3.12.1 \(\int (d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1101]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 = {656, 623} \begin {gather*} \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)} \end {gather*}

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 623

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)]

Rule 656

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]

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*}

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Mathematica [A]
time = 0.03, size = 32, normalized size = 0.74 \begin {gather*} \frac {(d+e x) \left (c (d+e x)^2\right )^{1+p}}{c e (3+2 p)} \end {gather*}

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 = 0.64, size = 41, normalized size = 0.95

method	result	size
gosper	\(\frac {\left (e x +d \right )^{3} \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{p}}{e \left (3+2 p \right )}\)	\(41\)
risch	\(\frac {\left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right ) \left (\left (e x +d \right )^{2} c \right )^{p}}{e \left (3+2 p \right )}\)	\(50\)
norman	\(\frac {d^{3} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{e \left (3+2 p \right )}+\frac {e^{2} x^{3} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{3+2 p}+\frac {3 d^{2} x \,{\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{3+2 p}+\frac {3 d e \,x^{2} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{3+2 p}\)	\(153\)

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,method=_RETURNVERBOSE)

[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] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (43) = 86\).
time = 0.31, size = 185, normalized size = 4.30 \begin {gather*} \frac {{\left (c^{p} x e + c^{p} d\right )} d^{2} e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{2 \, p + 1} + \frac {{\left (c^{p} {\left (2 \, p + 1\right )} x^{2} e^{2} + 2 \, c^{p} d p x e - c^{p} d^{2}\right )} d e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{2 \, p^{2} + 3 \, p + 1} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} c^{p} x^{3} e^{3} + {\left (2 \, p^{2} + p\right )} c^{p} d x^{2} e^{2} - 2 \, c^{p} d^{2} p x e + c^{p} d^{3}\right )} e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3} \end {gather*}

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

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Fricas [A]
time = 2.18, size = 58, normalized size = 1.35 \begin {gather*} \frac {{\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} e^{\left (-1\right )}}{2 \, p + 3} \end {gather*}

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]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {d^{2} x}{\left (c d^{2}\right )^{\frac {3}{2}}} & \text {for}\: e = 0 \wedge p = - \frac {3}{2} \\d^{2} x \left (c d^{2}\right )^{p} & \text {for}\: e = 0 \\\int \frac {\left (d + e x\right )^{2}}{\left (c \left (d + e x\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\\frac {d^{3} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 3 e} + \frac {3 d^{2} e x \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 3 e} + \frac {3 d e^{2} x^{2} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 3 e} + \frac {e^{3} x^{3} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 3 e} & \text {otherwise} \end {cases} \end {gather*}

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]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (43) = 86\).
time = 1.26, size = 128, normalized size = 2.98 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.45, size = 79, normalized size = 1.84 \begin {gather*} {\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p\,\left (\frac {3\,d^2\,x}{2\,p+3}+\frac {d^3}{e\,\left (2\,p+3\right )}+\frac {e^2\,x^3}{2\,p+3}+\frac {3\,d\,e\,x^2}{2\,p+3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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