3.1083 \(\int \frac{(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{1}{2 c^2 e (d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0215117, antiderivative size = 41, 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, 607} \[ -\frac{1}{2 c^2 e (d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 607

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*(a + b*x + c*x^2)^(p + 1))/((2*p + 1)*(b + 2
*c*x)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0096567, size = 28, normalized size = 0.68 \[ -\frac{d+e x}{2 c e \left (c (d+e x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 35, normalized size = 0.9 \begin{align*} -{\frac{ \left ( ex+d \right ) ^{3}}{2\,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)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.27339, size = 167, normalized size = 4.07 \begin{align*} -\frac{c^{2} d^{2} e^{4}}{4 \, \left (c e^{2}\right )^{\frac{9}{2}}{\left (x + \frac{d}{e}\right )}^{4}} + \frac{2 \, c d e^{3}}{3 \, \left (c e^{2}\right )^{\frac{7}{2}}{\left (x + \frac{d}{e}\right )}^{3}} - \frac{2 \, d}{3 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c e} - \frac{e^{2}}{2 \, \left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{d^{2}}{4 \, \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)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-1/4*c^2*d^2*e^4/((c*e^2)^(9/2)*(x + d/e)^4) + 2/3*c*d*e^3/((c*e^2)^(7/2)*(x + d/e)^3) - 2/3*d/((c*e^2*x^2 + 2
*c*d*e*x + c*d^2)^(3/2)*c*e) - 1/2*e^2/((c*e^2)^(5/2)*(x + d/e)^2) + 1/4*d^2/((c*e^2)^(5/2)*(x + d/e)^4)

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Fricas [A]  time = 2.38008, size = 140, normalized size = 3.41 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{2 \,{\left (c^{3} e^{4} x^{3} + 3 \, c^{3} d e^{3} x^{2} + 3 \, c^{3} d^{2} e^{2} x + c^{3} d^{3} e\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)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{\left (c \left (d + e x\right )^{2}\right )^{\frac{5}{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)**(5/2),x)

[Out]

Integral((d + e*x)**2/(c*(d + e*x)**2)**(5/2), x)

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Giac [A]  time = 1.53967, size = 95, normalized size = 2.32 \begin{align*} \frac{4 \, C_{0} d^{3} e^{\left (-3\right )} +{\left (12 \, C_{0} d^{2} e^{\left (-2\right )} + 4 \,{\left (3 \, C_{0} d e^{\left (-1\right )} + C_{0} x\right )} x - \frac{1}{c}\right )} x - \frac{d e^{\left (-1\right )}}{c}}{2 \,{\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)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

1/2*(4*C_0*d^3*e^(-3) + (12*C_0*d^2*e^(-2) + 4*(3*C_0*d*e^(-1) + C_0*x)*x - 1/c)*x - d*e^(-1)/c)/(c*x^2*e^2 +
2*c*d*x*e + c*d^2)^(3/2)