### 3.1746 $$\int (d+e x) (a^2+2 a b x+b^2 x^2)^p \, dx$$

Optimal. Leaf size=76 $\frac{(a+b x) (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+1)}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b^2 (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0238942, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {640, 609} $\frac{(a+b x) (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+1)}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b^2 (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 640

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

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

Mathematica [A]  time = 0.0340394, size = 54, normalized size = 0.71 $\frac{(a+b x) \left ((a+b x)^2\right )^p (-a e+2 b d (p+1)+b e (2 p+1) x)}{2 b^2 (p+1) (2 p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 65, normalized size = 0.9 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( -2\,bepx-2\,bdp-bxe+ae-2\,bd \right ) \left ( bx+a \right ) }{2\,{b}^{2} \left ( 2\,{p}^{2}+3\,p+1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.08004, size = 105, normalized size = 1.38 \begin{align*} \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{2 \, p} 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} e}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)*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)*e/((2*p^
2 + 3*p + 1)*b^2)

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Fricas [A]  time = 1.68205, size = 204, normalized size = 2.68 \begin{align*} \frac{{\left (2 \, a b d p + 2 \, a b d - a^{2} e +{\left (2 \, b^{2} e p + b^{2} e\right )} x^{2} + 2 \,{\left (b^{2} d +{\left (b^{2} d + a 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} + 3 \, b^{2} p + b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.15629, size = 308, normalized size = 4.05 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} p x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d p x + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b p x e +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} x^{2} e + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d p + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d x + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d -{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} e}{2 \,{\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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