\(\int \sqrt {e x} (c+d x) (a+b x^2)^p \, dx\) [61]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int \sqrt {e x} (c+d x) \left (a+b x^2\right )^p \, dx=\frac {2 c (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )}{3 e}+\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 e^2} \] Output:

2/3*c*(e*x)^(3/2)*(b*x^2+a)^p*hypergeom([3/4, -p],[7/4],-b*x^2/a)/e/((1+b* 
x^2/a)^p)+2/5*d*(e*x)^(5/2)*(b*x^2+a)^p*hypergeom([5/4, -p],[9/4],-b*x^2/a 
)/e^2/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \sqrt {e x} (c+d x) \left (a+b x^2\right )^p \, dx=\frac {2}{15} x \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (5 c \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )+3 d x \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )\right ) \] Input:

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

Output:

(2*x*Sqrt[e*x]*(a + b*x^2)^p*(5*c*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^2 
)/a)] + 3*d*x*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^2)/a)]))/(15*(1 + (b* 
x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {557, 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.

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

\(\Big \downarrow \) 557

\(\displaystyle c \int \sqrt {e x} \left (b x^2+a\right )^pdx+\frac {d \int (e x)^{3/2} \left (b x^2+a\right )^pdx}{e}\)

\(\Big \downarrow \) 279

\(\displaystyle c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \sqrt {e x} \left (\frac {b x^2}{a}+1\right )^pdx+\frac {d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int (e x)^{3/2} \left (\frac {b x^2}{a}+1\right )^pdx}{e}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {2 c (e x)^{3/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )}{3 e}+\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 e^2}\)

Input:

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

Output:

(2*c*(e*x)^(3/2)*(a + b*x^2)^p*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^2)/a 
)])/(3*e*(1 + (b*x^2)/a)^p) + (2*d*(e*x)^(5/2)*(a + b*x^2)^p*Hypergeometri 
c2F1[5/4, -p, 9/4, -((b*x^2)/a)])/(5*e^2*(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 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 25.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \sqrt {e x} (c+d x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + \frac {a^{p} d \sqrt {e} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

a**p*c*sqrt(e)*x**(3/2)*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**2*exp_pol 
ar(I*pi)/a)/(2*gamma(7/4)) + a**p*d*sqrt(e)*x**(5/2)*gamma(5/4)*hyper((5/4 
, -p), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(9/4))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(e)*(16*sqrt(x)*(a + b*x**2)**p*a*d*p**2 + 12*sqrt(x)*(a + b*x**2)* 
*p*a*d*p + 16*sqrt(x)*(a + b*x**2)**p*b*c*p**2*x + 24*sqrt(x)*(a + b*x**2) 
**p*b*c*p*x + 5*sqrt(x)*(a + b*x**2)**p*b*c*x + 16*sqrt(x)*(a + b*x**2)**p 
*b*d*p**2*x**2 + 16*sqrt(x)*(a + b*x**2)**p*b*d*p*x**2 + 3*sqrt(x)*(a + b* 
x**2)**p*b*d*x**2 - 512*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 144*a 
*p**2*x + 92*a*p*x + 15*a*x + 64*b*p**3*x**3 + 144*b*p**2*x**3 + 92*b*p*x* 
*3 + 15*b*x**3),x)*a**2*d*p**5 - 1536*int((sqrt(x)*(a + b*x**2)**p)/(64*a* 
p**3*x + 144*a*p**2*x + 92*a*p*x + 15*a*x + 64*b*p**3*x**3 + 144*b*p**2*x* 
*3 + 92*b*p*x**3 + 15*b*x**3),x)*a**2*d*p**4 - 1600*int((sqrt(x)*(a + b*x* 
*2)**p)/(64*a*p**3*x + 144*a*p**2*x + 92*a*p*x + 15*a*x + 64*b*p**3*x**3 + 
 144*b*p**2*x**3 + 92*b*p*x**3 + 15*b*x**3),x)*a**2*d*p**3 - 672*int((sqrt 
(x)*(a + b*x**2)**p)/(64*a*p**3*x + 144*a*p**2*x + 92*a*p*x + 15*a*x + 64* 
b*p**3*x**3 + 144*b*p**2*x**3 + 92*b*p*x**3 + 15*b*x**3),x)*a**2*d*p**2 - 
90*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 144*a*p**2*x + 92*a*p*x + 
15*a*x + 64*b*p**3*x**3 + 144*b*p**2*x**3 + 92*b*p*x**3 + 15*b*x**3),x)*a* 
*2*d*p + 2048*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3 + 144*a*p**2 + 92*a 
*p + 15*a + 64*b*p**3*x**2 + 144*b*p**2*x**2 + 92*b*p*x**2 + 15*b*x**2),x) 
*a*b*c*p**6 + 7680*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3 + 144*a*p**2 + 
 92*a*p + 15*a + 64*b*p**3*x**2 + 144*b*p**2*x**2 + 92*b*p*x**2 + 15*b*x** 
2),x)*a*b*c*p**5 + 10496*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3 + 144...