3.1745 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=127 \[ \frac{e (a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (p+1)}+\frac{(a+b x) (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+1)}+\frac{e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0611493, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {646, 43} \[ \frac{e (a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (p+1)}+\frac{(a+b x) (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+1)}+\frac{e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[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] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx &=\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (a b+b^2 x\right )^{2 p} (d+e x)^2 \, dx\\ &=\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (\frac{(b d-a e)^2 \left (a b+b^2 x\right )^{2 p}}{b^2}+\frac{2 e (b d-a e) \left (a b+b^2 x\right )^{1+2 p}}{b^3}+\frac{e^2 \left (a b+b^2 x\right )^{2+2 p}}{b^4}\right ) \, dx\\ &=\frac{(b d-a e)^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+2 p)}+\frac{e (b d-a e) (a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+p)}+\frac{e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (3+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0782135, size = 75, normalized size = 0.59 \[ \frac{(a+b x) \left ((a+b x)^2\right )^p \left (\frac{e (a+b x) (b d-a e)}{p+1}+\frac{(b d-a e)^2}{2 p+1}+\frac{e^2 (a+b x)^2}{2 p+3}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 175, normalized size = 1.4 \begin{align*}{\frac{ \left ( 2\,{b}^{2}{e}^{2}{p}^{2}{x}^{2}+4\,{b}^{2}de{p}^{2}x+3\,{b}^{2}{e}^{2}p{x}^{2}-2\,ab{e}^{2}px+2\,{b}^{2}{d}^{2}{p}^{2}+8\,{b}^{2}depx+{e}^{2}{x}^{2}{b}^{2}-2\,abdep-ab{e}^{2}x+5\,{b}^{2}{d}^{2}p+3\,x{b}^{2}de+{a}^{2}{e}^{2}-3\,abde+3\,{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{{b}^{3} \left ( 4\,{p}^{3}+12\,{p}^{2}+11\,p+3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)*(2*b^2*e^2*p^2*x^2+4*b^2*d*e*p^2*x+3*b^2*e^2*p*x^2-2*a*b*e^2*p*x+2*b^2*d^2*p^2+8*b^2*d*e*p*x+b^2*e^2*x
^2-2*a*b*d*e*p-a*b*e^2*x+5*b^2*d^2*p+3*b^2*d*e*x+a^2*e^2-3*a*b*d*e+3*b^2*d^2)*(b^2*x^2+2*a*b*x+a^2)^p/b^3/(4*p
^3+12*p^2+11*p+3)

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Maxima [A]  time = 1.12736, size = 212, normalized size = 1.67 \begin{align*} \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{2 \, p} d^{2}}{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 e}{{\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^{2}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65259, size = 509, normalized size = 4.01 \begin{align*} \frac{{\left (2 \, a b^{2} d^{2} p^{2} + 3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2} +{\left (2 \, b^{3} e^{2} p^{2} + 3 \, b^{3} e^{2} p + b^{3} e^{2}\right )} x^{3} +{\left (3 \, b^{3} d e + 2 \,{\left (2 \, b^{3} d e + a b^{2} e^{2}\right )} p^{2} +{\left (8 \, b^{3} d e + a b^{2} e^{2}\right )} p\right )} x^{2} +{\left (5 \, a b^{2} d^{2} - 2 \, a^{2} b d e\right )} p +{\left (3 \, b^{3} d^{2} + 2 \,{\left (b^{3} d^{2} + 2 \, a b^{2} d e\right )} p^{2} +{\left (5 \, b^{3} d^{2} + 6 \, a b^{2} d e - 2 \, a^{2} b e^{2}\right )} p\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*b^2*d^2*p^2 + 3*a*b^2*d^2 - 3*a^2*b*d*e + a^3*e^2 + (2*b^3*e^2*p^2 + 3*b^3*e^2*p + b^3*e^2)*x^3 + (3*b^3*
d*e + 2*(2*b^3*d*e + a*b^2*e^2)*p^2 + (8*b^3*d*e + a*b^2*e^2)*p)*x^2 + (5*a*b^2*d^2 - 2*a^2*b*d*e)*p + (3*b^3*
d^2 + 2*(b^3*d^2 + 2*a*b^2*d*e)*p^2 + (5*b^3*d^2 + 6*a*b^2*d*e - 2*a^2*b*e^2)*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^
p/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.14645, size = 821, normalized size = 6.46 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p^{2} x^{3} e^{2} + 4 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p^{2} x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p^{2} x + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p^{2} x^{2} e^{2} + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p x^{3} e^{2} + 4 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p^{2} x e + 8 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p^{2} + 5 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p x +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p x^{2} e^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} x^{3} e^{2} + 6 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p x e + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d x^{2} e + 5 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} x - 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b p x e^{2} - 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d p e + 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} - 3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d e +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{3} e^{2}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*p^2*x^3*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p^2*x^2*e + 2*(b^2*x^2 +
2*a*b*x + a^2)^p*b^3*d^2*p^2*x + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*p^2*x^2*e^2 + 3*(b^2*x^2 + 2*a*b*x + a^2)
^p*b^3*p*x^3*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p^2*x*e + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p*x^2*e
+ 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2*p^2 + 5*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d^2*p*x + (b^2*x^2 + 2*a*b*
x + a^2)^p*a*b^2*p*x^2*e^2 + (b^2*x^2 + 2*a*b*x + a^2)^p*b^3*x^3*e^2 + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p
*x*e + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*x^2*e + 5*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2*p + 3*(b^2*x^2 + 2*
a*b*x + a^2)^p*b^3*d^2*x - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*p*x*e^2 - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d
*p*e + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2 - 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*e + (b^2*x^2 + 2*a*b*x
+ a^2)^p*a^3*e^2)/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)