3.1082 \(\int \frac{(d+e x)^3}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{c^2 e \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0246531, antiderivative size = 32, 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} \[ -\frac{1}{c^2 e \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.0132627, size = 21, normalized size = 0.66 \[ -\frac{1}{c^2 e \sqrt{c (d+e x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-(1/(c^2*e*Sqrt[c*(d + e*x)^2]))

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Maple [A]  time = 0.04, size = 35, normalized size = 1.1 \begin{align*} -{\frac{ \left ( ex+d \right ) ^{4}}{e} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-(e*x+d)^4/e/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)

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Maxima [B]  time = 1.57334, size = 215, normalized size = 6.72 \begin{align*} -\frac{c^{2} d^{3} e^{4}}{\left (c e^{2}\right )^{\frac{9}{2}}{\left (x + \frac{d}{e}\right )}^{4}} - \frac{e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} + \frac{8 \, c d^{2} e^{3}}{3 \, \left (c e^{2}\right )^{\frac{7}{2}}{\left (x + \frac{d}{e}\right )}^{3}} - \frac{5 \, d^{2}}{3 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c e} - \frac{2 \, d e^{2}}{\left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{d^{3}}{\left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

-c^2*d^3*e^4/((c*e^2)^(9/2)*(x + d/e)^4) - e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 8/3*c*d^2*e^3/((c
*e^2)^(7/2)*(x + d/e)^3) - 5/3*d^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 2*d*e^2/((c*e^2)^(5/2)*(x + d
/e)^2) + d^3/((c*e^2)^(5/2)*(x + d/e)^4)

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Fricas [A]  time = 2.40682, size = 108, normalized size = 3.38 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{c^{3} e^{3} x^{2} + 2 \, c^{3} d e^{2} x + c^{3} d^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.52782, size = 70, normalized size = 2.19 \begin{align*} \begin{cases} - \frac{\sqrt{c d^{2} + 2 c d e x + c e^{2} x^{2}}}{c^{3} d^{2} e + 2 c^{3} d e^{2} x + c^{3} e^{3} x^{2}} & \text{for}\: e \neq 0 \\\frac{d^{3} x}{\left (c d^{2}\right )^{\frac{5}{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)**(5/2),x)

[Out]

Piecewise((-sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/(c**3*d**2*e + 2*c**3*d*e**2*x + c**3*e**3*x**2), Ne(e, 0))
, (d**3*x/(c*d**2)**(5/2), True))

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Giac [B]  time = 1.40174, size = 107, normalized size = 3.34 \begin{align*} \frac{2 \, C_{0} d^{3} e^{\left (-3\right )} +{\left (6 \, C_{0} d^{2} e^{\left (-2\right )} +{\left (6 \, C_{0} d e^{\left (-1\right )} + 2 \, C_{0} x - \frac{e}{c}\right )} x - \frac{2 \, d}{c}\right )} x - \frac{d^{2} e^{\left (-1\right )}}{c}}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac{3}{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)^(5/2),x, algorithm="giac")

[Out]

(2*C_0*d^3*e^(-3) + (6*C_0*d^2*e^(-2) + (6*C_0*d*e^(-1) + 2*C_0*x - e/c)*x - 2*d/c)*x - d^2*e^(-1)/c)/(c*x^2*e
^2 + 2*c*d*x*e + c*d^2)^(3/2)