\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{x^2} \, dx\) [694]

Optimal result

Integrand size = 24, antiderivative size = 58 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^2} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2 p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^2} \, dx=-\frac {\left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2 p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x} \] Input:

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

Output:

-((((a + b*x^2)^2)^p*Hypergeometric2F1[-1/2, -2*p, 1/2, -((b*x^2)/a)])/(x* 
(1 + (b*x^2)/a)^(2*p)))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(((a + b*x**2)**2)**p/x**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^2} \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}+16 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 b p \,x^{4}-b \,x^{4}+4 a p \,x^{2}-a \,x^{2}}d x \right ) a \,p^{2} x -4 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 b p \,x^{4}-b \,x^{4}+4 a p \,x^{2}-a \,x^{2}}d x \right ) a p x}{x \left (4 p -1\right )} \] Input:

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

Output:

((a**2 + 2*a*b*x**2 + b**2*x**4)**p + 16*int((a**2 + 2*a*b*x**2 + b**2*x** 
4)**p/(4*a*p*x**2 - a*x**2 + 4*b*p*x**4 - b*x**4),x)*a*p**2*x - 4*int((a** 
2 + 2*a*b*x**2 + b**2*x**4)**p/(4*a*p*x**2 - a*x**2 + 4*b*p*x**4 - b*x**4) 
,x)*a*p*x)/(x*(4*p - 1))