\(\int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [95]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 62 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {799, 794, 223, 209} \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2}-\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2} \]

[In]

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

[Out]

-1/2*((2*d - e*x)*Sqrt[d^2 - e^2*x^2])/e^2 - (d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^2)

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x \left (d^2 e-d e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d e} \\ & = -\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e} \\ & = -\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \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} \\ & = -\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {(-2 d+e x) \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03

method result size
risch \(-\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}-\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e \sqrt {e^{2}}}\) \(64\)
default \(\frac {\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}}{e}-\frac {d \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{2}}\) \(135\)

[In]

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

[Out]

-1/2*(-e*x+2*d)*(-e^2*x^2+d^2)^(1/2)/e^2-1/2*d^2/e/(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 = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {2 \, d^{2} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x - 2 \, d\right )}}{2 \, e^{2}} \]

[In]

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

[Out]

1/2*(2*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^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e {\left | e \right |}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {x}{e} - \frac {2 \, d}{e^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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