### 3.1030 $$\int (d+e x)^2 \sqrt{c d^2+2 c d e x+c e^2 x^2} \, dx$$

Optimal. Leaf size=39 $\frac{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e}$

[Out]

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

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Rubi [A]  time = 0.0215713, antiderivative size = 39, normalized size of antiderivative = 1., 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 = {642, 609} $\frac{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 \sqrt{c d^2+2 c d e x+c e^2 x^2} \, dx &=\frac{\int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx}{c}\\ &=\frac{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e}\\ \end{align*}

Mathematica [A]  time = 0.0140682, size = 28, normalized size = 0.72 $\frac{(d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 c e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 62, normalized size = 1.6 \begin{align*}{\frac{x \left ({e}^{3}{x}^{3}+4\,d{e}^{2}{x}^{2}+6\,{d}^{2}ex+4\,{d}^{3} \right ) }{4\,ex+4\,d}\sqrt{c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2}}} \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)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05364, size = 134, normalized size = 3.44 \begin{align*} \frac{{\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \,{\left (e x + d\right )}} \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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \left (d + e x\right )^{2}} \left (d + e x\right )^{2}\, dx \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)**(1/2),x)

[Out]

Integral(sqrt(c*(d + e*x)**2)*(d + e*x)**2, x)

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Giac [A]  time = 1.20574, size = 69, normalized size = 1.77 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} e^{2} + 2 \, c d x e + c d^{2}}{\left (d^{3} e^{\left (-1\right )} +{\left (3 \, d^{2} +{\left (x e^{2} + 3 \, d e\right )} x\right )} x\right )} \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)^(1/2),x, algorithm="giac")

[Out]

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