3.17 \(\int \frac{x^2 (d+e x)}{(d^2-e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{d (d+e x)}{e^3 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^3}-\frac{d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.0418788, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {797, 641, 217, 203, 637} \[ \frac{d (d+e x)}{e^3 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^3}-\frac{d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 797

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

Mathematica [A]  time = 0.0359309, size = 77, normalized size = 1.05 \[ \frac{-d \sqrt{d^2-e^2 x^2} \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+2 d^2+d e x-e^2 x^2}{e^3 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 99, normalized size = 1.4 \begin{align*} -{\frac{{x}^{2}}{e}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+2\,{\frac{{d}^{2}}{{e}^{3}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}+{\frac{dx}{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{d}{{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.63252, size = 123, normalized size = 1.68 \begin{align*} -\frac{x^{2}}{\sqrt{-e^{2} x^{2} + d^{2}} e} + \frac{d x}{\sqrt{-e^{2} x^{2} + d^{2}} e^{2}} - \frac{d \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}} e^{2}} + \frac{2 \, d^{2}}{\sqrt{-e^{2} x^{2} + d^{2}} e^{3}} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.08667, size = 176, normalized size = 2.41 \begin{align*} \frac{2 \, d e x - 2 \, d^{2} + 2 \,{\left (d e x - d^{2}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt{-e^{2} x^{2} + d^{2}}{\left (e x - 2 \, d\right )}}{e^{4} x - d e^{3}} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 7.27369, size = 185, normalized size = 2.53 \begin{align*} d \left (\begin{cases} \frac{i \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{e^{3}} - \frac{i x}{d e^{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{\operatorname{asin}{\left (\frac{e x}{d} \right )}}{e^{3}} + \frac{x}{d e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) + e \left (\begin{cases} \tilde{\infty } x^{4} & \text{for}\: \left (d = 0 \vee d = - \sqrt{e^{2} x^{2}} \vee d = \sqrt{e^{2} x^{2}}\right ) \wedge \left (d = - \sqrt{e^{2} x^{2}} \vee d = \sqrt{e^{2} x^{2}} \vee e = 0\right ) \\\frac{x^{4}}{4 \left (d^{2}\right )^{\frac{3}{2}}} & \text{for}\: e = 0 \\\frac{2 d^{2}}{e^{4} \sqrt{d^{2} - e^{2} x^{2}}} - \frac{x^{2}}{e^{2} \sqrt{d^{2} - e^{2} x^{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)**(3/2),x)

[Out]

d*Piecewise((I*acosh(e*x/d)/e**3 - I*x/(d*e**2*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (-as
in(e*x/d)/e**3 + x/(d*e**2*sqrt(1 - e**2*x**2/d**2)), True)) + e*Piecewise((zoo*x**4, (Eq(d, 0) | Eq(d, sqrt(e
**2*x**2)) | Eq(d, -sqrt(e**2*x**2))) & (Eq(e, 0) | Eq(d, sqrt(e**2*x**2)) | Eq(d, -sqrt(e**2*x**2)))), (x**4/
(4*(d**2)**(3/2)), Eq(e, 0)), (2*d**2/(e**4*sqrt(d**2 - e**2*x**2)) - x**2/(e**2*sqrt(d**2 - e**2*x**2)), True
))

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Giac [A]  time = 1.17666, size = 89, normalized size = 1.22 \begin{align*} -d \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left (2 \, d^{2} e^{\left (-3\right )} -{\left (x e^{\left (-1\right )} - d e^{\left (-2\right )}\right )} x\right )}}{x^{2} e^{2} - d^{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)^(3/2),x, algorithm="giac")

[Out]

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