Integrand size = 15, antiderivative size = 59 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 a^2 \left (a+b x^2\right )^{5/3}}{10 b^3}-\frac {3 a \left (a+b x^2\right )^{8/3}}{8 b^3}+\frac {3 \left (a+b x^2\right )^{11/3}}{22 b^3} \] Output:
3/10*a^2*(b*x^2+a)^(5/3)/b^3-3/8*a*(b*x^2+a)^(8/3)/b^3+3/22*(b*x^2+a)^(11/ 3)/b^3
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 \left (a+b x^2\right )^{5/3} \left (9 a^2-15 a b x^2+20 b^2 x^4\right )}{440 b^3} \] Input:
Integrate[x^5*(a + b*x^2)^(2/3),x]
Output:
(3*(a + b*x^2)^(5/3)*(9*a^2 - 15*a*b*x^2 + 20*b^2*x^4))/(440*b^3)
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int x^5 \left (a+b x^2\right )^{2/3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^4 \left (b x^2+a\right )^{2/3}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right )^{8/3}}{b^2}-\frac {2 a \left (b x^2+a\right )^{5/3}}{b^2}+\frac {a^2 \left (b x^2+a\right )^{2/3}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {3 a^2 \left (a+b x^2\right )^{5/3}}{5 b^3}+\frac {3 \left (a+b x^2\right )^{11/3}}{11 b^3}-\frac {3 a \left (a+b x^2\right )^{8/3}}{4 b^3}\right )\) |
Input:
Int[x^5*(a + b*x^2)^(2/3),x]
Output:
((3*a^2*(a + b*x^2)^(5/3))/(5*b^3) - (3*a*(a + b*x^2)^(8/3))/(4*b^3) + (3* (a + b*x^2)^(11/3))/(11*b^3))/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 = 0.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}} \left (20 b^{2} x^{4}-15 a b \,x^{2}+9 a^{2}\right )}{440 b^{3}}\) | \(36\) |
pseudoelliptic | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}} \left (20 b^{2} x^{4}-15 a b \,x^{2}+9 a^{2}\right )}{440 b^{3}}\) | \(36\) |
orering | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}} \left (20 b^{2} x^{4}-15 a b \,x^{2}+9 a^{2}\right )}{440 b^{3}}\) | \(36\) |
trager | \(\frac {3 \left (20 b^{3} x^{6}+5 a \,b^{2} x^{4}-6 a^{2} b \,x^{2}+9 a^{3}\right ) \left (b \,x^{2}+a \right )^{\frac {2}{3}}}{440 b^{3}}\) | \(47\) |
risch | \(\frac {3 \left (20 b^{3} x^{6}+5 a \,b^{2} x^{4}-6 a^{2} b \,x^{2}+9 a^{3}\right ) \left (b \,x^{2}+a \right )^{\frac {2}{3}}}{440 b^{3}}\) | \(47\) |
Input:
int(x^5*(b*x^2+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
3/440*(b*x^2+a)^(5/3)*(20*b^2*x^4-15*a*b*x^2+9*a^2)/b^3
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 \, {\left (20 \, b^{3} x^{6} + 5 \, a b^{2} x^{4} - 6 \, a^{2} b x^{2} + 9 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{440 \, b^{3}} \] Input:
integrate(x^5*(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
3/440*(20*b^3*x^6 + 5*a*b^2*x^4 - 6*a^2*b*x^2 + 9*a^3)*(b*x^2 + a)^(2/3)/b ^3
Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (54) = 108\).
Time = 1.27 (sec) , antiderivative size = 700, normalized size of antiderivative = 11.86 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {27 a^{\frac {35}{3}} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} - \frac {27 a^{\frac {35}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {63 a^{\frac {32}{3}} b x^{2} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} - \frac {81 a^{\frac {32}{3}} b x^{2}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {42 a^{\frac {29}{3}} b^{2} x^{4} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} - \frac {81 a^{\frac {29}{3}} b^{2} x^{4}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {78 a^{\frac {26}{3}} b^{3} x^{6} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} - \frac {27 a^{\frac {26}{3}} b^{3} x^{6}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {207 a^{\frac {23}{3}} b^{4} x^{8} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {195 a^{\frac {20}{3}} b^{5} x^{10} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} + \frac {60 a^{\frac {17}{3}} b^{6} x^{12} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {2}{3}}}{440 a^{8} b^{3} + 1320 a^{7} b^{4} x^{2} + 1320 a^{6} b^{5} x^{4} + 440 a^{5} b^{6} x^{6}} \] Input:
integrate(x**5*(b*x**2+a)**(2/3),x)
Output:
27*a**(35/3)*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) - 27*a**(35/3)/(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) + 63*a**(3 2/3)*b*x**2*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1 320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) - 81*a**(32/3)*b*x**2/(440*a**8*b **3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) + 42 *a**(29/3)*b**2*x**4*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 1320*a**7*b**4 *x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) - 81*a**(29/3)*b**2*x**4 /(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b** 6*x**6) + 78*a**(26/3)*b**3*x**6*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 13 20*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) - 27*a**(26/ 3)*b**3*x**6/(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) + 207*a**(23/3)*b**4*x**8*(1 + b*x**2/a)**(2/3)/(440*a **8*b**3 + 1320*a**7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) + 195*a**(20/3)*b**5*x**10*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 1320*a* *7*b**4*x**2 + 1320*a**6*b**5*x**4 + 440*a**5*b**6*x**6) + 60*a**(17/3)*b* *6*x**12*(1 + b*x**2/a)**(2/3)/(440*a**8*b**3 + 1320*a**7*b**4*x**2 + 1320 *a**6*b**5*x**4 + 440*a**5*b**6*x**6)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 \, {\left (b x^{2} + a\right )}^{\frac {11}{3}}}{22 \, b^{3}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}} a}{8 \, b^{3}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a^{2}}{10 \, b^{3}} \] Input:
integrate(x^5*(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
3/22*(b*x^2 + a)^(11/3)/b^3 - 3/8*(b*x^2 + a)^(8/3)*a/b^3 + 3/10*(b*x^2 + a)^(5/3)*a^2/b^3
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 \, {\left (20 \, {\left (b x^{2} + a\right )}^{\frac {11}{3}} - 55 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}} a + 44 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a^{2}\right )}}{440 \, b^{3}} \] Input:
integrate(x^5*(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
3/440*(20*(b*x^2 + a)^(11/3) - 55*(b*x^2 + a)^(8/3)*a + 44*(b*x^2 + a)^(5/ 3)*a^2)/b^3
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx={\left (b\,x^2+a\right )}^{2/3}\,\left (\frac {3\,x^6}{22}+\frac {27\,a^3}{440\,b^3}+\frac {3\,a\,x^4}{88\,b}-\frac {9\,a^2\,x^2}{220\,b^2}\right ) \] Input:
int(x^5*(a + b*x^2)^(2/3),x)
Output:
(a + b*x^2)^(2/3)*((3*x^6)/22 + (27*a^3)/(440*b^3) + (3*a*x^4)/(88*b) - (9 *a^2*x^2)/(220*b^2))
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.37 \[ \int x^5 \left (a+b x^2\right )^{2/3} \, dx=\frac {3 \sqrt {b \,x^{2}+a}\, \left (\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}\right )^{\frac {1}{3}} \left (20 b^{3} x^{6}+5 a \,b^{2} x^{4}-6 a^{2} b \,x^{2}+9 a^{3}\right )}{440 \left (\sqrt {b \,x^{2}+a}+\sqrt {b}\, x \right )^{\frac {1}{3}} b^{3}} \] Input:
int(x^5*(b*x^2+a)^(2/3),x)
Output:
(3*a**(7/6)*sqrt(a + b*x**2)*(sqrt(b)*sqrt(a + b*x**2)*x + a + b*x**2)**(1 /3)*(9*a**3 - 6*a**2*b*x**2 + 5*a*b**2*x**4 + 20*b**3*x**6))/(440*a**(1/6) *(sqrt(a + b*x**2) + sqrt(b)*x)**(1/3)*a*b**3)