\(\int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [96]

   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 = 24, antiderivative size = 46 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{e}+\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {679, 223, 209} \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {\sqrt {d^2-e^2 x^2}}{e} \]

[In]

Int[Sqrt[d^2 - e^2*x^2]/(d + e*x),x]

[Out]

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

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 679

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] - Dist[2*c*d*(p/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{e}-\frac {d \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{\sqrt {-e^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07

method result size
risch \(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e}+\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}\) \(49\)
default \(\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}}{e}\) \(78\)

[In]

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

[Out]

(-e^2*x^2+d^2)^(1/2)/e+d/(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 = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {2 \, d \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - \sqrt {-e^{2} x^{2} + d^{2}}}{e} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d \arcsin \left (\frac {e x}{d}\right )}{e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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