3.1100 \(\int (d+e x)^3 (c d^2+2 c d e x+c e^2 x^2)^p \, dx\)

Optimal. Leaf size=39 \[ \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{p+2}}{2 c^2 e (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.0256012, antiderivative size = 39, 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 = {643, 629} \[ \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{p+2}}{2 c^2 e (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p,x]

[Out]

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

Rule 643

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/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 - 1)/2]

Rule 629

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

Rubi steps

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

Mathematica [A]  time = 0.0166515, size = 30, normalized size = 0.77 \[ \frac{(d+e x)^4 \left (c (d+e x)^2\right )^p}{2 e (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p,x]

[Out]

((d + e*x)^4*(c*(d + e*x)^2)^p)/(2*e*(2 + p))

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Maple [A]  time = 0.04, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{4} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{2\,e \left ( 2+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e*x+d)^4/e/(2+p)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p

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Maxima [B]  time = 1.97006, size = 423, normalized size = 10.85 \begin{align*} \frac{{\left (c^{p} e x + c^{p} d\right )}{\left (e x + d\right )}^{2 \, p} d^{3}}{e{\left (2 \, p + 1\right )}} + \frac{3 \,{\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^{2}}{2 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} e} + \frac{3 \,{\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} d}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} e} + \frac{{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} c^{p} e^{4} x^{4} + 2 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} c^{p} d e^{3} x^{3} - 3 \,{\left (2 \, p^{2} + p\right )} c^{p} d^{2} e^{2} x^{2} + 6 \, c^{p} d^{3} e p x - 3 \, c^{p} d^{4}\right )}{\left (e x + d\right )}^{2 \, p}}{2 \,{\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(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^3/(e*(2*p + 1)) + 3/2*(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/((2*p^2 + 3*p + 1)*e) + 3*((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)*d/((4*p^3 + 12*p^2 + 11*p + 3)*e) + 1/2*((4*p^3 + 12*p^2 + 11*p + 3)*c^p
*e^4*x^4 + 2*(2*p^3 + 3*p^2 + p)*c^p*d*e^3*x^3 - 3*(2*p^2 + p)*c^p*d^2*e^2*x^2 + 6*c^p*d^3*e*p*x - 3*c^p*d^4)*
(e*x + d)^(2*p)/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*e)

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Fricas [A]  time = 2.29006, size = 147, normalized size = 3.77 \begin{align*} \frac{{\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \,{\left (e p + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p/(e*p + 2*e)

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Sympy [A]  time = 1.06861, size = 233, normalized size = 5.97 \begin{align*} \begin{cases} \frac{x}{c^{2} d} & \text{for}\: e = 0 \wedge p = -2 \\d^{3} x \left (c d^{2}\right )^{p} & \text{for}\: e = 0 \\\frac{\log{\left (\frac{d}{e} + x \right )}}{c^{2} e} & \text{for}\: p = -2 \\\frac{d^{4} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac{4 d^{3} e x \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac{6 d^{2} e^{2} x^{2} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac{4 d e^{3} x^{3} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac{e^{4} x^{4} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(c*e**2*x**2+2*c*d*e*x+c*d**2)**p,x)

[Out]

Piecewise((x/(c**2*d), Eq(e, 0) & Eq(p, -2)), (d**3*x*(c*d**2)**p, Eq(e, 0)), (log(d/e + x)/(c**2*e), Eq(p, -2
)), (d**4*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 4*e) + 4*d**3*e*x*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)*
*p/(2*e*p + 4*e) + 6*d**2*e**2*x**2*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 4*e) + 4*d*e**3*x**3*(c*d**
2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 4*e) + e**4*x**4*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 4*e),
True))

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Giac [B]  time = 1.28653, size = 216, normalized size = 5.54 \begin{align*} \frac{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} x^{4} e^{4} + 4 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d x^{3} e^{3} + 6 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d^{2} x^{2} e^{2} + 4 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d^{3} x e +{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} d^{4}}{2 \,{\left (p e + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*x^4*e^4 + 4*(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*d*x^3*e^3 + 6*(c*x^2*e^2
+ 2*c*d*x*e + c*d^2)^p*d^2*x^2*e^2 + 4*(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*d^3*x*e + (c*x^2*e^2 + 2*c*d*x*e + c*
d^2)^p*d^4)/(p*e + 2*e)