\(\int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [124]

Optimal result

Integrand size = 27, antiderivative size = 54 \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{d^2 (d+e x)}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {871, 12, 272, 65, 214} \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{d^2 (d+e x)}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \]

[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 12

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

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 871

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]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {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 {\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 )}{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] (verified)

Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2}}{d+e x}-\sqrt {d^2} \log (x)+\sqrt {d^2} \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.63

method	result	size
default	\(-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d \sqrt {d^{2}}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e \,d^{2} \left (x +\frac {d}{e}\right )}\)	\(88\)

[In]

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

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {e x + {\left (e x + d\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + d + \sqrt {-e^{2} x^{2} + d^{2}}}{d^{2} e x + d^{3}} \]

[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)

Sympy [F]

\[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )} x} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{2} {\left | e \right |}} - \frac {2 \, e}{d^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x\,\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \]

[In]

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

[Out]

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