Integrand size = 15, antiderivative size = 51 \[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right ) \] Output:
2/7*x^(7/2)*(b*x^2+a)^p*hypergeom([7/4, -p],[11/4],-b*x^2/a)/((1+b*x^2/a)^ p)
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right ) \] Input:
Integrate[x^(5/2)*(a + b*x^2)^p,x]
Output:
(2*x^(7/2)*(a + b*x^2)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)])/( 7*(1 + (b*x^2)/a)^p)
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {279, 278}
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.
\(\displaystyle \int x^{5/2} \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int x^{5/2} \left (\frac {b x^2}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {2}{7} x^{7/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right )\) |
Input:
Int[x^(5/2)*(a + b*x^2)^p,x]
Output:
(2*x^(7/2)*(a + b*x^2)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)])/( 7*(1 + (b*x^2)/a)^p)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int x^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int(x^(5/2)*(b*x^2+a)^p,x)
Output:
int(x^(5/2)*(b*x^2+a)^p,x)
\[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{\frac {5}{2}} \,d x } \] Input:
integrate(x^(5/2)*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*x^(5/2), x)
Timed out. \[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate(x**(5/2)*(b*x**2+a)**p,x)
Output:
Timed out
\[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{\frac {5}{2}} \,d x } \] Input:
integrate(x^(5/2)*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*x^(5/2), x)
\[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{\frac {5}{2}} \,d x } \] Input:
integrate(x^(5/2)*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*x^(5/2), x)
Timed out. \[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\int x^{5/2}\,{\left (b\,x^2+a\right )}^p \,d x \] Input:
int(x^(5/2)*(a + b*x^2)^p,x)
Output:
int(x^(5/2)*(a + b*x^2)^p, x)
\[ \int x^{5/2} \left (a+b x^2\right )^p \, dx=\frac {8 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} a p x +8 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b p \,x^{3}+6 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b \,x^{3}-192 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+40 b p \,x^{2}+16 a \,p^{2}+21 b \,x^{2}+40 a p +21 a}d x \right ) a^{2} p^{3}-480 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+40 b p \,x^{2}+16 a \,p^{2}+21 b \,x^{2}+40 a p +21 a}d x \right ) a^{2} p^{2}-252 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+40 b p \,x^{2}+16 a \,p^{2}+21 b \,x^{2}+40 a p +21 a}d x \right ) a^{2} p}{b \left (16 p^{2}+40 p +21\right )} \] Input:
int(x^(5/2)*(b*x^2+a)^p,x)
Output:
(2*(4*sqrt(x)*(a + b*x**2)**p*a*p*x + 4*sqrt(x)*(a + b*x**2)**p*b*p*x**3 + 3*sqrt(x)*(a + b*x**2)**p*b*x**3 - 96*int((sqrt(x)*(a + b*x**2)**p)/(16*a *p**2 + 40*a*p + 21*a + 16*b*p**2*x**2 + 40*b*p*x**2 + 21*b*x**2),x)*a**2* p**3 - 240*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2 + 40*a*p + 21*a + 16*b *p**2*x**2 + 40*b*p*x**2 + 21*b*x**2),x)*a**2*p**2 - 126*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2 + 40*a*p + 21*a + 16*b*p**2*x**2 + 40*b*p*x**2 + 2 1*b*x**2),x)*a**2*p))/(b*(16*p**2 + 40*p + 21))