\(\int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [125]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \sqrt {d^2-e^2 x^2}}{d^3 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x (d+e x)}+\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 81, 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, 821, 272, 65, 214} \[ \int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x (d+e x)}-\frac {2 \sqrt {d^2-e^2 x^2}}{d^3 x} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {-\frac {d (d+2 e x) \sqrt {d^2-e^2 x^2}}{x (d+e x)}+\sqrt {d^2} e \log (x)-\sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{d^4} \]

[In]

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

[Out]

(-((d*(d + 2*e*x)*Sqrt[d^2 - e^2*x^2])/(x*(d + e*x))) + Sqrt[d^2]*e*Log[x] - Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[
d^2 - e^2*x^2]])/d^4

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.33

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{3} x}+\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{3} \left (x +\frac {d}{e}\right )}\)	\(108\)
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{3} x}+\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{3} \left (x +\frac {d}{e}\right )}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {e^{2} x^{2} + d e x + {\left (e^{2} x^{2} + d e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (2 \, e x + d\right )}}{d^{3} e x^{2} + d^{4} x} \]

[In]

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

[Out]

-(e^2*x^2 + d*e*x + (e^2*x^2 + d*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(2*e*x + d))/(
d^3*e*x^2 + d^4*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/x^2/(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^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (75) = 150\).

Time = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.19 \[ \int \frac {1}{x^2 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{3} {\left | e \right |}} + \frac {{\left (e^{2} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x}\right )} e^{2} x}{2 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, d^{3} x {\left | e \right |}} \]

[In]

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

[Out]

e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^3*abs(e)) + 1/2*(e^2 + 5*(d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))/x)*e^2*x/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))
/(e^2*x) + 1)*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(d^3*x*abs(e))

Mupad [F(-1)]

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

[In]

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

[Out]

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