Integrand size = 13, antiderivative size = 72 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {a^2 \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac {a \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {\left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)} \] Output:
1/2*a^2*(b*x^2+a)^(p+1)/b^3/(p+1)-a*(b*x^2+a)^(2+p)/b^3/(2+p)+1/2*(b*x^2+a )^(3+p)/b^3/(3+p)
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^{1+p} \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{2 b^3 (1+p) (2+p) (3+p)} \] Input:
Integrate[x^5*(a + b*x^2)^p,x]
Output:
((a + b*x^2)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4) )/(2*b^3*(1 + p)*(2 + p)*(3 + p))
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {243, 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.
\(\displaystyle \int x^5 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^4 \left (b x^2+a\right )^pdx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^2 \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}-\frac {2 a \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {\left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )\) |
Input:
Int[x^5*(a + b*x^2)^p,x]
Output:
((a^2*(a + b*x^2)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x^2)^(2 + p))/(b^3* (2 + p)) + (a + b*x^2)^(3 + p)/(b^3*(3 + p)))/2
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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.11
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right )^{p +1} \left (b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+2 b^{2} x^{4}-2 a b p \,x^{2}-2 a b \,x^{2}+2 a^{2}\right )}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(80\) |
orering | \(\frac {\left (b \,x^{2}+a \right ) \left (b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+2 b^{2} x^{4}-2 a b p \,x^{2}-2 a b \,x^{2}+2 a^{2}\right ) \left (b \,x^{2}+a \right )^{p}}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(85\) |
risch | \(\frac {\left (b^{3} p^{2} x^{6}+3 b^{3} p \,x^{6}+a \,b^{2} p^{2} x^{4}+2 b^{3} x^{6}+a \,b^{2} p \,x^{4}-2 a^{2} b p \,x^{2}+2 a^{3}\right ) \left (b \,x^{2}+a \right )^{p}}{2 \left (2+p \right ) \left (3+p \right ) \left (p +1\right ) b^{3}}\) | \(93\) |
norman | \(\frac {a^{3} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}+\frac {x^{6} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{6+2 p}+\frac {a p \,x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+5 p +6\right )}-\frac {p \,a^{2} x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{2} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(125\) |
parallelrisch | \(\frac {x^{6} \left (b \,x^{2}+a \right )^{p} b^{3} p^{2}+3 x^{6} \left (b \,x^{2}+a \right )^{p} b^{3} p +2 x^{6} \left (b \,x^{2}+a \right )^{p} b^{3}+x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} p^{2}+x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} p -2 x^{2} \left (b \,x^{2}+a \right )^{p} a^{2} b p +2 \left (b \,x^{2}+a \right )^{p} a^{3}}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(147\) |
Input:
int(x^5*(b*x^2+a)^p,x,method=_RETURNVERBOSE)
Output:
1/2/b^3*(b*x^2+a)^(p+1)/(p^3+6*p^2+11*p+6)*(b^2*p^2*x^4+3*b^2*p*x^4+2*b^2* x^4-2*a*b*p*x^2-2*a*b*x^2+2*a^2)
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {{\left ({\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} + {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{4} + 2 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \] Input:
integrate(x^5*(b*x^2+a)^p,x, algorithm="fricas")
Output:
1/2*((b^3*p^2 + 3*b^3*p + 2*b^3)*x^6 - 2*a^2*b*p*x^2 + (a*b^2*p^2 + a*b^2* p)*x^4 + 2*a^3)*(b*x^2 + a)^p/(b^3*p^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 920 vs. \(2 (58) = 116\).
Time = 1.30 (sec) , antiderivative size = 920, normalized size of antiderivative = 12.78 \[ \int x^5 \left (a+b x^2\right )^p \, dx =\text {Too large to display} \] Input:
integrate(x**5*(b*x**2+a)**p,x)
Output:
Piecewise((a**p*x**6/6, Eq(b, 0)), (2*a**2*log(x - sqrt(-a/b))/(4*a**2*b** 3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4* b**5*x**4) + 4*a*b*x**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x** 2 + 4*b**5*x**4) + 4*a*b*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x* *4) + 2*b**2*x**4*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b** 5*x**4), Eq(p, -3)), (-2*a**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a* b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(x - sqrt(-a/b))/(2*b**3) + a**2 *log(x + sqrt(-a/b))/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 2*a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12* b**3*p**2 + 22*b**3*p + 12*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3* p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 3*b**3*p*x**6*(a + b*x**2)...
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} + {\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \] Input:
integrate(x^5*(b*x^2+a)^p,x, algorithm="maxima")
Output:
1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3 )*(b*x^2 + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3)
Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.83 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {{\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} p - 2 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a p + 2 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} - 6 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a}{2 \, {\left (b^{3} p^{2} + 5 \, b^{3} p + 6 \, b^{3}\right )}} + \frac {{\left (b x^{2} + a\right )}^{p + 1} a^{2}}{2 \, b^{3} {\left (p + 1\right )}} \] Input:
integrate(x^5*(b*x^2+a)^p,x, algorithm="giac")
Output:
1/2*((b*x^2 + a)^3*(b*x^2 + a)^p*p - 2*(b*x^2 + a)^2*(b*x^2 + a)^p*a*p + 2 *(b*x^2 + a)^3*(b*x^2 + a)^p - 6*(b*x^2 + a)^2*(b*x^2 + a)^p*a)/(b^3*p^2 + 5*b^3*p + 6*b^3) + 1/2*(b*x^2 + a)^(p + 1)*a^2/(b^3*(p + 1))
Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.62 \[ \int x^5 \left (a+b x^2\right )^p \, dx={\left (b\,x^2+a\right )}^p\,\left (\frac {a^3}{b^3\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {x^6\,\left (p^2+3\,p+2\right )}{2\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {a^2\,p\,x^2}{b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^4\,\left (p+1\right )}{2\,b\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:
int(x^5*(a + b*x^2)^p,x)
Output:
(a + b*x^2)^p*(a^3/(b^3*(11*p + 6*p^2 + p^3 + 6)) + (x^6*(3*p + p^2 + 2))/ (2*(11*p + 6*p^2 + p^3 + 6)) - (a^2*p*x^2)/(b^2*(11*p + 6*p^2 + p^3 + 6)) + (a*p*x^4*(p + 1))/(2*b*(11*p + 6*p^2 + p^3 + 6)))
Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a+b x^2\right )^p \, dx=\frac {\left (b \,x^{2}+a \right )^{p} \left (b^{3} p^{2} x^{6}+3 b^{3} p \,x^{6}+a \,b^{2} p^{2} x^{4}+2 b^{3} x^{6}+a \,b^{2} p \,x^{4}-2 a^{2} b p \,x^{2}+2 a^{3}\right )}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^5*(b*x^2+a)^p,x)
Output:
((a + b*x**2)**p*(2*a**3 - 2*a**2*b*p*x**2 + a*b**2*p**2*x**4 + a*b**2*p*x **4 + b**3*p**2*x**6 + 3*b**3*p*x**6 + 2*b**3*x**6))/(2*b**3*(p**3 + 6*p** 2 + 11*p + 6))