3.124 \(\int \frac{1}{x (d+e x) \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=54 \[ \frac{\sqrt{d^2-e^2 x^2}}{d^2 (d+e x)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.0443948, antiderivative size = 54, normalized size of antiderivative = 1., 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 = {857, 12, 266, 63, 208} \[ \frac{\sqrt{d^2-e^2 x^2}}{d^2 (d+e x)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

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

Rule 857

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0394156, size = 52, normalized size = 0.96 \[ \frac{\frac{\sqrt{d^2-e^2 x^2}}{d+e x}-\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0.056, size = 88, normalized size = 1.6 \begin{align*} -{\frac{1}{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{1}{e{d}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64702, size = 131, normalized size = 2.43 \begin{align*} \frac{e x +{\left (e x + d\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + d + \sqrt{-e^{2} x^{2} + d^{2}}}{d^{2} e x + d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError