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

Optimal. Leaf size=52 \[ \frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {793, 217, 203} \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d^2 - e^2*x^2]/(e^2*(d + e*x)) + ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.94 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2}}{d+e x}+\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 71, normalized size = 1.37 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)}+\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 67, normalized size = 1.29 \begin {gather*} \frac {e x - 2 \, {\left (e x + d\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d + \sqrt {-e^{2} x^{2} + d^{2}}}{e^{3} x + d e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x - 2*(e*x + d)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + d + sqrt(-e^2*x^2 + d^2))/(e^3*x + d*e^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)^2
+2*exp(2)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp
(1)^4+exp(2)^2)/exp(1)^2

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maple [A]  time = 0.01, size = 74, normalized size = 1.42 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{\left (x +\frac {d}{e}\right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/e^3/(x+d/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [A]  time = 0.97, size = 40, normalized size = 0.77 \begin {gather*} \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{3} x + d e^{2}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-e^2*x^2 + d^2)/(e^3*x + d*e^2) + arcsin(e*x/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}\,\left (d+e\,x\right )} \,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 )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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