\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{x^3} \, dx\) [690]

Optimal result

Integrand size = 24, antiderivative size = 64 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^3} \, dx=\frac {b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2 (1+p),1+\frac {b x^2}{a}\right )}{2 a^2 (1+2 p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^3} \, dx=\frac {b \left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2+2 p,1+\frac {b x^2}{a}\right )}{2 a^2 (1+2 p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1385, 243, 75}

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.

\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^3} \, dx\)

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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]

\[\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{x^{3}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="fricas")
 

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)
 

Sympy [F]

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

integrate((b**2*x**4+2*a*b*x**2+a**2)**p/x**3,x)
 

Output:

Integral(((a + b*x**2)**2)**p/x**3, x)
 

Maxima [F]

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

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="maxima")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)
 

Giac [F]

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

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="giac")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^3} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p}{x^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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