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

Optimal. Leaf size=47 \[ \frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {\sqrt {d^2-e^2 x^2}}{e} \]

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {641, 217, 203} \begin {gather*} \frac {d \tan ^{-1}\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[(d + e*x)/Sqrt[d^2 - e^2*x^2],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 641

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

Rubi steps

\begin {align*} \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, 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 = 47, normalized size = 1.00 \begin {gather*} \frac {d \tan ^{-1}\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]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 66, normalized size = 1.40 \begin {gather*} \frac {d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^2}-\frac {\sqrt {d^2-e^2 x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)/Sqrt[d^2 - e^2*x^2],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 = 50, normalized size = 1.06 \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*x+d)/(-e^2*x^2+d^2)^(1/2),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 = 0.24, size = 32, normalized size = 0.68 \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*x+d)/(-e^2*x^2+d^2)^(1/2),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 = 0.04, size = 50, normalized size = 1.06 \begin {gather*} \frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 32, normalized size = 0.68 \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*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.93, size = 54, normalized size = 1.15 \begin {gather*} \frac {d\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{\sqrt {-e^2}}-\frac {\sqrt {d^2-e^2\,x^2}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.14, size = 42, normalized size = 0.89 \begin {gather*} \begin {cases} \frac {d \left (\begin {cases} \operatorname {asin}{\left (e x \sqrt {\frac {1}{d^{2}}} \right )} & \text {for}\: d^{2} > 0 \end {cases}\right ) - \sqrt {d^{2} - e^{2} x^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {d x}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((d*Piecewise((asin(e*x*sqrt(d**(-2))), d**2 > 0)) - sqrt(d**2 - e**2*x**2))/e, Ne(e, 0)), (d*x/sqrt
(d**2), True))

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