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

Optimal. Leaf size=58 \[ \frac{x^2 (d+e x)}{3 d e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2}{3 e^3 \sqrt{d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0262578, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {796, 12, 261} \[ \frac{x^2 (d+e x)}{3 d e \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2}{3 e^3 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 796

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(x^2*(a*g - c*f*x)*(a + c*x^2)^(p
+ 1))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[x*Simp[2*a*g - c*f*(2*p + 5)*x, x]*(a + c*x^2)^(p + 1
), x], x] /; FreeQ[{a, c, f, g}, x] && EqQ[a*g^2 + f^2*c, 0] && LtQ[p, -2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0246986, size = 52, normalized size = 0.9 \[ \frac{-2 d^2+2 d e x+e^2 x^2}{3 d e^3 (d-e x) \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 55, normalized size = 1. \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( ex+d \right ) ^{2} \left ( -{x}^{2}{e}^{2}-2\,dex+2\,{d}^{2} \right ) }{3\,d{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.0003, size = 119, normalized size = 2.05 \begin{align*} \frac{x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e} + \frac{d x}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{2}} - \frac{2 \, d^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{3}} - \frac{x}{3 \, \sqrt{-e^{2} x^{2} + d^{2}} d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.14648, size = 204, normalized size = 3.52 \begin{align*} -\frac{2 \, e^{3} x^{3} - 2 \, d e^{2} x^{2} - 2 \, d^{2} e x + 2 \, d^{3} -{\left (e^{2} x^{2} + 2 \, d e x - 2 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \,{\left (d e^{6} x^{3} - d^{2} e^{5} x^{2} - d^{3} e^{4} x + d^{4} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 9.63715, size = 233, normalized size = 4.02 \begin{align*} d \left (\begin{cases} \frac{i x^{3}}{- 3 d^{5} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}} + 3 d^{3} e^{2} x^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\- \frac{x^{3}}{- 3 d^{5} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}} + 3 d^{3} e^{2} x^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) + e \left (\begin{cases} \frac{2 d^{2}}{- 3 d^{2} e^{4} \sqrt{d^{2} - e^{2} x^{2}} + 3 e^{6} x^{2} \sqrt{d^{2} - e^{2} x^{2}}} - \frac{3 e^{2} x^{2}}{- 3 d^{2} e^{4} \sqrt{d^{2} - e^{2} x^{2}} + 3 e^{6} x^{2} \sqrt{d^{2} - e^{2} x^{2}}} & \text{for}\: e \neq 0 \\\frac{x^{4}}{4 \left (d^{2}\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

d*Piecewise((I*x**3/(-3*d**5*sqrt(-1 + e**2*x**2/d**2) + 3*d**3*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2
*x**2)/Abs(d**2) > 1), (-x**3/(-3*d**5*sqrt(1 - e**2*x**2/d**2) + 3*d**3*e**2*x**2*sqrt(1 - e**2*x**2/d**2)),
True)) + e*Piecewise((2*d**2/(-3*d**2*e**4*sqrt(d**2 - e**2*x**2) + 3*e**6*x**2*sqrt(d**2 - e**2*x**2)) - 3*e*
*2*x**2/(-3*d**2*e**4*sqrt(d**2 - e**2*x**2) + 3*e**6*x**2*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**4/(4*(d**2)
**(5/2)), True))

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Giac [A]  time = 1.17522, size = 69, normalized size = 1.19 \begin{align*} \frac{{\left (x^{2}{\left (\frac{x}{d} + 3 \, e^{\left (-1\right )}\right )} - 2 \, d^{2} e^{\left (-3\right )}\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{3 \,{\left (x^{2} e^{2} - d^{2}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

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