3.36 \(\int \frac{x (d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{d (5 d+3 e x) \sqrt{d^2-e^2 x^2}}{3 e^2}-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^2} \]

[Out]

-(x^2*Sqrt[d^2 - e^2*x^2])/3 - (d*(5*d + 3*e*x)*Sqrt[d^2 - e^2*x^2])/(3*e^2) + (d^3*ArcTan[(e*x)/Sqrt[d^2 - e^
2*x^2]])/e^2

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Rubi [A]  time = 0.084863, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1809, 780, 217, 203} \[ -\frac{d (5 d+3 e x) \sqrt{d^2-e^2 x^2}}{3 e^2}-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*Sqrt[d^2 - e^2*x^2])/3 - (d*(5*d + 3*e*x)*Sqrt[d^2 - e^2*x^2])/(3*e^2) + (d^3*ArcTan[(e*x)/Sqrt[d^2 - e^
2*x^2]])/e^2

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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])

Rubi steps

\begin{align*} \int \frac{x (d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx &=-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{x \left (-5 d^2 e^2-6 d e^3 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{3 e^2}\\ &=-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}-\frac{d (5 d+3 e x) \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{d^3 \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e}\\ &=-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}-\frac{d (5 d+3 e x) \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{d^3 \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{e}\\ &=-\frac{1}{3} x^2 \sqrt{d^2-e^2 x^2}-\frac{d (5 d+3 e x) \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^2}\\ \end{align*}

Mathematica [A]  time = 0.0621549, size = 69, normalized size = 0.83 \[ \frac{3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\sqrt{d^2-e^2 x^2} \left (5 d^2+3 d e x+e^2 x^2\right )}{3 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(5*d^2 + 3*d*e*x + e^2*x^2)) + 3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(3*e^2)

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Maple [A]  time = 0.057, size = 98, normalized size = 1.2 \begin{align*} -{\frac{{x}^{2}}{3}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{5\,{d}^{2}}{3\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{dx}{e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{d}^{3}}{e}\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*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.47, size = 122, normalized size = 1.47 \begin{align*} -\frac{1}{3} \, \sqrt{-e^{2} x^{2} + d^{2}} x^{2} + \frac{d^{3} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}} e} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} d x}{e} - \frac{5 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.81565, size = 150, normalized size = 1.81 \begin{align*} -\frac{6 \, d^{3} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (e^{2} x^{2} + 3 \, d e x + 5 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.57588, size = 219, normalized size = 2.64 \begin{align*} d^{2} \left (\begin{cases} \frac{x^{2}}{2 \sqrt{d^{2}}} & \text{for}\: e^{2} = 0 \\- \frac{\sqrt{d^{2} - e^{2} x^{2}}}{e^{2}} & \text{otherwise} \end{cases}\right ) + 2 d e \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{i d x \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{d x}{2 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{x^{3}}{2 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{2 d^{2} \sqrt{d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac{x^{2} \sqrt{d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text{for}\: e \neq 0 \\\frac{x^{4}}{4 \sqrt{d^{2}}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)**2/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e**2, True)) + 2*d*e*Piecewise((-I
*d**2*acosh(e*x/d)/(2*e**3) - I*d*x*sqrt(-1 + e**2*x**2/d**2)/(2*e**2), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*a
sin(e*x/d)/(2*e**3) - d*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2*d*sqrt(1 - e**2*x**2/d**2)), True)) + e*
*2*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4
/(4*sqrt(d**2)), True))

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Giac [A]  time = 1.1695, size = 66, normalized size = 0.8 \begin{align*} d^{3} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-2\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{3} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left (5 \, d^{2} e^{\left (-2\right )} +{\left (3 \, d e^{\left (-1\right )} + x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

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