3.2160 \(\int (a+b x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx\)

Optimal. Leaf size=83 \[ \frac{(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac{e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \]

[Out]

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

_______________________________________________________________________________________

Rubi [A]  time = 0.135148, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103 \[ \frac{(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac{e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \]

Antiderivative was successfully verified.

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

[Out]

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

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 18.5941, size = 71, normalized size = 0.86 \[ \frac{\left (d + e x\right ) \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p + 1}}{b \left (2 p + 3\right )} - \frac{\left (a e - b d\right ) \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p + 1}}{2 b^{2} \left (p + 1\right ) \left (2 p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)*(e*x+d)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

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

_______________________________________________________________________________________

Mathematica [A]  time = 0.060216, size = 51, normalized size = 0.61 \[ \frac{\left ((a+b x)^2\right )^{p+1} (-a e+b d (2 p+3)+2 b e (p+1) x)}{2 b^2 (p+1) (2 p+3)} \]

Antiderivative was successfully verified.

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

[Out]

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

_______________________________________________________________________________________

Maple [A]  time = 0.005, size = 67, normalized size = 0.8 \[ -{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( -2\,bepx-2\,bdp-2\,bex+ae-3\,bd \right ) \left ( bx+a \right ) ^{2}}{2\,{b}^{2} \left ( 2\,{p}^{2}+5\,p+3 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x+a)*(e*x+d)*(b^2*x^2+2*a*b*x+a^2)^p,x)

[Out]

-1/2*(b^2*x^2+2*a*b*x+a^2)^p*(-2*b*e*p*x-2*b*d*p-2*b*e*x+a*e-3*b*d)*(b*x+a)^2/b^
2/(2*p^2+5*p+3)

_______________________________________________________________________________________

Maxima [A]  time = 0.744823, size = 278, normalized size = 3.35 \[ \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{2 \, p} a d}{b{\left (2 \, p + 1\right )}} + \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )}{\left (b x + a\right )}^{2 \, p} d}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b} + \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )}{\left (b x + a\right )}^{2 \, p} a e}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{3} +{\left (2 \, p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + a^{3}\right )}{\left (b x + a\right )}^{2 \, p} e}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)*(e*x + d)*(b^2*x^2 + 2*a*b*x + a^2)^p,x, algorithm="maxima")

[Out]

(b*x + a)*(b*x + a)^(2*p)*a*d/(b*(2*p + 1)) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x
- a^2)*(b*x + a)^(2*p)*d/((2*p^2 + 3*p + 1)*b) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b
*p*x - a^2)*(b*x + a)^(2*p)*a*e/((2*p^2 + 3*p + 1)*b^2) + ((2*p^2 + 3*p + 1)*b^3
*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a)^(2*p)*e/((4*p^3 + 12
*p^2 + 11*p + 3)*b^2)

_______________________________________________________________________________________

Fricas [A]  time = 0.298766, size = 192, normalized size = 2.31 \[ \frac{{\left (2 \, a^{2} b d p + 3 \, a^{2} b d - a^{3} e + 2 \,{\left (b^{3} e p + b^{3} e\right )} x^{3} +{\left (3 \, b^{3} d + 3 \, a b^{2} e + 2 \,{\left (b^{3} d + 2 \, a b^{2} e\right )} p\right )} x^{2} + 2 \,{\left (3 \, a b^{2} d +{\left (2 \, a b^{2} d + a^{2} b e\right )} p\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)*(e*x + d)*(b^2*x^2 + 2*a*b*x + a^2)^p,x, algorithm="fricas")

[Out]

1/2*(2*a^2*b*d*p + 3*a^2*b*d - a^3*e + 2*(b^3*e*p + b^3*e)*x^3 + (3*b^3*d + 3*a*
b^2*e + 2*(b^3*d + 2*a*b^2*e)*p)*x^2 + 2*(3*a*b^2*d + (2*a*b^2*d + a^2*b*e)*p)*x
)*(b^2*x^2 + 2*a*b*x + a^2)^p/(2*b^2*p^2 + 5*b^2*p + 3*b^2)

_______________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x+a)*(e*x+d)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Exception raised: TypeError

_______________________________________________________________________________________

GIAC/XCAS [A]  time = 0.287302, size = 509, normalized size = 6.13 \[ \frac{2 \, b^{3} p x^{3} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )} + 2 \, b^{3} d p x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} + 4 \, a b^{2} p x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )} + 2 \, b^{3} x^{3} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )} + 4 \, a b^{2} d p x e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} + 3 \, b^{3} d x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} + 2 \, a^{2} b p x e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )} + 3 \, a b^{2} x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )} + 2 \, a^{2} b d p e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} + 6 \, a b^{2} d x e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} + 3 \, a^{2} b d e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} - a^{3} e^{\left (p{\rm ln}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) + 1\right )}}{2 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)*(e*x + d)*(b^2*x^2 + 2*a*b*x + a^2)^p,x, algorithm="giac")

[Out]

1/2*(2*b^3*p*x^3*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2) + 1) + 2*b^3*d*p*x^2*e^(p*ln(b
^2*x^2 + 2*a*b*x + a^2)) + 4*a*b^2*p*x^2*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2) + 1) +
2*b^3*x^3*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2) + 1) + 4*a*b^2*d*p*x*e^(p*ln(b^2*x^2
+ 2*a*b*x + a^2)) + 3*b^3*d*x^2*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2)) + 2*a^2*b*p*x
*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2) + 1) + 3*a*b^2*x^2*e^(p*ln(b^2*x^2 + 2*a*b*x +
a^2) + 1) + 2*a^2*b*d*p*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2)) + 6*a*b^2*d*x*e^(p*ln
(b^2*x^2 + 2*a*b*x + a^2)) + 3*a^2*b*d*e^(p*ln(b^2*x^2 + 2*a*b*x + a^2)) - a^3*e
^(p*ln(b^2*x^2 + 2*a*b*x + a^2) + 1))/(2*b^2*p^2 + 5*b^2*p + 3*b^2)