\(\int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [122]

Optimal result

Integrand size = 25, antiderivative size = 52 \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {807, 223, 209} \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25 \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}-\frac {2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[In]

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

[Out]

Sqrt[d^2 - e^2*x^2]/(e^2*(d + e*x)) - (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 = 74, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{3} \left (x +\frac {d}{e}\right )}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {e x - 2 \, {\left (e x + d\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d + \sqrt {-e^{2} x^{2} + d^{2}}}{e^{3} x + d e^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(x/(e*x+d)/(-e**2*x**2+d**2)**(1/2),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 = 40, normalized size of antiderivative = 0.77 \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{3} x + d e^{2}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \frac {x}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e {\left | e \right |}} - \frac {2}{e {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)