3.11.100 \(\int (d+e x)^3 (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1100]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 39, 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 = {657, 643} \begin {gather*} \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{p+2}}{2 c^2 e (p+2)} \end {gather*}

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_))*((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]

Rule 657

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]

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*}

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Mathematica [A]
time = 0.02, size = 30, normalized size = 0.77 \begin {gather*} \frac {(d+e x)^4 \left (c (d+e x)^2\right )^p}{2 e (2+p)} \end {gather*}

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 = 0.64, size = 40, normalized size = 1.03

method	result	size
gosper	\(\frac {\left (e x +d \right )^{4} \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{p}}{2 e \left (2+p \right )}\)	\(40\)
risch	\(\frac {\left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right ) \left (\left (e x +d \right )^{2} c \right )^{p}}{2 e \left (2+p \right )}\)	\(60\)
norman	\(\frac {2 d^{3} x \,{\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{2+p}+\frac {d^{4} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{2 e \left (2+p \right )}+\frac {e^{3} x^{4} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{4+2 p}+\frac {2 d \,e^{2} x^{3} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{2+p}+\frac {3 d^{2} e \,x^{2} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{2+p}\)	\(187\)

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,method=_RETURNVERBOSE)

[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] Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (36) = 72\).
time = 0.31, size = 315, normalized size = 8.08 \begin {gather*} \frac {{\left (c^{p} x e + c^{p} d\right )} d^{3} e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{2 \, p + 1} + \frac {3 \, {\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^{2} e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )}} + \frac {3 \, {\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 )} d e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3} + \frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} c^{p} x^{4} e^{4} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} c^{p} d x^{3} e^{3} - 3 \, {\left (2 \, p^{2} + p\right )} c^{p} d^{2} x^{2} e^{2} + 6 \, c^{p} d^{3} p x e - 3 \, c^{p} d^{4}\right )} e^{\left (2 \, p \log \left (x e + d\right ) - 1\right )}}{2 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )}} \end {gather*}

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

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Fricas [A]
time = 2.82, size = 67, normalized size = 1.72 \begin {gather*} \frac {{\left (x^{4} e^{4} + 4 \, d x^{3} e^{3} + 6 \, d^{2} x^{2} e^{2} + 4 \, d^{3} x e + d^{4}\right )} {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} e^{\left (-1\right )}}{2 \, {\left (p + 2\right )}} \end {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (34) = 68\).
time = 0.32, size = 233, normalized size = 5.97 \begin {gather*} \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 {gather*}

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 [A]
time = 1.91, size = 56, normalized size = 1.44 \begin {gather*} \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{2} {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{p} e^{\left (-1\right )}}{2 \, c^{2} {\left (p + 2\right )}} \end {gather*}

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

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Mupad [B]
time = 0.48, size = 90, normalized size = 2.31 \begin {gather*} {\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p\,\left (\frac {d^4}{2\,e\,\left (p+2\right )}+\frac {e^3\,x^4}{2\,\left (p+2\right )}+\frac {2\,d^3\,x}{p+2}+\frac {3\,d^2\,e\,x^2}{p+2}+\frac {2\,d\,e^2\,x^3}{p+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3*(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*(d^4/(2*e*(p + 2)) + (e^3*x^4)/(2*(p + 2)) + (2*d^3*x)/(p + 2) + (3*d^2*e*x^
2)/(p + 2) + (2*d*e^2*x^3)/(p + 2))

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