\(\int \frac {(a+b x^2+c x^4)^p}{x} \, dx\) [1096]

Optimal result

Integrand size = 18, antiderivative size = 152 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x} \, dx=\frac {4^{-1+p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^2},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{p} \] Output:

4^(-1+p)*(c*x^4+b*x^2+a)^p*AppellF1(-2*p,-p,-p,1-2*p,-1/2*(b-(-4*a*c+b^2)^ 
(1/2))/c/x^2,-1/2*(b+(-4*a*c+b^2)^(1/2))/c/x^2)/p/(((b-(-4*a*c+b^2)^(1/2)+ 
2*c*x^2)/c/x^2)^p)/(((b+(-4*a*c+b^2)^(1/2)+2*c*x^2)/c/x^2)^p)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x} \, dx=\frac {4^{-1+p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{p} \] Input:

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

Output:

(4^(-1 + p)*(a + b*x^2 + c*x^4)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b 
+ Sqrt[b^2 - 4*a*c])/(c*x^2), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x^2)])/(p*((b 
- Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^ 
2)/(c*x^2))^p)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1434, 1178, 150}

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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1178

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2^{2 p-2} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^2},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{p}\)

Input:

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

Output:

(2^(-2 + 2*p)*(a + b*x^2 + c*x^4)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*( 
b - Sqrt[b^2 - 4*a*c])/(c*x^2), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^2)])/(p* 
((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p*((b + Sqrt[b^2 - 4*a*c] + 2* 
c*x^2)/(c*x^2))^p)
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1178

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(1/(d + e*x))^(2*p))*((a + 
b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q + 2*c* 
x)/(2*c*(d + e*x))))^p))   Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b 
 - q)/(2*c)))*x, x]^p*Simp[1 - (d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d 
 + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[m, 0]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x} \, dx=\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) a p -2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{c \,x^{4}+b \,x^{2}+a}d x \right ) c p}{2 p} \] Input:

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

Output:

((a + b*x**2 + c*x**4)**p + 2*int((a + b*x**2 + c*x**4)**p/(a*x + b*x**3 + 
 c*x**5),x)*a*p - 2*int(((a + b*x**2 + c*x**4)**p*x**3)/(a + b*x**2 + c*x* 
*4),x)*c*p)/(2*p)