3.1080 \(\int \frac{(d+e x)^5}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=31 \[ \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{c^3 e} \]

[Out]

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

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Rubi [A]  time = 0.0237814, 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} \[ \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{c^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0050517, size = 23, normalized size = 0.74 \[ \frac{x (d+e x)}{c^2 \sqrt{c (d+e x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 32, normalized size = 1. \begin{align*}{ \left ( ex+d \right ) ^{5}x \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)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.25439, size = 266, normalized size = 8.58 \begin{align*} \frac{e^{3} x^{4}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} - \frac{4 \, c^{2} d^{5} e^{4}}{\left (c e^{2}\right )^{\frac{9}{2}}{\left (x + \frac{d}{e}\right )}^{4}} - \frac{6 \, d^{2} e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} + \frac{32 \, c d^{4} e^{3}}{3 \, \left (c e^{2}\right )^{\frac{7}{2}}{\left (x + \frac{d}{e}\right )}^{3}} - \frac{17 \, d^{4}}{3 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c e} - \frac{8 \, d^{3} e^{2}}{\left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{4 \, d^{5}}{\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)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

e^3*x^4/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) - 4*c^2*d^5*e^4/((c*e^2)^(9/2)*(x + d/e)^4) - 6*d^2*e*x^2/((
c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 32/3*c*d^4*e^3/((c*e^2)^(7/2)*(x + d/e)^3) - 17/3*d^4/((c*e^2*x^2 +
2*c*d*e*x + c*d^2)^(3/2)*c*e) - 8*d^3*e^2/((c*e^2)^(5/2)*(x + d/e)^2) + 4*d^5/((c*e^2)^(5/2)*(x + d/e)^4)

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Fricas [A]  time = 2.38364, 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^{3} e x + c^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(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)*x/(c^3*e*x + c^3*d)

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Sympy [A]  time = 1.70706, 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^{3} e} & \text{for}\: e \neq 0 \\\frac{d^{5} 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)**5/(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*e), Ne(e, 0)), (d**5*x/(c*d**2)**(5/2), True))

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Giac [B]  time = 1.45984, size = 126, normalized size = 4.06 \begin{align*} \frac{2 \, C_{0} d^{3} e^{\left (-3\right )} - \frac{3 \, d^{4} e^{\left (-1\right )}}{c} +{\left (6 \, C_{0} d^{2} e^{\left (-2\right )} - \frac{8 \, d^{3}}{c} +{\left (6 \, C_{0} d e^{\left (-1\right )} +{\left (2 \, C_{0} + \frac{x e^{3}}{c}\right )} x - \frac{6 \, d^{2} e}{c}\right )} x\right )} x}{{\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)^5/(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) - 3*d^4*e^(-1)/c + (6*C_0*d^2*e^(-2) - 8*d^3/c + (6*C_0*d*e^(-1) + (2*C_0 + x*e^3/c)*x - 6*d
^2*e/c)*x)*x)/(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)