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

Optimal. Leaf size=103 \[ -\frac{d^2 \sqrt{d^2-e^2 x^2}}{e^3}-\frac{d x \sqrt{d^2-e^2 x^2}}{2 e^2}+\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.0534872, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {797, 641, 195, 217, 203} \[ -\frac{d^2 \sqrt{d^2-e^2 x^2}}{e^3}-\frac{d x \sqrt{d^2-e^2 x^2}}{2 e^2}+\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2*Sqrt[d^2 - e^2*x^2])/e^3) - (d*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + (d^2 - e^2*x^2)^(3/2)/(3*e^3) + (d^3*Ar
cTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.045859, size = 70, normalized size = 0.68 \[ \frac{3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\sqrt{d^2-e^2 x^2} \left (4 d^2+3 d e x+2 e^2 x^2\right )}{6 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(4*d^2 + 3*d*e*x + 2*e^2*x^2)) + 3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(6*e^3)

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Maple [A]  time = 0.053, size = 102, normalized size = 1. \begin{align*} -{\frac{{x}^{2}}{3\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{2\,{d}^{2}}{3\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{dx}{2\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{d}^{3}}{2\,{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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.58034, size = 127, normalized size = 1.23 \begin{align*} -\frac{\sqrt{-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e} + \frac{d^{3} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}} e^{2}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} - \frac{2 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{2}}{3 \, 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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8527, size = 153, normalized size = 1.49 \begin{align*} -\frac{6 \, d^{3} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (2 \, e^{2} x^{2} + 3 \, d e x + 4 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \, 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 5.4229, size = 178, normalized size = 1.73 \begin{align*} d \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 \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**2*(e*x+d)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d*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*asin(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*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 = 1.19594, size = 73, normalized size = 0.71 \begin{align*} \frac{1}{2} \, d^{3} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{6} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left (4 \, d^{2} e^{\left (-3\right )} +{\left (2 \, x e^{\left (-1\right )} + 3 \, d e^{\left (-2\right )}\right )} x\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)^(1/2),x, algorithm="giac")

[Out]

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