### 3.1072 $$\int \frac{(d+e x)^3}{(c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=31 $\frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{c^2 e}$

[Out]

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

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Rubi [A]  time = 0.0238644, antiderivative size = 31, 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 = {643, 629} $\frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{c^2 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0073226, size = 22, normalized size = 0.71 $\frac{x (d+e x)^3}{\left (c (d+e x)^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*(d + e*x)^3)/(c*(d + e*x)^2)^(3/2)

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Maple [A]  time = 0.04, size = 32, normalized size = 1. \begin{align*}{ \left ( ex+d \right ) ^{3}x \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.28817, size = 86, normalized size = 2.77 \begin{align*} \frac{e x^{2}}{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c} - \frac{d^{2}}{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c 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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.42828, size = 77, normalized size = 2.48 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} x}{c^{2} e x + c^{2} d} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.91692, size = 42, normalized size = 1.35 \begin{align*} \begin{cases} \frac{\sqrt{c d^{2} + 2 c d e x + c e^{2} x^{2}}}{c^{2} e} & \text{for}\: e \neq 0 \\\frac{d^{3} x}{\left (c d^{2}\right )^{\frac{3}{2}}} & \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)**(3/2),x)

[Out]

Piecewise((sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/(c**2*e), Ne(e, 0)), (d**3*x/(c*d**2)**(3/2), True))

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Giac [A]  time = 1.33103, size = 72, normalized size = 2.32 \begin{align*} \frac{2 \, C_{0} d e^{\left (-1\right )} +{\left (2 \, C_{0} + \frac{x e}{c}\right )} x - \frac{d^{2} e^{\left (-1\right )}}{c}}{\sqrt{c x^{2} e^{2} + 2 \, c d x e + c d^{2}}} \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)^(3/2),x, algorithm="giac")

[Out]

(2*C_0*d*e^(-1) + (2*C_0 + x*e/c)*x - d^2*e^(-1)/c)/sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2)