\(\int \frac {(c+d x) (a+b x^2)^p}{\sqrt {e x}} \, dx\) [62]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {2 c \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{e}+\frac {2 d (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )}{3 e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {2 x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (3 c \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 \sqrt {e x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 113, 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.

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x + 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 = 23.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {a^{p} c \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {2 \sqrt {e}\, \left (4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} c p +3 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} c +4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{p} d p x +\sqrt {x}\, \left (b \,x^{2}+a \right )^{p} d x +128 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+16 b p \,x^{3}+16 a \,p^{2} x +3 b \,x^{3}+16 a p x +3 a x}d x \right ) a c \,p^{4}+224 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+16 b p \,x^{3}+16 a \,p^{2} x +3 b \,x^{3}+16 a p x +3 a x}d x \right ) a c \,p^{3}+120 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+16 b p \,x^{3}+16 a \,p^{2} x +3 b \,x^{3}+16 a p x +3 a x}d x \right ) a c \,p^{2}+18 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{3}+16 b p \,x^{3}+16 a \,p^{2} x +3 b \,x^{3}+16 a p x +3 a x}d x \right ) a c p +128 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a d \,p^{4}+160 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a d \,p^{3}+56 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a d \,p^{2}+6 \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a d p \right )}{e \left (16 p^{2}+16 p +3\right )} \] Input:

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

Output:

(2*sqrt(e)*(4*sqrt(x)*(a + b*x**2)**p*c*p + 3*sqrt(x)*(a + b*x**2)**p*c + 
4*sqrt(x)*(a + b*x**2)**p*d*p*x + sqrt(x)*(a + b*x**2)**p*d*x + 128*int((s 
qrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 16*a*p*x + 3*a*x + 16*b*p**2*x**3 + 
 16*b*p*x**3 + 3*b*x**3),x)*a*c*p**4 + 224*int((sqrt(x)*(a + b*x**2)**p)/( 
16*a*p**2*x + 16*a*p*x + 3*a*x + 16*b*p**2*x**3 + 16*b*p*x**3 + 3*b*x**3), 
x)*a*c*p**3 + 120*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2*x + 16*a*p*x + 
3*a*x + 16*b*p**2*x**3 + 16*b*p*x**3 + 3*b*x**3),x)*a*c*p**2 + 18*int((sqr 
t(x)*(a + b*x**2)**p)/(16*a*p**2*x + 16*a*p*x + 3*a*x + 16*b*p**2*x**3 + 1 
6*b*p*x**3 + 3*b*x**3),x)*a*c*p + 128*int((sqrt(x)*(a + b*x**2)**p)/(16*a* 
p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2),x)*a*d*p**4 
 + 160*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2 
*x**2 + 16*b*p*x**2 + 3*b*x**2),x)*a*d*p**3 + 56*int((sqrt(x)*(a + b*x**2) 
**p)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2), 
x)*a*d*p**2 + 6*int((sqrt(x)*(a + b*x**2)**p)/(16*a*p**2 + 16*a*p + 3*a + 
16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2),x)*a*d*p))/(e*(16*p**2 + 16*p + 3 
))