Integrand size = 24, antiderivative size = 63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (1,1+2 p,2 (1+p),1+\frac {b x^2}{a}\right )}{2 a (1+2 p)} \] Output:
-1/2*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p*hypergeom([1, 1+2*p],[2*p+2],1+b* x^2/a)/a/(1+2*p)
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=-\frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \operatorname {Hypergeometric2F1}\left (1,1+2 p,2+2 p,1+\frac {b x^2}{a}\right )}{2 a (1+2 p)} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/x,x]
Output:
-1/2*((a + b*x^2)*((a + b*x^2)^2)^p*Hypergeometric2F1[1, 1 + 2*p, 2 + 2*p, 1 + (b*x^2)/a])/(a*(1 + 2*p))
Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int \frac {\left (\frac {b x^2}{a}+1\right )^{2 p}}{x}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int \frac {\left (\frac {b x^2}{a}+1\right )^{2 p}}{x^2}dx^2\) |
\(\Big \downarrow \) 75 |
\(\displaystyle -\frac {\left (\frac {b x^2}{a}+1\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (1,2 p+1,2 (p+1),\frac {b x^2}{a}+1\right )}{2 (2 p+1)}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/x,x]
Output:
-1/2*((1 + (b*x^2)/a)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[1, 1 + 2*p, 2*(1 + p), 1 + (b*x^2)/a])/(1 + 2*p)
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])
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]
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)]
\[\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{x}d x\]
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^p/x,x)
Output:
int((b^2*x^4+2*a*b*x^2+a^2)^p/x,x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x} \,d x } \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x,x, algorithm="fricas")
Output:
integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x, x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{p}}{x}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**p/x,x)
Output:
Integral(((a + b*x**2)**2)**p/x, x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x} \,d x } \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x,x, algorithm="maxima")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x, x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x} \,d x } \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x,x, algorithm="giac")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x, x)
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p}{x} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^p/x,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^p/x, x)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x} \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}+4 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{b \,x^{3}+a x}d x \right ) a p}{4 p} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^p/x,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)*a*p)/(4*p)