p^{2} x^{6}+3 a^{2} p \,x^{6} b^{2}-6 a^{3} p \,x^{3} b +6 a^{4}\right ) \left (b \,x^{3}+a \right )^{p}}{3 \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (p +1\right ) b^{4}}\) | \(150\) |
parallelrisch | \(\frac {x^{12} \left (b \,x^{3}+a \right )^{p} b^{4} p^{3}+6 x^{12} \left (b \,x^{3}+a \right )^{p} b^{4} p^{2}+11 x^{12} \left (b \,x^{3}+a \right )^{p} b^{4} p +6 x^{12} \left (b \,x^{3}+a \right )^{p} b^{4}+x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3} p^{3}+3 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3} p^{2}+2 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3} p -3 x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{2} p^{2}-3 x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{2} p +6 x^{3} \left (b \,x^{3}+a \right )^{p} a^{3} b p -6 a^{4} \left (b \,x^{3}+a \right )^{p}}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(238\) |
Input:
int(x^11*(b*x^3+a)^p,x,method=_RETURNVERBOSE)
Output:
-1/3/b^4*(b*x^3+a)^(p+1)/(p^4+10*p^3+35*p^2+50*p+24)*(-b^3*p^3*x^9-6*b^3*p ^2*x^9-11*b^3*p*x^9-6*b^3*x^9+3*a*b^2*p^2*x^6+9*a*b^2*p*x^6+6*a*b^2*x^6-6* a^2*b*p*x^3-6*a^2*b*x^3+6*a^3)
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.56 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=\frac {{\left ({\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{12} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x^{9} + 6 \, a^{3} b p x^{3} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{6} - 6 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )}} \] Input:
integrate(x^11*(b*x^3+a)^p,x, algorithm="fricas")
Output:
1/3*((b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^12 + (a*b^3*p^3 + 3*a*b^3* p^2 + 2*a*b^3*p)*x^9 + 6*a^3*b*p*x^3 - 3*(a^2*b^2*p^2 + a^2*b^2*p)*x^6 - 6 *a^4)*(b*x^3 + a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4 )
Leaf count of result is larger than twice the leaf count of optimal. 2467 vs. \(2 (78) = 156\).
Time = 22.49 (sec) , antiderivative size = 2467, normalized size of antiderivative = 25.97 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=\text {Too large to display} \] Input:
integrate(x**11*(b*x**3+a)**p,x)
Output:
Piecewise((a**p*x**12/12, 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) - 3 6*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)/(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/(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 + 18...
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=\frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{12} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{9} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 6 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \] Input:
integrate(x^11*(b*x^3+a)^p,x, algorithm="maxima")
Output:
1/3*((p^3 + 6*p^2 + 11*p + 6)*b^4*x^12 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^9 - 3 *(p^2 + p)*a^2*b^2*x^6 + 6*a^3*b*p*x^3 - 6*a^4)*(b*x^3 + a)^p/((p^4 + 10*p ^3 + 35*p^2 + 50*p + 24)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (91) = 182\).
Time = 0.13 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.74 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=\frac {{\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} p^{2} - 3 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a p^{2} + 3 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} p^{2} + 5 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} p - 18 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a p + 21 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} p + 6 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} - 24 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a + 36 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2}}{3 \, {\left (b^{4} p^{3} + 9 \, b^{4} p^{2} + 26 \, b^{4} p + 24 \, b^{4}\right )}} - \frac {{\left (b x^{3} + a\right )}^{p + 1} a^{3}}{3 \, b^{4} {\left (p + 1\right )}} \] Input:
integrate(x^11*(b*x^3+a)^p,x, algorithm="giac")
Output:
1/3*((b*x^3 + a)^4*(b*x^3 + a)^p*p^2 - 3*(b*x^3 + a)^3*(b*x^3 + a)^p*a*p^2 + 3*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2*p^2 + 5*(b*x^3 + a)^4*(b*x^3 + a)^p*p - 18*(b*x^3 + a)^3*(b*x^3 + a)^p*a*p + 21*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2 *p + 6*(b*x^3 + a)^4*(b*x^3 + a)^p - 24*(b*x^3 + a)^3*(b*x^3 + a)^p*a + 36 *(b*x^3 + a)^2*(b*x^3 + a)^p*a^2)/(b^4*p^3 + 9*b^4*p^2 + 26*b^4*p + 24*b^4 ) - 1/3*(b*x^3 + a)^(p + 1)*a^3/(b^4*(p + 1))
Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.93 \[ \int x^{11} \left (a+b x^3\right )^p \, dx={\left (b\,x^3+a\right )}^p\,\left (\frac {x^{12}\,\left (p^3+6\,p^2+11\,p+6\right )}{3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {2\,a^4}{b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {2\,a^3\,p\,x^3}{b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^9\,\left (p^2+3\,p+2\right )}{3\,b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {a^2\,p\,x^6\,\left (p+1\right )}{b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right ) \] Input:
int(x^11*(a + b*x^3)^p,x)
Output:
(a + b*x^3)^p*((x^12*(11*p + 6*p^2 + p^3 + 6))/(3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (2*a^4)/(b^4*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (2*a^3* p*x^3)/(b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^9*(3*p + p^2 + 2 ))/(3*b*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (a^2*p*x^6*(p + 1))/(b^2*(5 0*p + 35*p^2 + 10*p^3 + p^4 + 24)))
Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int x^{11} \left (a+b x^3\right )^p \, dx=\frac {\left (b \,x^{3}+a \right )^{p} \left (b^{4} p^{3} x^{12}+6 b^{4} p^{2} x^{12}+11 b^{4} p \,x^{12}+a \,b^{3} p^{3} x^{9}+6 b^{4} x^{12}+3 a \,b^{3} p^{2} x^{9}+2 a \,b^{3} p \,x^{9}-3 a^{2} b^{2} p^{2} x^{6}-3 a^{2} b^{2} p \,x^{6}+6 a^{3} b p \,x^{3}-6 a^{4}\right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:
int(x^11*(b*x^3+a)^p,x)
Output:
((a + b*x**3)**p*( - 6*a**4 + 6*a**3*b*p*x**3 - 3*a**2*b**2*p**2*x**6 - 3* a**2*b**2*p*x**6 + a*b**3*p**3*x**9 + 3*a*b**3*p**2*x**9 + 2*a*b**3*p*x**9 + b**4*p**3*x**12 + 6*b**4*p**2*x**12 + 11*b**4*p*x**12 + 6*b**4*x**12))/ (3*b**4*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))