Integrand size = 15, antiderivative size = 59 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 a^2 \sqrt [3]{a+b x^2}}{2 b^3}-\frac {3 a \left (a+b x^2\right )^{4/3}}{4 b^3}+\frac {3 \left (a+b x^2\right )^{7/3}}{14 b^3} \] Output:
3/2*a^2*(b*x^2+a)^(1/3)/b^3-3/4*a*(b*x^2+a)^(4/3)/b^3+3/14*(b*x^2+a)^(7/3) /b^3
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 \sqrt [3]{a+b x^2} \left (9 a^2-3 a b x^2+2 b^2 x^4\right )}{28 b^3} \] Input:
Integrate[x^5/(a + b*x^2)^(2/3),x]
Output:
(3*(a + b*x^2)^(1/3)*(9*a^2 - 3*a*b*x^2 + 2*b^2*x^4))/(28*b^3)
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{\left (b x^2+a\right )^{2/3}}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^2}{b^2 \left (b x^2+a\right )^{2/3}}-\frac {2 \sqrt [3]{b x^2+a} a}{b^2}+\frac {\left (b x^2+a\right )^{4/3}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {3 a^2 \sqrt [3]{a+b x^2}}{b^3}+\frac {3 \left (a+b x^2\right )^{7/3}}{7 b^3}-\frac {3 a \left (a+b x^2\right )^{4/3}}{2 b^3}\right )\) |
Input:
Int[x^5/(a + b*x^2)^(2/3),x]
Output:
((3*a^2*(a + b*x^2)^(1/3))/b^3 - (3*a*(a + b*x^2)^(4/3))/(2*b^3) + (3*(a + b*x^2)^(7/3))/(7*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 = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (2 b^{2} x^{4}-3 a b \,x^{2}+9 a^{2}\right )}{28 b^{3}}\) | \(36\) |
trager | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (2 b^{2} x^{4}-3 a b \,x^{2}+9 a^{2}\right )}{28 b^{3}}\) | \(36\) |
risch | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (2 b^{2} x^{4}-3 a b \,x^{2}+9 a^{2}\right )}{28 b^{3}}\) | \(36\) |
pseudoelliptic | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (2 b^{2} x^{4}-3 a b \,x^{2}+9 a^{2}\right )}{28 b^{3}}\) | \(36\) |
orering | \(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (2 b^{2} x^{4}-3 a b \,x^{2}+9 a^{2}\right )}{28 b^{3}}\) | \(36\) |
Input:
int(x^5/(b*x^2+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
3/28*(b*x^2+a)^(1/3)*(2*b^2*x^4-3*a*b*x^2+9*a^2)/b^3
Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.59 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 \, {\left (2 \, b^{2} x^{4} - 3 \, a b x^{2} + 9 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{28 \, b^{3}} \] Input:
integrate(x^5/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
3/28*(2*b^2*x^4 - 3*a*b*x^2 + 9*a^2)*(b*x^2 + a)^(1/3)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 631 vs. \(2 (54) = 108\).
Time = 1.02 (sec) , antiderivative size = 631, normalized size of antiderivative = 10.69 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {27 a^{\frac {31}{3}} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} - \frac {27 a^{\frac {31}{3}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} + \frac {72 a^{\frac {28}{3}} b x^{2} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} - \frac {81 a^{\frac {28}{3}} b x^{2}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} + \frac {60 a^{\frac {25}{3}} b^{2} x^{4} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} - \frac {81 a^{\frac {25}{3}} b^{2} x^{4}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} + \frac {18 a^{\frac {22}{3}} b^{3} x^{6} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} - \frac {27 a^{\frac {22}{3}} b^{3} x^{6}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} + \frac {9 a^{\frac {19}{3}} b^{4} x^{8} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} + \frac {6 a^{\frac {16}{3}} b^{5} x^{10} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{28 a^{8} b^{3} + 84 a^{7} b^{4} x^{2} + 84 a^{6} b^{5} x^{4} + 28 a^{5} b^{6} x^{6}} \] Input:
integrate(x**5/(b*x**2+a)**(2/3),x)
Output:
27*a**(31/3)*(1 + b*x**2/a)**(1/3)/(28*a**8*b**3 + 84*a**7*b**4*x**2 + 84* a**6*b**5*x**4 + 28*a**5*b**6*x**6) - 27*a**(31/3)/(28*a**8*b**3 + 84*a**7 *b**4*x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) + 72*a**(28/3)*b*x**2* (1 + b*x**2/a)**(1/3)/(28*a**8*b**3 + 84*a**7*b**4*x**2 + 84*a**6*b**5*x** 4 + 28*a**5*b**6*x**6) - 81*a**(28/3)*b*x**2/(28*a**8*b**3 + 84*a**7*b**4* x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) + 60*a**(25/3)*b**2*x**4*(1 + b*x**2/a)**(1/3)/(28*a**8*b**3 + 84*a**7*b**4*x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) - 81*a**(25/3)*b**2*x**4/(28*a**8*b**3 + 84*a**7*b**4* x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) + 18*a**(22/3)*b**3*x**6*(1 + b*x**2/a)**(1/3)/(28*a**8*b**3 + 84*a**7*b**4*x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) - 27*a**(22/3)*b**3*x**6/(28*a**8*b**3 + 84*a**7*b**4* x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) + 9*a**(19/3)*b**4*x**8*(1 + b*x**2/a)**(1/3)/(28*a**8*b**3 + 84*a**7*b**4*x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6) + 6*a**(16/3)*b**5*x**10*(1 + b*x**2/a)**(1/3)/(28*a**8 *b**3 + 84*a**7*b**4*x**2 + 84*a**6*b**5*x**4 + 28*a**5*b**6*x**6)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}}}{14 \, b^{3}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} a}{4 \, b^{3}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{2}}{2 \, b^{3}} \] Input:
integrate(x^5/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
3/14*(b*x^2 + a)^(7/3)/b^3 - 3/4*(b*x^2 + a)^(4/3)*a/b^3 + 3/2*(b*x^2 + a) ^(1/3)*a^2/b^3
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{2}}{2 \, b^{3}} + \frac {3 \, {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} - 7 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} a\right )}}{28 \, b^{3}} \] Input:
integrate(x^5/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
3/2*(b*x^2 + a)^(1/3)*a^2/b^3 + 3/28*(2*(b*x^2 + a)^(7/3) - 7*(b*x^2 + a)^ (4/3)*a)/b^3
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61 \[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx={\left (b\,x^2+a\right )}^{1/3}\,\left (\frac {27\,a^2}{28\,b^3}+\frac {3\,x^4}{14\,b}-\frac {9\,a\,x^2}{28\,b^2}\right ) \] Input:
int(x^5/(a + b*x^2)^(2/3),x)
Output:
(a + b*x^2)^(1/3)*((27*a^2)/(28*b^3) + (3*x^4)/(14*b) - (9*a*x^2)/(28*b^2) )
\[ \int \frac {x^5}{\left (a+b x^2\right )^{2/3}} \, dx=\int \frac {x^{5}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \] Input:
int(x^5/(b*x^2+a)^(2/3),x)
Output:
int(x**5/(a + b*x**2)**(2/3),x)