\(\int (e x)^{-1+2 n} (a+b (c+d x^n)^2)^p \, dx\) [341]

Optimal result

Integrand size = 25, antiderivative size = 222 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\frac {x^{-2 n} (e x)^{2 n} \left (a+b c^2+2 b c d x^n+b d^2 x^{2 n}\right )^{1+p}}{2 b d^2 e n (1+p)}+\frac {2^p c x^{-2 n} (e x)^{2 n} \left (\frac {\sqrt {-a}-\sqrt {b} c-\sqrt {b} d x^n}{\sqrt {-a}}\right )^{-1-p} \left (a+b c^2+2 b c d x^n+b d^2 x^{2 n}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {\sqrt {-a}+\sqrt {b} c+\sqrt {b} d x^n}{2 \sqrt {-a}}\right )}{\sqrt {-a} \sqrt {b} d^2 e n (1+p)} \] Output:

1/2*(e*x)^(2*n)*(a+b*c^2+2*b*c*d*x^n+b*d^2*x^(2*n))^(p+1)/b/d^2/e/n/(p+1)/ 
(x^(2*n))+2^p*c*(e*x)^(2*n)*(((-a)^(1/2)-b^(1/2)*c-b^(1/2)*d*x^n)/(-a)^(1/ 
2))^(-1-p)*(a+b*c^2+2*b*c*d*x^n+b*d^2*x^(2*n))^(p+1)*hypergeom([-p, p+1],[ 
2+p],1/2*((-a)^(1/2)+b^(1/2)*c+b^(1/2)*d*x^n)/(-a)^(1/2))/(-a)^(1/2)/b^(1/ 
2)/d^2/e/n/(p+1)/(x^(2*n))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.18 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.84 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\frac {(e x)^{2 n} \left (\frac {\sqrt {-a b d^2}-b d \left (c+d x^n\right )}{-b c d+\sqrt {-a b d^2}}\right )^{-p} \left (\frac {\sqrt {-a b d^2}+b d \left (c+d x^n\right )}{b c d+\sqrt {-a b d^2}}\right )^{-p} \left (a+b \left (c+d x^n\right )^2\right )^p \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {b d^2 x^n}{b c d+\sqrt {-a b d^2}},\frac {b d^2 x^n}{-b c d+\sqrt {-a b d^2}}\right )}{2 e n} \] Input:

Integrate[(e*x)^(-1 + 2*n)*(a + b*(c + d*x^n)^2)^p,x]
 

Output:

((e*x)^(2*n)*(a + b*(c + d*x^n)^2)^p*AppellF1[2, -p, -p, 3, -((b*d^2*x^n)/ 
(b*c*d + Sqrt[-(a*b*d^2)])), (b*d^2*x^n)/(-(b*c*d) + Sqrt[-(a*b*d^2)])])/( 
2*e*n*((Sqrt[-(a*b*d^2)] - b*d*(c + d*x^n))/(-(b*c*d) + Sqrt[-(a*b*d^2)])) 
^p*((Sqrt[-(a*b*d^2)] + b*d*(c + d*x^n))/(b*c*d + Sqrt[-(a*b*d^2)]))^p)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2086, 1694, 1693, 1160, 1096}

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.

\(\displaystyle \int (e x)^{2 n-1} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx\)

\(\Big \downarrow \) 2086

\(\displaystyle \int (e x)^{2 n-1} \left (a+b c^2+2 b c d x^n+b d^2 x^{2 n}\right )^pdx\)

\(\Big \downarrow \) 1694

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int x^{2 n-1} \left (2 b c d x^n+b d^2 x^{2 n}+b c^2+a\right )^pdx}{e}\)

\(\Big \downarrow \) 1693

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int x^n \left (2 b c d x^n+b d^2 x^{2 n}+b c^2+a\right )^pdx^n}{e n}\)

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1096

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

Input:

Int[(e*x)^(-1 + 2*n)*(a + b*(c + d*x^n)^2)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 1096

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 
 - 4*a*c, 2]}, Simp[(-(a + b*x + c*x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x) 
/(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q) 
], x]] /; FreeQ[{a, b, c, p}, x] &&  !IntegerQ[4*p] &&  !IntegerQ[3*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1693

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, 
x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ 
[Simplify[(m + 1)/n]]
 

rule 1694

Int[((d_)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x 
_Symbol] :> Simp[d^IntPart[m]*((d*x)^FracPart[m]/x^FracPart[m])   Int[x^m*( 
a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[ 
n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2086

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, 
x]^p, x] /; FreeQ[{d, m, p}, x] && TrinomialQ[u, x] &&  !TrinomialMatchQ[u, 
 x]
 

Maple [F]

\[\int \left (e x \right )^{-1+2 n} {\left (a +b \left (c +d \,x^{n}\right )^{2}\right )}^{p}d x\]

Input:

int((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x)
 

Output:

int((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x)
 

Fricas [F]

\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\int { {\left ({\left (d x^{n} + c\right )}^{2} b + a\right )}^{p} \left (e x\right )^{2 \, n - 1} \,d x } \] Input:

integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*d^2*x^(2*n) + 2*b*c*d*x^n + b*c^2 + a)^p*(e*x)^(2*n - 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((e*x)**(-1+2*n)*(a+b*(c+d*x**n)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\int { {\left ({\left (d x^{n} + c\right )}^{2} b + a\right )}^{p} \left (e x\right )^{2 \, n - 1} \,d x } \] Input:

integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x, algorithm="maxima")
 

Output:

integrate(((d*x^n + c)^2*b + a)^p*(e*x)^(2*n - 1), x)
 

Giac [F]

\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\int { {\left ({\left (d x^{n} + c\right )}^{2} b + a\right )}^{p} \left (e x\right )^{2 \, n - 1} \,d x } \] Input:

integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x, algorithm="giac")
 

Output:

integrate(((d*x^n + c)^2*b + a)^p*(e*x)^(2*n - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\int {\left (e\,x\right )}^{2\,n-1}\,{\left (a+b\,{\left (c+d\,x^n\right )}^2\right )}^p \,d x \] Input:

int((e*x)^(2*n - 1)*(a + b*(c + d*x^n)^2)^p,x)
 

Output:

int((e*x)^(2*n - 1)*(a + b*(c + d*x^n)^2)^p, x)
 

Reduce [F]

\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )^2\right )^p \, dx=\frac {e^{2 n} \left (2 x^{2 n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p} b \,d^{2} p +x^{2 n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p} b \,d^{2}+2 x^{n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p} b c d p -\left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p} a -\left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p} b \,c^{2}+8 \left (\int \frac {x^{2 n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p}}{2 x^{2 n} b \,d^{2} p x +x^{2 n} b \,d^{2} x +4 x^{n} b c d p x +2 x^{n} b c d x +2 a p x +a x +2 b \,c^{2} p x +b \,c^{2} x}d x \right ) a b \,d^{2} n \,p^{3}+12 \left (\int \frac {x^{2 n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p}}{2 x^{2 n} b \,d^{2} p x +x^{2 n} b \,d^{2} x +4 x^{n} b c d p x +2 x^{n} b c d x +2 a p x +a x +2 b \,c^{2} p x +b \,c^{2} x}d x \right ) a b \,d^{2} n \,p^{2}+4 \left (\int \frac {x^{2 n} \left (x^{2 n} b \,d^{2}+2 x^{n} b c d +a +b \,c^{2}\right )^{p}}{2 x^{2 n} b \,d^{2} p x +x^{2 n} b \,d^{2} x +4 x^{n} b c d p x +2 x^{n} b c d x +2 a p x +a x +2 b \,c^{2} p x +b \,c^{2} x}d x \right ) a b \,d^{2} n p \right )}{2 b \,d^{2} e n \left (2 p^{2}+3 p +1\right )} \] Input:

int((e*x)^(-1+2*n)*(a+b*(c+d*x^n)^2)^p,x)
 

Output:

(e**(2*n)*(2*x**(2*n)*(x**(2*n)*b*d**2 + 2*x**n*b*c*d + a + b*c**2)**p*b*d 
**2*p + x**(2*n)*(x**(2*n)*b*d**2 + 2*x**n*b*c*d + a + b*c**2)**p*b*d**2 + 
 2*x**n*(x**(2*n)*b*d**2 + 2*x**n*b*c*d + a + b*c**2)**p*b*c*d*p - (x**(2* 
n)*b*d**2 + 2*x**n*b*c*d + a + b*c**2)**p*a - (x**(2*n)*b*d**2 + 2*x**n*b* 
c*d + a + b*c**2)**p*b*c**2 + 8*int((x**(2*n)*(x**(2*n)*b*d**2 + 2*x**n*b* 
c*d + a + b*c**2)**p)/(2*x**(2*n)*b*d**2*p*x + x**(2*n)*b*d**2*x + 4*x**n* 
b*c*d*p*x + 2*x**n*b*c*d*x + 2*a*p*x + a*x + 2*b*c**2*p*x + b*c**2*x),x)*a 
*b*d**2*n*p**3 + 12*int((x**(2*n)*(x**(2*n)*b*d**2 + 2*x**n*b*c*d + a + b* 
c**2)**p)/(2*x**(2*n)*b*d**2*p*x + x**(2*n)*b*d**2*x + 4*x**n*b*c*d*p*x + 
2*x**n*b*c*d*x + 2*a*p*x + a*x + 2*b*c**2*p*x + b*c**2*x),x)*a*b*d**2*n*p* 
*2 + 4*int((x**(2*n)*(x**(2*n)*b*d**2 + 2*x**n*b*c*d + a + b*c**2)**p)/(2* 
x**(2*n)*b*d**2*p*x + x**(2*n)*b*d**2*x + 4*x**n*b*c*d*p*x + 2*x**n*b*c*d* 
x + 2*a*p*x + a*x + 2*b*c**2*p*x + b*c**2*x),x)*a*b*d**2*n*p))/(2*b*d**2*e 
*n*(2*p**2 + 3*p + 1))