\(\int \frac {(a+b x^2)^p (c+d x^2)}{\sqrt {e x}} \, dx\) [521]

Optimal result

Integrand size = 24, antiderivative size = 105 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\frac {2 d \sqrt {e x} \left (a+b x^2\right )^{1+p}}{b e (5+4 p)}+\frac {2 \left (c-\frac {a d}{5 b+4 b p}\right ) \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{e} \] Output:

2*d*(e*x)^(1/2)*(b*x^2+a)^(p+1)/b/e/(5+4*p)+2*(c-a*d/(4*b*p+5*b))*(e*x)^(1 
/2)*(b*x^2+a)^p*hypergeom([1/4, -p],[5/4],-b*x^2/a)/e/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\frac {2 x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+(-a d+b c (5+4 p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{b (5+4 p) \sqrt {e x}} \] Input:

Integrate[((a + b*x^2)^p*(c + d*x^2))/Sqrt[e*x],x]
 

Output:

(2*x*(a + b*x^2)^p*(d*(a + b*x^2)*(1 + (b*x^2)/a)^p + (-(a*d) + b*c*(5 + 4 
*p))*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^2)/a)]))/(b*(5 + 4*p)*Sqrt[e*x 
]*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 105, 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.

\(\displaystyle \int \frac {\left (c+d x^2\right ) \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx\)

\(\Big \downarrow \) 363

\(\displaystyle \left (c-\frac {a d}{4 b p+5 b}\right ) \int \frac {\left (b x^2+a\right )^p}{\sqrt {e x}}dx+\frac {2 d \sqrt {e x} \left (a+b x^2\right )^{p+1}}{b e (4 p+5)}\)

\(\Big \downarrow \) 279

\(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c-\frac {a d}{4 b p+5 b}\right ) \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{\sqrt {e x}}dx+\frac {2 d \sqrt {e x} \left (a+b x^2\right )^{p+1}}{b e (4 p+5)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {2 \sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c-\frac {a d}{4 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{e}+\frac {2 d \sqrt {e x} \left (a+b x^2\right )^{p+1}}{b e (4 p+5)}\)

Input:

Int[((a + b*x^2)^p*(c + d*x^2))/Sqrt[e*x],x]
 

Output:

(2*d*Sqrt[e*x]*(a + b*x^2)^(1 + p))/(b*e*(5 + 4*p)) + (2*(c - (a*d)/(5*b + 
 4*b*p))*Sqrt[e*x]*(a + b*x^2)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^2) 
/a)])/(e*(1 + (b*x^2)/a)^p)
 

Defintions of rubi rules used

rule 278

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])
 

rule 279

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])
 

rule 363

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]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p} \left (x^{2} d +c \right )}{\sqrt {e x}}d x\]

Input:

int((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x)
 

Output:

int((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x^{2} + a\right )}^{p}}{\sqrt {e x}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x, algorithm="fricas")
 

Output:

integral((d*x^2 + c)*sqrt(e*x)*(b*x^2 + a)^p/(e*x), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 28.63 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\frac {a^{p} c \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \] Input:

integrate((b*x**2+a)**p*(d*x**2+c)/(e*x)**(1/2),x)
 

Output:

a**p*c*sqrt(x)*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**2*exp_polar(I*pi)/ 
a)/(2*sqrt(e)*gamma(5/4)) + a**p*d*x**(5/2)*gamma(5/4)*hyper((5/4, -p), (9 
/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(9/4))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x^{2} + a\right )}^{p}}{\sqrt {e x}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(b*x^2 + a)^p/sqrt(e*x), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x^{2} + a\right )}^{p}}{\sqrt {e x}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(b*x^2 + a)^p/sqrt(e*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,\left (d\,x^2+c\right )}{\sqrt {e\,x}} \,d x \] Input:

int(((a + b*x^2)^p*(c + d*x^2))/(e*x)^(1/2),x)
 

Output:

int(((a + b*x^2)^p*(c + d*x^2))/(e*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{\sqrt {e x}} \, dx=\frac {2 \sqrt {e}\, \left (4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} a d p +4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b c p +5 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b c +4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b d p \,x^{2}+\sqrt {x}\, \left (b \,x^{2}+a \right )^{p} b d \,x^{2}-32 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a^{2} d \,p^{3}-48 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a^{2} d \,p^{2}-10 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a^{2} d p +128 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a b c \,p^{4}+352 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a b c \,p^{3}+280 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a b c \,p^{2}+50 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+24 b p \,x^{3}+16 a \,p^{2} x +5 b \,x^{3}+24 a p x +5 a x}d x \right ) a b c p \right )}{b e \left (16 p^{2}+24 p +5\right )} \] Input:

int((b*x^2+a)^p*(d*x^2+c)/(e*x)^(1/2),x)
 

Output:

(2*sqrt(e)*(4*sqrt(x)*(a + b*x**2)**p*a*d*p + 4*sqrt(x)*(a + b*x**2)**p*b* 
c*p + 5*sqrt(x)*(a + b*x**2)**p*b*c + 4*sqrt(x)*(a + b*x**2)**p*b*d*p*x**2 
 + sqrt(x)*(a + b*x**2)**p*b*d*x**2 - 32*int((sqrt(x)*(a + b*x**2)**p)/(16 
*a*p**2*x + 24*a*p*x + 5*a*x + 16*b*p**2*x**3 + 24*b*p*x**3 + 5*b*x**3),x) 
*a**2*d*p**3 - 48*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 24*a*p*x + 
5*a*x + 16*b*p**2*x**3 + 24*b*p*x**3 + 5*b*x**3),x)*a**2*d*p**2 - 10*int(( 
sqrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 24*a*p*x + 5*a*x + 16*b*p**2*x**3 
+ 24*b*p*x**3 + 5*b*x**3),x)*a**2*d*p + 128*int((sqrt(x)*(a + b*x**2)**p)/ 
(16*a*p**2*x + 24*a*p*x + 5*a*x + 16*b*p**2*x**3 + 24*b*p*x**3 + 5*b*x**3) 
,x)*a*b*c*p**4 + 352*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 24*a*p*x 
 + 5*a*x + 16*b*p**2*x**3 + 24*b*p*x**3 + 5*b*x**3),x)*a*b*c*p**3 + 280*in 
t((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 24*a*p*x + 5*a*x + 16*b*p**2*x* 
*3 + 24*b*p*x**3 + 5*b*x**3),x)*a*b*c*p**2 + 50*int((sqrt(x)*(a + b*x**2)* 
*p)/(16*a*p**2*x + 24*a*p*x + 5*a*x + 16*b*p**2*x**3 + 24*b*p*x**3 + 5*b*x 
**3),x)*a*b*c*p))/(b*e*(16*p**2 + 24*p + 5))