Integrand size = 13, antiderivative size = 49 \[ \int x^4 \left (a+b x^2\right )^p \, dx=\frac {1}{5} x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \] Output:
1/5*x^5*(b*x^2+a)^p*hypergeom([5/2, -p],[7/2],-b*x^2/a)/((1+b*x^2/a)^p)
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int x^4 \left (a+b x^2\right )^p \, dx=\frac {1}{5} x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \] Input:
Integrate[x^4*(a + b*x^2)^p,x]
Output:
(x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, 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.
\(\displaystyle \int x^4 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int x^4 \left (\frac {b x^2}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\) |
Input:
Int[x^4*(a + b*x^2)^p,x]
Output:
(x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(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])
\[\int x^{4} \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int(x^4*(b*x^2+a)^p,x)
Output:
int(x^4*(b*x^2+a)^p,x)
\[ \int x^4 \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*x^4, x)
Result contains complex when optimal does not.
Time = 3.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.53 \[ \int x^4 \left (a+b x^2\right )^p \, dx=\frac {a^{p} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:
integrate(x**4*(b*x**2+a)**p,x)
Output:
a**p*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
\[ \int x^4 \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*x^4, x)
\[ \int x^4 \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*x^4, x)
Timed out. \[ \int x^4 \left (a+b x^2\right )^p \, dx=\int x^4\,{\left (b\,x^2+a\right )}^p \,d x \] Input:
int(x^4*(a + b*x^2)^p,x)
Output:
int(x^4*(a + b*x^2)^p, x)
\[ \int x^4 \left (a+b x^2\right )^p \, dx=\frac {-6 \left (b \,x^{2}+a \right )^{p} a^{2} p x +4 \left (b \,x^{2}+a \right )^{p} a b \,p^{2} x^{3}+2 \left (b \,x^{2}+a \right )^{p} a b p \,x^{3}+4 \left (b \,x^{2}+a \right )^{p} b^{2} p^{2} x^{5}+8 \left (b \,x^{2}+a \right )^{p} b^{2} p \,x^{5}+3 \left (b \,x^{2}+a \right )^{p} b^{2} x^{5}+48 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{8 b \,p^{3} x^{2}+36 b \,p^{2} x^{2}+8 a \,p^{3}+46 b p \,x^{2}+36 a \,p^{2}+15 b \,x^{2}+46 a p +15 a}d x \right ) a^{3} p^{4}+216 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{8 b \,p^{3} x^{2}+36 b \,p^{2} x^{2}+8 a \,p^{3}+46 b p \,x^{2}+36 a \,p^{2}+15 b \,x^{2}+46 a p +15 a}d x \right ) a^{3} p^{3}+276 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{8 b \,p^{3} x^{2}+36 b \,p^{2} x^{2}+8 a \,p^{3}+46 b p \,x^{2}+36 a \,p^{2}+15 b \,x^{2}+46 a p +15 a}d x \right ) a^{3} p^{2}+90 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{8 b \,p^{3} x^{2}+36 b \,p^{2} x^{2}+8 a \,p^{3}+46 b p \,x^{2}+36 a \,p^{2}+15 b \,x^{2}+46 a p +15 a}d x \right ) a^{3} p}{b^{2} \left (8 p^{3}+36 p^{2}+46 p +15\right )} \] Input:
int(x^4*(b*x^2+a)^p,x)
Output:
( - 6*(a + b*x**2)**p*a**2*p*x + 4*(a + b*x**2)**p*a*b*p**2*x**3 + 2*(a + b*x**2)**p*a*b*p*x**3 + 4*(a + b*x**2)**p*b**2*p**2*x**5 + 8*(a + b*x**2)* *p*b**2*p*x**5 + 3*(a + b*x**2)**p*b**2*x**5 + 48*int((a + b*x**2)**p/(8*a *p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b* p*x**2 + 15*b*x**2),x)*a**3*p**4 + 216*int((a + b*x**2)**p/(8*a*p**3 + 36* a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15 *b*x**2),x)*a**3*p**3 + 276*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46 *a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x) *a**3*p**2 + 90*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*p)/(b* *2*(8*p**3 + 36*p**2 + 46*p + 15))