3.1.99 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx\) [99]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 = {864, 849, 821, 272, 65, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sqrt[d^2 - e^2*x^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 65

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 214

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

Rule 272

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 821

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 849

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 864

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]))

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 \text {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 {\text {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*}

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Mathematica [A]
time = 0.17, size = 75, normalized size = 0.91 \begin {gather*} \frac {-\frac {(d-2 e x) \sqrt {d^2-e^2 x^2}}{x^2}+2 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^2} \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(72)=144\).
time = 0.07, size = 326, normalized size = 3.98

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-2 e x +d \right )}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d \sqrt {d^{2}}}\)	\(75\)
default	\(-\frac {e^{2} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d^{3}}+\frac {-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{2 d^{2}}}{d}-\frac {e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}\right )}{d^{2}}+\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{d^{3}}\)	\(326\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(-x^2*e^2 + d^2)/((x*e + d)*x^3), x)

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Fricas [A]
time = 3.54, size = 61, normalized size = 0.74 \begin {gather*} \frac {x^{2} e^{2} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + \sqrt {-x^{2} e^{2} + d^{2}} {\left (2 \, x e - d\right )}}{2 \, d^{2} x^{2}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{3} \left (d + e x\right )}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (69) = 138\).
time = 1.32, size = 172, normalized size = 2.10 \begin {gather*} -\frac {x^{2} {\left (\frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}}{x} - e^{2}\right )} e^{4}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{2}} - \frac {e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{2 \, d^{2}} - \frac {\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{2} e^{\left (-2\right )}}{x^{2}} - \frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{2}}{x}}{8 \, d^{4}} \end {gather*}

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]

-1/8*x^2*(4*(d*e + sqrt(-x^2*e^2 + d^2)*e)/x - e^2)*e^4/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^2) - 1/2*e^2*log(1
/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^2 - 1/8*((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^2*e^(-2
)/x^2 - 4*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^2/x)/d^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}}{x^3\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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