### 3.1757 $$\int (d+e x)^{-3-2 p} (a^2+2 a b x+b^2 x^2)^p \, dx$$

Optimal. Leaf size=115 $\frac{b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 p-1}}{2 (p+1) (2 p+1) (b d-a e)^2}+\frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)}$

[Out]

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

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Rubi [A]  time = 0.0482377, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {646, 45, 37} $\frac{b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 p-1}}{2 (p+1) (2 p+1) (b d-a e)^2}+\frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{-2 (p+1)}}{2 (p+1) (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(b*(a + b*x)*(d + e*x)^(-1 - 2*p)*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*(b*d - a*e)^2*(1 + p)*(1 + 2*p)) + ((a + b*x
)*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*(b*d - a*e)*(1 + p)*(d + e*x)^(2*(1 + 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 45

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] - Dist[(d*Simplify[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] && NeQ[b*c - a*d, 0] && 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] ||  !SumSimplerQ[n, 1])

Rule 37

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] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0399814, size = 72, normalized size = 0.63 $\frac{(a+b x) \left ((a+b x)^2\right )^p (d+e x)^{-2 (p+1)} (-a e (2 p+1)+2 b d (p+1)+b e x)}{2 (p+1) (2 p+1) (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((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)))

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Maple [A]  time = 0.044, size = 139, normalized size = 1.2 \begin{align*} -{\frac{ \left ( ex+d \right ) ^{-2-2\,p} \left ( bx+a \right ) \left ( 2\,aep-2\,bdp-bxe+ae-2\,bd \right ) \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{4\,{a}^{2}{e}^{2}{p}^{2}-8\,abde{p}^{2}+4\,{b}^{2}{d}^{2}{p}^{2}+6\,{a}^{2}{e}^{2}p-12\,abdep+6\,{b}^{2}{d}^{2}p+2\,{a}^{2}{e}^{2}-4\,abde+2\,{b}^{2}{d}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.92459, size = 451, normalized size = 3.92 \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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