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

Optimal. Leaf size=62 \[ -\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {785, 780, 217, 203} \begin {gather*} -\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.92 \begin {gather*} \frac {(e x-2 d) \sqrt {d^2-e^2 x^2}-d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.22, size = 80, normalized size = 1.29 \begin {gather*} \frac {(e x-2 d) \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 60, normalized size = 0.97 \begin {gather*} \frac {2 \, d^{2} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x - 2 \, d\right )}}{2 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 43, normalized size = 0.69 \begin {gather*} -\frac {1}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\relax (d) + \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e^{\left (-1\right )} - 2 \, d e^{\left (-2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 140, normalized size = 2.26 \begin {gather*} -\frac {d^{2} \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}}\, e}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x}{2 e}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/e*x*(-e^2*x^2+d^2)^(1/2)+1/2/e*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-d/e^2*(2*(x+d/e)
*d*e-(x+d/e)^2*e^2)^(1/2)-d^2/e/(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 = 56, normalized size = 0.90 \begin {gather*} -\frac {d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{d+e\,x} \,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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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 )}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x), x)

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