\(\int \sqrt {d x} (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [697]

Optimal result

Integrand size = 28, antiderivative size = 67 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2 (d x)^{3/2} \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-2 p,\frac {7}{4},-\frac {b x^2}{a}\right )}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2}{3} x \sqrt {d x} \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-2 p,\frac {7}{4},-\frac {b x^2}{a}\right ) \] Input:

Integrate[Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1385, 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.

\(\displaystyle \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\)

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 278

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

Input:

Int[Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

(2*(d*x)^(3/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[3/4, -2*p, 
7/4, -((b*x^2)/a)])/(3*d*(1 + (b*x^2)/a)^(2*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 1385

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* 
FracPart[p]))   Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, 
 p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p] && NeQ[u, 
 x^(n - 1)] && NeQ[u, x^(2*n - 1)]
 

Maple [F]

\[\int \sqrt {d x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}d x\]

Input:

int((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Output:

int((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Fricas [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { \sqrt {d x} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")
 

Output:

integral(sqrt(d*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)
 

Sympy [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int \sqrt {d x} \left (\left (a + b x^{2}\right )^{2}\right )^{p}\, dx \] Input:

integrate((d*x)**(1/2)*(b**2*x**4+2*a*b*x**2+a**2)**p,x)
 

Output:

Integral(sqrt(d*x)*((a + b*x**2)**2)**p, x)
 

Maxima [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { \sqrt {d x} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")
 

Output:

integrate(sqrt(d*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)
 

Giac [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { \sqrt {d x} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")
 

Output:

integrate(sqrt(d*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int \sqrt {d\,x}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \,d x \] Input:

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)
 

Output:

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^p, x)
 

Reduce [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2 \sqrt {d}\, \left (\sqrt {x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} x +32 \left (\int \frac {\sqrt {x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{8 b p \,x^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a \,p^{2}+12 \left (\int \frac {\sqrt {x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{8 b p \,x^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a p \right )}{8 p +3} \] Input:

int((d*x)^(1/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Output:

(2*sqrt(d)*(sqrt(x)*(a**2 + 2*a*b*x**2 + b**2*x**4)**p*x + 32*int((sqrt(x) 
*(a**2 + 2*a*b*x**2 + b**2*x**4)**p)/(8*a*p + 3*a + 8*b*p*x**2 + 3*b*x**2) 
,x)*a*p**2 + 12*int((sqrt(x)*(a**2 + 2*a*b*x**2 + b**2*x**4)**p)/(8*a*p + 
3*a + 8*b*p*x**2 + 3*b*x**2),x)*a*p))/(8*p + 3)