3.1.95 \(\int \frac {x \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [95]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {799, 794, 223, 209} \begin {gather*} -\frac {d^2 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2}-\frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*d - e*x)*Sqrt[d^2 - e^2*x^2])/e^2 - (d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 794

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 799

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 \text {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 = 0.12, size = 77, normalized size = 1.24 \begin {gather*} \frac {e (-2 d+e x) \sqrt {d^2-e^2 x^2}-d^2 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-2*d + e*x)*Sqrt[d^2 - e^2*x^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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(134\) vs. \(2(54)=108\).
time = 0.06, size = 135, normalized size = 2.18

method	result	size
risch	\(-\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}-\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e \sqrt {e^{2}}}\)	\(64\)
default	\(\frac {\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}}{e}-\frac {d \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{2}}\)	\(135\)

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,method=_RETURNVERBOSE)

[Out]

1/e*(1/2*x*(-e^2*x^2+d^2)^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))-d/e^2*((-(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 = 0.48, size = 52, normalized size = 0.84 \begin {gather*} -\frac {1}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} + \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} x e^{\left (-1\right )} - \sqrt {-x^{2} e^{2} + d^{2}} d e^{\left (-2\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="maxima")

[Out]

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

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Fricas [A]
time = 5.40, size = 57, normalized size = 0.92 \begin {gather*} \frac {1}{2} \, {\left (2 \, d^{2} \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}} {\left (x e - 2 \, d\right )}\right )} e^{\left (-2\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="fricas")

[Out]

1/2*(2*d^2*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + sqrt(-x^2*e^2 + d^2)*(x*e - 2*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 )}}{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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Giac [A]
time = 1.49, 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}\left (d\right ) + \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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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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