\(\int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [121]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1653, 12, 807, 223, 209} \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{e^3} \]

[In]

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

[Out]

-(Sqrt[d^2 - e^2*x^2]/e^3) - (d*Sqrt[d^2 - e^2*x^2])/(e^3*(d + e*x)) - (d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 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 {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {\int \frac {d e^3 x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx}{e^4} \\ & = -\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx}{e} \\ & = -\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2} \\ & = -\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \text {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 {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {(-2 d-e x) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}+\frac {2 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} \left (x +\frac {d}{e}\right )}\)	\(97\)
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} \left (x +\frac {d}{e}\right )}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \, d e x + 2 \, d^{2} - 2 \, {\left (d e x + d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + 2 \, d\right )}}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{4} x + d e^{3}} - \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{2} {\left | e \right |}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{3}} + \frac {2 \, d}{e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)