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

Optimal. Leaf size=86 \[ \frac{d (2 d-e x) \sqrt{d^2-e^2 x^2}}{2 e^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.11044, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1639, 12, 785, 780, 217, 203} \[ \frac{d (2 d-e x) \sqrt{d^2-e^2 x^2}}{2 e^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1639

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

Rule 12

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

Rule 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 0]

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^2 \sqrt{d^2-e^2 x^2}}{d+e x} \, dx &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{\int \frac{3 d e^3 x \sqrt{d^2-e^2 x^2}}{d+e x} \, dx}{3 e^4}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d \int \frac{x \sqrt{d^2-e^2 x^2}}{d+e x} \, dx}{e}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{\int \frac{x \left (d^2 e-d e^2 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{e^2}\\ &=\frac{d (2 d-e x) \sqrt{d^2-e^2 x^2}}{2 e^3}-\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 (2 d-e x) \sqrt{d^2-e^2 x^2}}{2 e^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{3 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 d-e x) \sqrt{d^2-e^2 x^2}}{2 e^3}-\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 = 0.0854072, size = 69, normalized size = 0.8 \[ \frac{\sqrt{d^2-e^2 x^2} \left (4 d^2-3 d e x+2 e^2 x^2\right )+3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{6 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[d^2 - e^2*x^2])/(d + e*x),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 [B]  time = 0.058, size = 160, normalized size = 1.9 \begin{align*} -{\frac{1}{3\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{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}}}}}+{\frac{{d}^{2}}{{e}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{{d}^{3}}{{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55548, size = 153, normalized size = 1.78 \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^2*x^2+d^2)^(1/2)/(e*x+d),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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x), x)

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Giac [A]  time = 1.1845, size = 73, normalized size = 0.85 \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^2*x^2+d^2)^(1/2)/(e*x+d),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)