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

Optimal. Leaf size=46 \[ \frac {\sqrt {d^2-e^2 x^2}}{e}+\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

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Rubi [A]  time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {665, 217, 203} \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

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 665

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*} \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx &=\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 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.93 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}+d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 65, normalized size = 1.41 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}}{e}+\frac {d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 52, normalized size = 1.13 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 31, normalized size = 0.67 \begin {gather*} d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) + \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 77, normalized size = 1.67 \begin {gather*} \frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 31, normalized size = 0.67 \begin {gather*} \frac {d \arcsin \left (\frac {e x}{d}\right )}{e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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