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]

-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

________________________________________________________________________________________

Rubi [A]  time = 0.11, 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 = {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 12

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

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 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 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 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 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]

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 = 0.07, size = 69, normalized size = 0.80 \[ \frac {\sqrt {d^2-e^2 x^2} \left (4 d^2-3 d e x+2 e^2 x^2\right )+3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{6 e^3} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 73, normalized size = 0.85 \[ -\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (2 \, e^{2} x^{2} - 3 \, d e x + 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.20, size = 54, normalized size = 0.63 \[ \frac {1}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) + \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (4 \, d^{2} e^{\left (-3\right )} + {\left (2 \, x e^{\left (-1\right )} - 3 \, d e^{\left (-2\right )}\right )} x\right )} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.01, size = 160, normalized size = 1.86 \[ \frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}-\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d x}{2 e^{2}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2}}{e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{3}} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 77, normalized size = 0.90 \[ \frac {d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{3 \, e^{3}} \]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________