\(\int (e x)^{-3-4 p} (a+b x^2+c x^4)^p \, dx\) [1107]

Optimal result

Integrand size = 24, antiderivative size = 179 \[ \int (e x)^{-3-4 p} \left (a+b x^2+c x^4\right )^p \, dx=-\frac {(e x)^{-2-4 p} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right ) \left (\frac {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}{1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,\frac {\sqrt {b^2-4 a c} x^2}{a \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}\right )}{2 e (1+2 p)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04 \[ \int (e x)^{-3-4 p} \left (a+b x^2+c x^4\right )^p \, dx=\frac {(e x)^{-4 p} \left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,\frac {2 \sqrt {b^2-4 a c} x^2}{2 a+\left (b+\sqrt {b^2-4 a c}\right ) x^2}\right )}{2 \left (-b+\sqrt {b^2-4 a c}\right ) e^3 (1+2 p) x^2} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1461, 394}

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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1461

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + R 
t[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2])))^F 
racPart[p]))   Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2 
*c*(x^2/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^{-3-4 p} \left (a+b x^2+c x^4\right )^p \, dx=\frac {-\left (c \,x^{4}+b \,x^{2}+a \right )^{p} b -2 \left (c \,x^{4}+b \,x^{2}+a \right )^{p} c \,x^{2}-16 x^{4 p} \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{2 x^{4 p} a p x +x^{4 p} a x +2 x^{4 p} b p \,x^{3}+x^{4 p} b \,x^{3}+2 x^{4 p} c p \,x^{5}+x^{4 p} c \,x^{5}}d x \right ) a c \,p^{2} x^{2}-8 x^{4 p} \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{2 x^{4 p} a p x +x^{4 p} a x +2 x^{4 p} b p \,x^{3}+x^{4 p} b \,x^{3}+2 x^{4 p} c p \,x^{5}+x^{4 p} c \,x^{5}}d x \right ) a c p \,x^{2}+4 x^{4 p} \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{2 x^{4 p} a p x +x^{4 p} a x +2 x^{4 p} b p \,x^{3}+x^{4 p} b \,x^{3}+2 x^{4 p} c p \,x^{5}+x^{4 p} c \,x^{5}}d x \right ) b^{2} p^{2} x^{2}+2 x^{4 p} \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{2 x^{4 p} a p x +x^{4 p} a x +2 x^{4 p} b p \,x^{3}+x^{4 p} b \,x^{3}+2 x^{4 p} c p \,x^{5}+x^{4 p} c \,x^{5}}d x \right ) b^{2} p \,x^{2}}{2 x^{4 p} e^{4 p} b \,e^{3} x^{2} \left (2 p +1\right )} \] Input:

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

Output:

( - (a + b*x**2 + c*x**4)**p*b - 2*(a + b*x**2 + c*x**4)**p*c*x**2 - 16*x* 
*(4*p)*int((a + b*x**2 + c*x**4)**p/(2*x**(4*p)*a*p*x + x**(4*p)*a*x + 2*x 
**(4*p)*b*p*x**3 + x**(4*p)*b*x**3 + 2*x**(4*p)*c*p*x**5 + x**(4*p)*c*x**5 
),x)*a*c*p**2*x**2 - 8*x**(4*p)*int((a + b*x**2 + c*x**4)**p/(2*x**(4*p)*a 
*p*x + x**(4*p)*a*x + 2*x**(4*p)*b*p*x**3 + x**(4*p)*b*x**3 + 2*x**(4*p)*c 
*p*x**5 + x**(4*p)*c*x**5),x)*a*c*p*x**2 + 4*x**(4*p)*int((a + b*x**2 + c* 
x**4)**p/(2*x**(4*p)*a*p*x + x**(4*p)*a*x + 2*x**(4*p)*b*p*x**3 + x**(4*p) 
*b*x**3 + 2*x**(4*p)*c*p*x**5 + x**(4*p)*c*x**5),x)*b**2*p**2*x**2 + 2*x** 
(4*p)*int((a + b*x**2 + c*x**4)**p/(2*x**(4*p)*a*p*x + x**(4*p)*a*x + 2*x* 
*(4*p)*b*p*x**3 + x**(4*p)*b*x**3 + 2*x**(4*p)*c*p*x**5 + x**(4*p)*c*x**5) 
,x)*b**2*p*x**2)/(2*x**(4*p)*e**(4*p)*b*e**3*x**2*(2*p + 1))