3.11.40 \(\int (d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)^{3/2} \, dx\) [1040]

Optimal. Leaf size=39 \[ \frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{6 c e} \]

[Out]

1/6*(e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)/c/e

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Rubi [A]
time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, 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 )^{5/2}}{6 c e} \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)^(3/2),x]

[Out]

((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2))/(6*c*e)

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 )^{3/2} \, dx &=\frac {\int \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx}{c}\\ &=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{6 c e}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 0.72 \begin {gather*} \frac {(d+e x) \left (c (d+e x)^2\right )^{5/2}}{6 c e} \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)^(3/2),x]

[Out]

((d + e*x)*(c*(d + e*x)^2)^(5/2))/(6*c*e)

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Maple [A]
time = 0.58, size = 35, normalized size = 0.90

method result size
risch \(\frac {c \left (e x +d \right )^{5} \sqrt {\left (e x +d \right )^{2} c}}{6 e}\) \(25\)
default \(\frac {\left (e x +d \right )^{3} \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {3}{2}}}{6 e}\) \(35\)
gosper \(\frac {x \left (e^{5} x^{5}+6 d \,e^{4} x^{4}+15 d^{2} e^{3} x^{3}+20 d^{3} e^{2} x^{2}+15 d^{4} e x +6 d^{5}\right ) \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {3}{2}}}{6 \left (e x +d \right )^{3}}\) \(84\)
trager \(\frac {c x \left (e^{5} x^{5}+6 d \,e^{4} x^{4}+15 d^{2} e^{3} x^{3}+20 d^{3} e^{2} x^{2}+15 d^{4} e x +6 d^{5}\right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{6 e x +6 d}\) \(85\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 59, normalized size = 1.51 \begin {gather*} \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {5}{2}} d e^{\left (-1\right )}}{6 \, c} + \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {5}{2}} x}{6 \, c} \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)^(3/2),x, algorithm="maxima")

[Out]

1/6*(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(5/2)*d*e^(-1)/c + 1/6*(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(5/2)*x/c

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (35) = 70\).
time = 3.23, size = 89, normalized size = 2.28 \begin {gather*} \frac {{\left (c x^{6} e^{5} + 6 \, c d x^{5} e^{4} + 15 \, c d^{2} x^{4} e^{3} + 20 \, c d^{3} x^{3} e^{2} + 15 \, c d^{4} x^{2} e + 6 \, c d^{5} x\right )} \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{6 \, {\left (x e + d\right )}} \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)^(3/2),x, algorithm="fricas")

[Out]

1/6*(c*x^6*e^5 + 6*c*d*x^5*e^4 + 15*c*d^2*x^4*e^3 + 20*c*d^3*x^3*e^2 + 15*c*d^4*x^2*e + 6*c*d^5*x)*sqrt(c*x^2*
e^2 + 2*c*d*x*e + c*d^2)/(x*e + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \left (d + e x\right )^{2}\right )^{\frac {3}{2}} \left (d + e x\right )^{2}\, dx \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)**(3/2),x)

[Out]

Integral((c*(d + e*x)**2)**(3/2)*(d + e*x)**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (35) = 70\).
time = 1.71, size = 82, normalized size = 2.10 \begin {gather*} \frac {1}{6} \, {\left (3 \, {\left (x^{2} e + 2 \, d x\right )} c d^{4} \mathrm {sgn}\left (x e + d\right ) + 3 \, {\left (x^{2} e + 2 \, d x\right )}^{2} c d^{2} e \mathrm {sgn}\left (x e + d\right ) + {\left (x^{2} e + 2 \, d x\right )}^{3} c e^{2} \mathrm {sgn}\left (x e + d\right )\right )} \sqrt {c} \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)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int {\left (d+e\,x\right )}^2\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2} \,d x \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)^(3/2),x)

[Out]

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

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