\(\int \frac {(a+b x^2+c x^4)^p}{x^3} \, dx\) [1097]

Optimal result

Integrand size = 18, antiderivative size = 166 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^3} \, dx=-\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),-\frac {b-\sqrt {b^2-4 a c}}{2 c x^2},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{(1-2 p) x^2} \] Output:

-2^(-1+2*p)*(c*x^4+b*x^2+a)^p*AppellF1(1-2*p,-p,-p,2-2*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)/(1-2*p)/x^2/(((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 = 0.27 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^3} \, dx=\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{(-1+2 p) x^2} \] Input:

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

Output:

(2^(-1 + 2*p)*(a + b*x^2 + c*x^4)^p*AppellF1[1 - 2*p, -p, -p, 2 - 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)])/(( 
-1 + 2*p)*x^2*((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 = 0.49 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1178

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^3} \, dx=\frac {-\left (c \,x^{4}+b \,x^{2}+a \right )^{p} b p +\left (c \,x^{4}+b \,x^{2}+a \right )^{p} b -\left (c \,x^{4}+b \,x^{2}+a \right )^{p} c \,x^{2}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{c p \,x^{5}-c \,x^{5}+b p \,x^{3}-b \,x^{3}+a p x -a x}d x \right ) b^{2} p^{3} x^{2}-4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{c p \,x^{5}-c \,x^{5}+b p \,x^{3}-b \,x^{3}+a p x -a x}d x \right ) b^{2} p^{2} x^{2}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{c p \,x^{5}-c \,x^{5}+b p \,x^{3}-b \,x^{3}+a p x -a x}d x \right ) b^{2} p \,x^{2}+4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{c p \,x^{4}-c \,x^{4}+b p \,x^{2}-b \,x^{2}+a p -a}d x \right ) c^{2} p^{2} x^{2}-4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{c p \,x^{4}-c \,x^{4}+b p \,x^{2}-b \,x^{2}+a p -a}d x \right ) c^{2} p \,x^{2}+4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x}{c p \,x^{4}-c \,x^{4}+b p \,x^{2}-b \,x^{2}+a p -a}d x \right ) b c \,p^{3} x^{2}-6 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x}{c p \,x^{4}-c \,x^{4}+b p \,x^{2}-b \,x^{2}+a p -a}d x \right ) b c \,p^{2} x^{2}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x}{c p \,x^{4}-c \,x^{4}+b p \,x^{2}-b \,x^{2}+a p -a}d x \right ) b c p \,x^{2}}{2 b \,x^{2} \left (p -1\right )} \] Input:

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

Output:

( - (a + b*x**2 + c*x**4)**p*b*p + (a + b*x**2 + c*x**4)**p*b - (a + b*x** 
2 + c*x**4)**p*c*x**2 + 2*int((a + b*x**2 + c*x**4)**p/(a*p*x - a*x + b*p* 
x**3 - b*x**3 + c*p*x**5 - c*x**5),x)*b**2*p**3*x**2 - 4*int((a + b*x**2 + 
 c*x**4)**p/(a*p*x - a*x + b*p*x**3 - b*x**3 + c*p*x**5 - c*x**5),x)*b**2* 
p**2*x**2 + 2*int((a + b*x**2 + c*x**4)**p/(a*p*x - a*x + b*p*x**3 - b*x** 
3 + c*p*x**5 - c*x**5),x)*b**2*p*x**2 + 4*int(((a + b*x**2 + c*x**4)**p*x* 
*3)/(a*p - a + b*p*x**2 - b*x**2 + c*p*x**4 - c*x**4),x)*c**2*p**2*x**2 - 
4*int(((a + b*x**2 + c*x**4)**p*x**3)/(a*p - a + b*p*x**2 - b*x**2 + c*p*x 
**4 - c*x**4),x)*c**2*p*x**2 + 4*int(((a + b*x**2 + c*x**4)**p*x)/(a*p - a 
 + b*p*x**2 - b*x**2 + c*p*x**4 - c*x**4),x)*b*c*p**3*x**2 - 6*int(((a + b 
*x**2 + c*x**4)**p*x)/(a*p - a + b*p*x**2 - b*x**2 + c*p*x**4 - c*x**4),x) 
*b*c*p**2*x**2 + 2*int(((a + b*x**2 + c*x**4)**p*x)/(a*p - a + b*p*x**2 - 
b*x**2 + c*p*x**4 - c*x**4),x)*b*c*p*x**2)/(2*b*x**2*(p - 1))