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

Optimal result

Integrand size = 27, antiderivative size = 86 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d (2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

[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] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1653, 12, 799, 794, 223, 209} \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}+\frac {d (2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (4 d^2-3 d e x+2 e^2 x^2\right )}{6 e^3}-\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87

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	\(-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{3}}-\frac {d \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{2}}+\frac {d^{2} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{3}}\)	\(157\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (2 \, e^{2} x^{2} - 3 \, d e x + 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

[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

Sympy [F]

\[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\int \frac {x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{3 \, e^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{2} {\left | e \right |}} + \frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x {\left (\frac {2 \, x}{e} - \frac {3 \, d}{e^{2}}\right )} + \frac {4 \, d^{2}}{e^{3}}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\int \frac {x^2\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \]

[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)