3.99 \(\int \frac{\sqrt{d^2-e^2 x^2}}{x^3 (d+e x)} \, dx\)

Optimal. Leaf size=82 \[ \frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0783041, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {850, 835, 807, 266, 63, 208} \[ \frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{d^2-e^2 x^2}}{x^3 (d+e x)} \, dx &=\int \frac{d-e x}{x^3 \sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}-\frac{\int \frac{2 d^2 e-d e^2 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{2 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}+\frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}+\frac{e^2 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{2 d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}+\frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}+\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{4 d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}+\frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{d^2}{e^2}-\frac{x^2}{e^2}} \, dx,x,\sqrt{d^2-e^2 x^2}\right )}{2 d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{2 d x^2}+\frac{e \sqrt{d^2-e^2 x^2}}{d^2 x}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.11493, size = 70, normalized size = 0.85 \[ -\frac{(d-2 e x) \sqrt{d^2-e^2 x^2}+e^2 x^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )-e^2 x^2 \log (x)}{2 d^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.066, size = 254, normalized size = 3.1 \begin{align*}{\frac{{e}^{2}}{2\,{d}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{e}^{2}}{2\,d}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{{e}^{2}}{{d}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{{e}^{3}}{{d}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{1}{2\,{d}^{3}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{e}{{d}^{4}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{3}x}{{d}^{4}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{e}^{3}}{{d}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^2/d^3*(-e^2*x^2+d^2)^(1/2)-1/2*e^2/d/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-e^2/d^
3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-e^3/d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))
^(1/2))-1/2/d^3/x^2*(-e^2*x^2+d^2)^(3/2)+e/d^4/x*(-e^2*x^2+d^2)^(3/2)+e^3/d^4*x*(-e^2*x^2+d^2)^(1/2)+e^3/d^2/(
e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84987, size = 128, normalized size = 1.56 \begin{align*} \frac{e^{2} x^{2} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt{-e^{2} x^{2} + d^{2}}{\left (2 \, e x - d\right )}}{2 \, d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{x^{3} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(1/2)/x**3/(e*x+d),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**3*(d + e*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError