3.1.94 \(\int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [94]

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]

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

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Rubi [A]
time = 0.07, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {1653, 12, 799, 794, 223, 209} \begin {gather*} \frac {d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}+\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} \end {gather*}

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 12

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

Rule 209

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

Rule 223

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 794

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 799

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 1653

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]

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 \text {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*}

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Mathematica [A]
time = 0.13, size = 89, normalized size = 1.03 \begin {gather*} \frac {e \sqrt {d^2-e^2 x^2} \left (4 d^2-3 d e x+2 e^2 x^2\right )+3 d^3 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{6 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(156\) vs. \(2(74)=148\).
time = 0.07, size = 157, normalized size = 1.83

method	result	size
risch	\(\frac {\left (2 e^{2} x^{2}-3 d e x +4 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{3}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\)	\(75\)
default	\(-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{3}}-\frac {d \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{2}}+\frac {d^{2} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{3}}\)	\(157\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 71, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{\left (-2\right )} + \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-3\right )} - \frac {1}{3} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} e^{\left (-3\right )} \end {gather*}

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]

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

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Fricas [A]
time = 5.37, size = 69, normalized size = 0.80 \begin {gather*} -\frac {1}{6} \, {\left (6 \, d^{3} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (2 \, x^{2} e^{2} - 3 \, d x e + 4 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-3\right )} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \end {gather*}

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 = 1.37, size = 54, normalized size = 0.63 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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