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

Optimal. Leaf size=52 \[ \frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[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)

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {807, 223, 209} \begin {gather*} \frac {\text {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)} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {\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*}

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Mathematica [A]
time = 0.16, size = 71, normalized size = 1.37 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

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

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Maple [A]
time = 0.07, size = 74, normalized size = 1.42

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.48, size = 37, normalized size = 0.71 \begin {gather*} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} + \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{x e^{3} + d e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.77, size = 70, normalized size = 1.35 \begin {gather*} -\frac {2 \, {\left (x e + d\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - x e - d - \sqrt {-x^{2} e^{2} + d^{2}}}{x e^{3} + d e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 1.34, size = 49, normalized size = 0.94 \begin {gather*} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\left (d\right ) - \frac {2 \, e^{\left (-2\right )}}{\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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