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

Optimal. Leaf size=51 \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d x} \]

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Rubi [A]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {850, 807, 266, 63, 208} \begin {gather*} \frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]))

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 53, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2}-e x \log \left (\sqrt {d^2-e^2 x^2}+d\right )+e x \log (x)}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.26, size = 103, normalized size = 2.02 \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2}}{d x}+\frac {e \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{d}-\frac {e \log \left (-d \sqrt {d^2-e^2 x^2}+d^2+d \sqrt {-e^2} x\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 50, normalized size = 0.98 \begin {gather*} -\frac {e x \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}}}{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 102, normalized size = 2.00 \begin {gather*} \frac {e \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 )}{d} + \frac {x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d + 1/2*x*e^3/((d*e + sqrt(-x^2*e^2 + d^2)*e)*
d) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/(d*x)

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maple [B]  time = 0.01, size = 222, normalized size = 4.35 \begin {gather*} \frac {e^{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}}\, d}-\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, d}+\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x}{d^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e}{d^{2}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e}{d^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^3/x*(-e^2*x^2+d^2)^(3/2)-1/d^3*e^2*x*(-e^2*x^2+d^2)^(1/2)-1/d*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^
2+d^2)^(1/2)*x)-e/d^2*(-e^2*x^2+d^2)^(1/2)+e/(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*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+e^2/d/(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}}{x^2\,\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^2*(d + e*x)),x)

[Out]

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

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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^{2} \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**2/(e*x+d),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**2*(d + e*x)), x)

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