3.1101 \(\int (d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)^p \, dx\)

Optimal. Leaf size=43 \[ \frac{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p+1}}{c e (2 p+3)} \]

[Out]

((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(1 + p))/(c*e*(3 + 2*p))

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Rubi [A]  time = 0.0259138, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {642, 609} \[ \frac{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p+1}}{c e (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(1 + p))/(c*e*(3 + 2*p))

Rule 642

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[
2*c*d - b*e, 0] && IntegerQ[m/2]

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

Mathematica [A]  time = 0.0185436, size = 32, normalized size = 0.74 \[ \frac{(d+e x) \left (c (d+e x)^2\right )^{p+1}}{c e (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*(c*(d + e*x)^2)^(1 + p))/(c*e*(3 + 2*p))

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Maple [A]  time = 0.039, size = 41, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{3} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{e \left ( 3+2\,p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x+d)^3/e/(3+2*p)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p

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Maxima [B]  time = 1.29265, size = 246, normalized size = 5.72 \begin{align*} \frac{{\left (c^{p} e x + c^{p} d\right )}{\left (e x + d\right )}^{2 \, p} d^{2}}{e{\left (2 \, p + 1\right )}} + \frac{{\left (c^{p} e^{2}{\left (2 \, p + 1\right )} x^{2} + 2 \, c^{p} d e p x - c^{p} d^{2}\right )}{\left (e x + d\right )}^{2 \, p} d}{{\left (2 \, p^{2} + 3 \, p + 1\right )} e} + \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} c^{p} e^{3} x^{3} +{\left (2 \, p^{2} + p\right )} c^{p} d e^{2} x^{2} - 2 \, c^{p} d^{2} e p x + c^{p} d^{3}\right )}{\left (e x + d\right )}^{2 \, p}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.61892, size = 123, normalized size = 2.86 \begin{align*} \frac{{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, e p + 3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p/(2*e*p + 3*e)

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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*(c*e**2*x**2+2*c*d*e*x+c*d**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.21143, size = 173, normalized size = 4.02 \begin{align*} \frac{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} x^{3} e^{3} + 3 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d x^{2} e^{2} + 3 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d^{2} x e +{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d^{3}}{2 \, p e + 3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*x^3*e^3 + 3*(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*d*x^2*e^2 + 3*(c*x^2*e^2 + 2*
c*d*x*e + c*d^2)^p*d^2*x*e + (c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*d^3)/(2*p*e + 3*e)