3.18.57 \(\int (d+e x)^{-3-2 p} (a^2+2 a b x+b^2 x^2)^p \, dx\) [1757]

3.18.57.1 Optimal result

Integrand size = 30, antiderivative size = 115 \[ \int (d+e x)^{-3-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {b (a+b x) (d+e x)^{-1-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p}{2 (b d-a e)^2 (1+p) (1+2 p)}+\frac {(a+b x) (d+e x)^{-2 (1+p)} \left (a^2+2 a b x+b^2 x^2\right )^p}{2 (b d-a e) (1+p)} \]

output

1/2*b*(b*x+a)*(e*x+d)^(-1-2*p)*(b^2*x^2+2*a*b*x+a^2)^p/(-a*e+b*d)^2/(p+1)/ 
(1+2*p)+1/2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^p/(-a*e+b*d)/(p+1)/((e*x+d)^(2+2 
*p))
 

3.18.57.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.63 \[ \int (d+e x)^{-3-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^p (d+e x)^{-2 (1+p)} (2 b d (1+p)-a e (1+2 p)+b e x)}{2 (b d-a e)^2 (1+p) (1+2 p)} \]

input

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

output

((a + b*x)*((a + b*x)^2)^p*(2*b*d*(1 + p) - a*e*(1 + 2*p) + b*e*x))/(2*(b* 
d - a*e)^2*(1 + p)*(1 + 2*p)*(d + e*x)^(2*(1 + p)))
 

3.18.57.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1102, 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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

input

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

output

((a^2 + 2*a*b*x + b^2*x^2)^p*(((a*b + b^2*x)^(1 + 2*p)*(d + e*x)^(-1 - 2*p 
))/(2*(b*d - a*e)^2*(1 + p)*(1 + 2*p)) + (a*b + b^2*x)^(1 + 2*p)/(2*b*(b*d 
 - a*e)*(1 + p)*(d + e*x)^(2*(1 + p)))))/(a*b + b^2*x)^(2*p)
 

3.18.57.3.1 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 1102

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F 
racPart[p]))   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, 
 d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
 

3.18.57.4 Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (e x +d \right )^{-2-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} \left (2 a e p -2 b d p -b e x +a e -2 b d \right )}{2 \left (2 a^{2} e^{2} p^{2}-4 a b d e \,p^{2}+2 b^{2} d^{2} p^{2}+3 a^{2} e^{2} p -6 a b d e p +3 b^{2} d^{2} p +a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(139\)
parallelrisch	\(\frac {x^{3} \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} b^{3} e^{3}-2 x^{2} \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} e^{3} p +2 x^{2} \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} b^{3} d \,e^{2} p +3 x^{2} \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} b^{3} d \,e^{2}-2 x \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b \,e^{3} p +2 x \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} b^{3} d^{2} e p -x \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b \,e^{3}+2 x \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} d \,e^{2}+2 x \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} b^{3} d^{2} e -2 \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b d \,e^{2} p +2 \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} d^{2} e p -\left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b d \,e^{2}+2 \left (e x +d \right )^{-3-2 p} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} d^{2} e}{2 b e \left (2 a^{2} e^{2} p^{2}-4 a b d e \,p^{2}+2 b^{2} d^{2} p^{2}+3 a^{2} e^{2} p -6 a b d e p +3 b^{2} d^{2} p +a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(609\)

input

int((e*x+d)^(-3-2*p)*(b^2*x^2+2*a*b*x+a^2)^p,x,method=_RETURNVERBOSE)
 

output

-1/2*(b*x+a)*(e*x+d)^(-2-2*p)*(b^2*x^2+2*a*b*x+a^2)^p*(2*a*e*p-2*b*d*p-b*e 
*x+a*e-2*b*d)/(2*a^2*e^2*p^2-4*a*b*d*e*p^2+2*b^2*d^2*p^2+3*a^2*e^2*p-6*a*b 
*d*e*p+3*b^2*d^2*p+a^2*e^2-2*a*b*d*e+b^2*d^2)
 

3.18.57.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.90 \[ \int (d+e x)^{-3-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b^{2} e^{2} x^{3} + 2 \, a b d^{2} - a^{2} d e + {\left (3 \, b^{2} d e + 2 \, {\left (b^{2} d e - a b e^{2}\right )} p\right )} x^{2} + 2 \, {\left (a b d^{2} - a^{2} d e\right )} p + {\left (2 \, b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} + 2 \, {\left (b^{2} d^{2} - a^{2} e^{2}\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 3}}{2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2} + 2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} p^{2} + 3 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} p\right )}} \]

input

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

output

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

3.18.57.6 Sympy [F]

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

input

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

output

Integral((d + e*x)**(-2*p - 3)*((a + b*x)**2)**p, x)
 

3.18.57.7 Maxima [F]

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

input

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

output

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

3.18.57.8 Giac [F]

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

input

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

output

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

3.18.57.9 Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.25 \[ \int (d+e x)^{-3-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, dx={\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {x\,\left (2\,b^2\,d^2-a^2\,e^2-2\,a^2\,e^2\,p+2\,b^2\,d^2\,p+2\,a\,b\,d\,e\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}+\frac {b^2\,e^2\,x^3}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}-\frac {a\,d\,\left (a\,e-2\,b\,d+2\,a\,e\,p-2\,b\,d\,p\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}+\frac {b\,e\,x^2\,\left (3\,b\,d-2\,a\,e\,p+2\,b\,d\,p\right )}{2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{2\,p+3}\,\left (2\,p^2+3\,p+1\right )}\right ) \]

input

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

output

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