\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{x^4} \, dx\) [695]

Optimal result

Integrand size = 24, antiderivative size = 60 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^4} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2 p,-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^4} \, dx=-\frac {\left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2 p,-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 60, 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.

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

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^4} \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}+16 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 b p \,x^{6}-3 b \,x^{6}+4 a p \,x^{4}-3 a \,x^{4}}d x \right ) a \,p^{2} x^{3}-12 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 b p \,x^{6}-3 b \,x^{6}+4 a p \,x^{4}-3 a \,x^{4}}d x \right ) a p \,x^{3}}{x^{3} \left (4 p -3\right )} \] Input:

int((b^2*x^4+2*a*b*x^2+a^2)^p/x^4,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**4 - 3*a*x**4 + 4*b*p*x**6 - 3*b*x**6),x)*a*p**2*x**3 - 12* 
int((a**2 + 2*a*b*x**2 + b**2*x**4)**p/(4*a*p*x**4 - 3*a*x**4 + 4*b*p*x**6 
 - 3*b*x**6),x)*a*p*x**3)/(x**3*(4*p - 3))