\(\int (d+e x)^{-4-2 p} (d^2-e^2 x^2)^p \, dx\) [380]

Optimal result

Integrand size = 26, antiderivative size = 142 \[ \int (d+e x)^{-4-2 p} \left (d^2-e^2 x^2\right )^p \, dx=-\frac {(d+e x)^{-3-2 p} \left (d^2-e^2 x^2\right )^{1+p}}{2 d^2 e (2+p) (3+p)}-\frac {(d+e x)^{-2 (1+p)} \left (d^2-e^2 x^2\right )^{1+p}}{4 d^3 e (1+p) (2+p) (3+p)}-\frac {(d+e x)^{-2 (2+p)} \left (d^2-e^2 x^2\right )^{1+p}}{2 d e (3+p)} \] Output:

-1/2*(e*x+d)^(-3-2*p)*(-e^2*x^2+d^2)^(p+1)/d^2/e/(2+p)/(3+p)-1/4*(-e^2*x^2 
+d^2)^(p+1)/d^3/e/(p+1)/(2+p)/(3+p)/((e*x+d)^(2*p+2))-1/2*(-e^2*x^2+d^2)^( 
p+1)/d/e/(3+p)/((e*x+d)^(4+2*p))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.58 \[ \int (d+e x)^{-4-2 p} \left (d^2-e^2 x^2\right )^p \, dx=-\frac {2^{-4-p} (d-e x) (d+e x)^{-2 p} \left (1+\frac {e x}{d}\right )^p \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1+p,4+p,2+p,\frac {d-e x}{2 d}\right )}{d^4 e (1+p)} \] Input:

Integrate[(d + e*x)^(-4 - 2*p)*(d^2 - e^2*x^2)^p,x]
 

Output:

-((2^(-4 - p)*(d - e*x)*(1 + (e*x)/d)^p*(d^2 - e^2*x^2)^p*Hypergeometric2F 
1[1 + p, 4 + p, 2 + p, (d - e*x)/(2*d)])/(d^4*e*(1 + p)*(d + e*x)^(2*p)))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {474, 473, 55, 55, 48}

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.

\(\displaystyle \int (d+e x)^{-2 p-4} \left (d^2-e^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {(d+e x)^{-2 p} \left (\frac {e x}{d}+1\right )^{2 p} \int \left (\frac {e x}{d}+1\right )^{-2 (p+2)} \left (d^2-e^2 x^2\right )^pdx}{d^4}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {(d+e x)^{-2 p} \left (\frac {e x}{d}+1\right )^{p-1} \left (d^2-d e x\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \int \left (\frac {e x}{d}+1\right )^{-p-4} \left (d^2-d e x\right )^pdx}{d^4}\)

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

\(\displaystyle \frac {(d+e x)^{-2 p} \left (\frac {e x}{d}+1\right )^{p-1} \left (d^2-d e x\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \left (\frac {-\frac {\left (d^2-d e x\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-2}}{2 d e (p+2)}-\frac {\left (d^2-d e x\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-1}}{4 d e (p+1) (p+2)}}{p+3}-\frac {\left (\frac {e x}{d}+1\right )^{-p-3} \left (d^2-d e x\right )^{p+1}}{2 d e (p+3)}\right )}{d^4}\)

Input:

Int[(d + e*x)^(-4 - 2*p)*(d^2 - e^2*x^2)^p,x]
 

Output:

((1 + (e*x)/d)^(-1 + p)*(d^2 - d*e*x)^(-1 - p)*(d^2 - e^2*x^2)^(1 + p)*(-1 
/2*((1 + (e*x)/d)^(-3 - p)*(d^2 - d*e*x)^(1 + p))/(d*e*(3 + p)) + (-1/2*(( 
1 + (e*x)/d)^(-2 - p)*(d^2 - d*e*x)^(1 + p))/(d*e*(2 + p)) - ((1 + (e*x)/d 
)^(-1 - p)*(d^2 - d*e*x)^(1 + p))/(4*d*e*(1 + p)*(2 + p)))/(3 + p)))/(d^4* 
(d + e*x)^(2*p))
 

Defintions of rubi rules used

rule 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n])   Int[(1 + d 
*(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + 
 a*d^2, 0] &&  !(IntegerQ[n] || GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{-3-2 p} \left (-e^{2} x^{2}+d^{2}\right )^{p} \left (2 d^{2} p^{2}+2 d e p x +e^{2} x^{2}+8 d^{2} p +4 d e x +7 d^{2}\right )}{4 d^{3} e \left (2+p \right ) \left (p^{2}+4 p +3\right )}\)	\(93\)
orering	\(-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{p} \left (e x +d \right )^{-4-2 p} \left (2 d^{2} p^{2}+2 d e p x +e^{2} x^{2}+8 d^{2} p +4 d e x +7 d^{2}\right ) \left (e x +d \right ) \left (-e x +d \right )}{4 e \left (p^{2}+4 p +3\right ) \left (2+p \right ) d^{3}}\)	\(98\)

Input:

int((e*x+d)^(-4-2*p)*(-e^2*x^2+d^2)^p,x,method=_RETURNVERBOSE)
 

Output:

-1/4/d^3/e/(2+p)*(-e*x+d)*(e*x+d)^(-3-2*p)/(p^2+4*p+3)*(-e^2*x^2+d^2)^p*(2 
*d^2*p^2+2*d*e*p*x+e^2*x^2+8*d^2*p+4*d*e*x+7*d^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int (d+e x)^{-4-2 p} \left (d^2-e^2 x^2\right )^p \, dx=\frac {{\left (e^{4} x^{4} - 2 \, d^{4} p^{2} - 8 \, d^{4} p - 7 \, d^{4} + 2 \, {\left (d e^{3} p + 2 \, d e^{3}\right )} x^{3} + 2 \, {\left (d^{2} e^{2} p^{2} + 4 \, d^{2} e^{2} p + 3 \, d^{2} e^{2}\right )} x^{2} - 2 \, {\left (d^{3} e p + 2 \, d^{3} e\right )} x\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4}}{4 \, {\left (d^{3} e p^{3} + 6 \, d^{3} e p^{2} + 11 \, d^{3} e p + 6 \, d^{3} e\right )}} \] Input:

integrate((e*x+d)^(-4-2*p)*(-e^2*x^2+d^2)^p,x, algorithm="fricas")
 

Output:

1/4*(e^4*x^4 - 2*d^4*p^2 - 8*d^4*p - 7*d^4 + 2*(d*e^3*p + 2*d*e^3)*x^3 + 2 
*(d^2*e^2*p^2 + 4*d^2*e^2*p + 3*d^2*e^2)*x^2 - 2*(d^3*e*p + 2*d^3*e)*x)*(- 
e^2*x^2 + d^2)^p*(e*x + d)^(-2*p - 4)/(d^3*e*p^3 + 6*d^3*e*p^2 + 11*d^3*e* 
p + 6*d^3*e)
 

Sympy [F]

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

integrate((e*x+d)**(-4-2*p)*(-e**2*x**2+d**2)**p,x)
 

Output:

Integral((-(-d + e*x)*(d + e*x))**p*(d + e*x)**(-2*p - 4), x)
 

Maxima [F]

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

integrate((e*x+d)^(-4-2*p)*(-e^2*x^2+d^2)^p,x, algorithm="maxima")
 

Output:

integrate((-e^2*x^2 + d^2)^p*(e*x + d)^(-2*p - 4), x)
 

Giac [F]

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

integrate((e*x+d)^(-4-2*p)*(-e^2*x^2+d^2)^p,x, algorithm="giac")
 

Output:

integrate((-e^2*x^2 + d^2)^p*(e*x + d)^(-2*p - 4), x)
 

Mupad [B] (verification not implemented)

Time = 6.60 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.55 \[ \int (d+e x)^{-4-2 p} \left (d^2-e^2 x^2\right )^p \, dx={\left (d^2-e^2\,x^2\right )}^p\,\left (\frac {e^3\,x^4}{4\,d^3\,{\left (d+e\,x\right )}^{2\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {d\,\left (2\,p^2+8\,p+7\right )}{4\,e\,{\left (d+e\,x\right )}^{2\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {x\,\left (p+2\right )}{2\,{\left (d+e\,x\right )}^{2\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {e\,x^2\,\left (p^2+4\,p+3\right )}{2\,d\,{\left (d+e\,x\right )}^{2\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {e^2\,x^3\,\left (p+2\right )}{2\,d^2\,{\left (d+e\,x\right )}^{2\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:

int((d^2 - e^2*x^2)^p/(d + e*x)^(2*p + 4),x)
 

Output:

(d^2 - e^2*x^2)^p*((e^3*x^4)/(4*d^3*(d + e*x)^(2*p + 4)*(11*p + 6*p^2 + p^ 
3 + 6)) - (d*(8*p + 2*p^2 + 7))/(4*e*(d + e*x)^(2*p + 4)*(11*p + 6*p^2 + p 
^3 + 6)) - (x*(p + 2))/(2*(d + e*x)^(2*p + 4)*(11*p + 6*p^2 + p^3 + 6)) + 
(e*x^2*(4*p + p^2 + 3))/(2*d*(d + e*x)^(2*p + 4)*(11*p + 6*p^2 + p^3 + 6)) 
 + (e^2*x^3*(p + 2))/(2*d^2*(d + e*x)^(2*p + 4)*(11*p + 6*p^2 + p^3 + 6)))
 

Reduce [F]

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

int((e*x+d)^(-4-2*p)*(-e^2*x^2+d^2)^p,x)
 

Output:

int((d**2 - e**2*x**2)**p/((d + e*x)**(2*p)*d**4 + 4*(d + e*x)**(2*p)*d**3 
*e*x + 6*(d + e*x)**(2*p)*d**2*e**2*x**2 + 4*(d + e*x)**(2*p)*d*e**3*x**3 
+ (d + e*x)**(2*p)*e**4*x**4),x)