### 3.1076 $$\int \frac{1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=31 $-\frac{1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0251365, 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} $-\frac{1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

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

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{1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx &=c \int \frac{d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx\\ &=-\frac{1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0198879, size = 20, normalized size = 0.65 $-\frac{1}{3 e \left (c (d+e x)^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

-1/(3*e*(c*(d + e*x)^2)^(3/2))

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Maple [A]  time = 0.046, size = 28, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,e} \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(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.13796, size = 63, normalized size = 2.03 \begin{align*} -\frac{1}{3 \,{\left (c^{\frac{3}{2}} e^{4} x^{3} + 3 \, c^{\frac{3}{2}} d e^{3} x^{2} + 3 \, c^{\frac{3}{2}} d^{2} e^{2} x + c^{\frac{3}{2}} d^{3} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.40767, size = 167, normalized size = 5.39 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \,{\left (c^{2} e^{5} x^{4} + 4 \, c^{2} d e^{4} x^{3} + 6 \, c^{2} d^{2} e^{3} x^{2} + 4 \, c^{2} d^{3} e^{2} x + c^{2} d^{4} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.66401, size = 42, normalized size = 1.35 \begin{align*} \begin{cases} - \frac{1}{3 e \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac{3}{2}}} & \text{for}\: e \neq 0 \\\frac{x}{d \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(1/(e*x+d)/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2),x)

[Out]

Piecewise((-1/(3*e*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(3/2)), Ne(e, 0)), (x/(d*(c*d**2)**(3/2)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x