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]

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

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.90 \[ \frac {-2 d^2+2 d e x+e^2 x^2}{3 d e^3 (d-e x) \sqrt {d^2-e^2 x^2}} \]

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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fricas [B]  time = 0.83, size = 104, normalized size = 1.79 \[ -\frac {2 \, e^{3} x^{3} - 2 \, d e^{2} x^{2} - 2 \, d^{2} e x + 2 \, d^{3} - {\left (e^{2} x^{2} + 2 \, d e x - 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (d e^{6} x^{3} - d^{2} e^{5} x^{2} - d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

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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giac [A]  time = 0.30, size = 51, normalized size = 0.88 \[ \frac {{\left (x^{2} {\left (\frac {x}{d} + 3 \, e^{\left (-1\right )}\right )} - 2 \, d^{2} e^{\left (-3\right )}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x^{2} e^{2} - d^{2}\right )}^{2}} \]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.95 \[ -\frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (-e^{2} x^{2}-2 d e x +2 d^{2}\right )}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,e^{3}} \]

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 = 0.44, size = 88, normalized size = 1.52 \[ \frac {x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} + \frac {d x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{2}} \]

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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mupad [B]  time = 2.59, size = 55, normalized size = 0.95 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (-2\,d^2+2\,d\,e\,x+e^2\,x^2\right )}{3\,d\,e^3\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.88, size = 231, normalized size = 3.98 \[ d \left (\begin {cases} \frac {i x^{3}}{- 3 d^{5} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 3 d^{3} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {x^{3}}{- 3 d^{5} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 3 d^{3} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} \frac {2 d^{2}}{- 3 d^{2} e^{4} \sqrt {d^{2} - e^{2} x^{2}} + 3 e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}}} - \frac {3 e^{2} x^{2}}{- 3 d^{2} e^{4} \sqrt {d^{2} - e^{2} x^{2}} + 3 e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) \]

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/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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