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

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {679, 223, 209} \begin {gather*} \frac {d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {\sqrt {d^2-e^2 x^2}}{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 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*} \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 \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*}

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Mathematica [A]
time = 0.11, size = 63, normalized size = 1.37 \begin {gather*} \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}} \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]/e - (d*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/Sqrt[-e^2]

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Maple [A]
time = 0.07, size = 78, normalized size = 1.70

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	\(\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}}{e}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 29, normalized size = 0.63 \begin {gather*} d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \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="maxima")

[Out]

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

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Fricas [A]
time = 4.63, size = 48, normalized size = 1.04 \begin {gather*} -{\left (2 \, d \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - \sqrt {-x^{2} e^{2} + d^{2}}\right )} 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="fricas")

[Out]

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

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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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Giac [A]
time = 2.34, size = 31, normalized size = 0.67 \begin {gather*} d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) + \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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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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