Integrand size = 24, antiderivative size = 107 \[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^{1+p}}{b e (9+4 p)}+\frac {2 \left (c-\frac {5 a d}{9 b+4 b p}\right ) (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 e} \] Output:
2*d*(e*x)^(5/2)*(b*x^2+a)^(p+1)/b/e/(9+4*p)+2/5*(c-5*a*d/(4*b*p+9*b))*(e*x )^(5/2)*(b*x^2+a)^p*hypergeom([5/4, -p],[9/4],-b*x^2/a)/e/((1+b*x^2/a)^p)
Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {2}{45} x (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (9 c \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )+5 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {9}{4},-p,\frac {13}{4},-\frac {b x^2}{a}\right )\right ) \] Input:
Integrate[(e*x)^(3/2)*(a + b*x^2)^p*(c + d*x^2),x]
Output:
(2*x*(e*x)^(3/2)*(a + b*x^2)^p*(9*c*Hypergeometric2F1[5/4, -p, 9/4, -((b*x ^2)/a)] + 5*d*x^2*Hypergeometric2F1[9/4, -p, 13/4, -((b*x^2)/a)]))/(45*(1 + (b*x^2)/a)^p)
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {363, 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.
| \(\displaystyle \int (e x)^{3/2} \left (c+d x^2\right ) \left (a+b x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \left (c-\frac {5 a d}{4 b p+9 b}\right ) \int (e x)^{3/2} \left (b x^2+a\right )^pdx+\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c-\frac {5 a d}{4 b p+9 b}\right ) \int (e x)^{3/2} \left (\frac {b x^2}{a}+1\right )^pdx+\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {2 (e x)^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c-\frac {5 a d}{4 b p+9 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 e}+\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}\) |
Input:
Int[(e*x)^(3/2)*(a + b*x^2)^p*(c + d*x^2),x]
Output:
(2*d*(e*x)^(5/2)*(a + b*x^2)^(1 + p))/(b*e*(9 + 4*p)) + (2*(c - (5*a*d)/(9 *b + 4*b*p))*(e*x)^(5/2)*(a + b*x^2)^p*Hypergeometric2F1[5/4, -p, 9/4, -(( b*x^2)/a)])/(5*e*(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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
\[\int \left (e x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{p} \left (x^{2} d +c \right )d x\]
Input:
int((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x)
Output:
int((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x)
\[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\int { {\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x, algorithm="fricas")
Output:
integral((d*e*x^3 + c*e*x)*sqrt(e*x)*(b*x^2 + a)^p, x)
Result contains complex when optimal does not.
Time = 173.53 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {a^{p} c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, - p \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} \] Input:
integrate((e*x)**(3/2)*(b*x**2+a)**p*(d*x**2+c),x)
Output:
a**p*c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**2*exp_po lar(I*pi)/a)/(2*gamma(9/4)) + a**p*d*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((9 /4, -p), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(13/4))
\[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\int { {\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)*(e*x)^(3/2)*(b*x^2 + a)^p, x)
\[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\int { {\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x, algorithm="giac")
Output:
integrate((d*x^2 + c)*(e*x)^(3/2)*(b*x^2 + a)^p, x)
Timed out. \[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\int {\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^p\,\left (d\,x^2+c\right ) \,d x \] Input:
int((e*x)^(3/2)*(a + b*x^2)^p*(c + d*x^2),x)
Output:
int((e*x)^(3/2)*(a + b*x^2)^p*(c + d*x^2), x)
\[ \int (e x)^{3/2} \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx =\text {Too large to display} \] Input:
int((e*x)^(3/2)*(b*x^2+a)^p*(d*x^2+c),x)
Output:
(2*sqrt(e)*e*( - 20*sqrt(x)*(a + b*x**2)**p*a**2*d*p + 16*sqrt(x)*(a + b*x **2)**p*a*b*c*p**2 + 36*sqrt(x)*(a + b*x**2)**p*a*b*c*p + 16*sqrt(x)*(a + b*x**2)**p*a*b*d*p**2*x**2 + 4*sqrt(x)*(a + b*x**2)**p*a*b*d*p*x**2 + 16*s qrt(x)*(a + b*x**2)**p*b**2*c*p**2*x**2 + 40*sqrt(x)*(a + b*x**2)**p*b**2* c*p*x**2 + 9*sqrt(x)*(a + b*x**2)**p*b**2*c*x**2 + 16*sqrt(x)*(a + b*x**2) **p*b**2*d*p**2*x**4 + 24*sqrt(x)*(a + b*x**2)**p*b**2*d*p*x**4 + 5*sqrt(x )*(a + b*x**2)**p*b**2*d*x**4 + 640*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p* *3*x + 240*a*p**2*x + 236*a*p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2*x** 3 + 236*b*p*x**3 + 45*b*x**3),x)*a**3*d*p**4 + 2400*int((sqrt(x)*(a + b*x* *2)**p)/(64*a*p**3*x + 240*a*p**2*x + 236*a*p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2*x**3 + 236*b*p*x**3 + 45*b*x**3),x)*a**3*d*p**3 + 2360*int((s qrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 240*a*p**2*x + 236*a*p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2*x**3 + 236*b*p*x**3 + 45*b*x**3),x)*a**3*d*p* *2 + 450*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 240*a*p**2*x + 236*a *p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2*x**3 + 236*b*p*x**3 + 45*b*x** 3),x)*a**3*d*p - 512*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 240*a*p* *2*x + 236*a*p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2*x**3 + 236*b*p*x** 3 + 45*b*x**3),x)*a**2*b*c*p**5 - 3072*int((sqrt(x)*(a + b*x**2)**p)/(64*a *p**3*x + 240*a*p**2*x + 236*a*p*x + 45*a*x + 64*b*p**3*x**3 + 240*b*p**2* x**3 + 236*b*p*x**3 + 45*b*x**3),x)*a**2*b*c*p**4 - 6208*int((sqrt(x)*(...