\(\int \frac {(c+d x)^2 (a+b x^2)^p}{\sqrt {e x}} \, dx\) [67]

Optimal result

Integrand size = 24, antiderivative size = 176 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {2 d^2 \sqrt {e x} \left (a+b x^2\right )^{1+p}}{b e (5+4 p)}-\frac {2 \left (a d^2-b c^2 (5+4 p)\right ) \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{b e (5+4 p)}+\frac {4 c d (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )}{3 e^2} \] Output:

2*d^2*(e*x)^(1/2)*(b*x^2+a)^(p+1)/b/e/(5+4*p)-2*(a*d^2-b*c^2*(5+4*p))*(e*x 
)^(1/2)*(b*x^2+a)^p*hypergeom([1/4, -p],[5/4],-b*x^2/a)/b/e/(5+4*p)/((1+b* 
x^2/a)^p)+4/3*c*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 = 0.00 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.62 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {2 x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (15 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )+d x \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^2}{a}\right )+3 d x \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )\right )\right )}{15 \sqrt {e x}} \] Input:

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

Output:

(2*x*(a + b*x^2)^p*(15*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^2)/a)] + 
 d*x*(10*c*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^2)/a)] + 3*d*x*Hypergeom 
etric2F1[5/4, -p, 9/4, -((b*x^2)/a)])))/(15*Sqrt[e*x]*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {559, 27, 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.

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

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 559

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( 
m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1))   Int[(e*x)^m*(a + b* 
x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 
)*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && IGtQ[n, 1] &&  !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 40.58 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^p}{\sqrt {e x}} \, dx=\frac {a^{p} c^{2} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} c d x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\sqrt {e} \Gamma \left (\frac {7}{4}\right )} + \frac {a^{p} d^{2} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

a**p*c**2*sqrt(x)*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**2*exp_polar(I*p 
i)/a)/(2*sqrt(e)*gamma(5/4)) + a**p*c*d*x**(3/2)*gamma(3/4)*hyper((3/4, -p 
), (7/4,), b*x**2*exp_polar(I*pi)/a)/(sqrt(e)*gamma(7/4)) + a**p*d**2*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]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(2*sqrt(e)*(16*sqrt(x)*(a + b*x**2)**p*a*d**2*p**2 + 12*sqrt(x)*(a + b*x** 
2)**p*a*d**2*p + 16*sqrt(x)*(a + b*x**2)**p*b*c**2*p**2 + 32*sqrt(x)*(a + 
b*x**2)**p*b*c**2*p + 15*sqrt(x)*(a + b*x**2)**p*b*c**2 + 32*sqrt(x)*(a + 
b*x**2)**p*b*c*d*p**2*x + 48*sqrt(x)*(a + b*x**2)**p*b*c*d*p*x + 10*sqrt(x 
)*(a + b*x**2)**p*b*c*d*x + 16*sqrt(x)*(a + b*x**2)**p*b*d**2*p**2*x**2 + 
16*sqrt(x)*(a + b*x**2)**p*b*d**2*p*x**2 + 3*sqrt(x)*(a + b*x**2)**p*b*d** 
2*x**2 - 512*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p**3*x + 144*a*p**2*x + 9 
2*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**2*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**2*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**2*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**2*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**2*p + 2048*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 + 1 
5*b*x**3),x)*a*b*c**2*p**6 + 8704*int((sqrt(x)*(a + b*x**2)**p)/(64*a*p...