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

Optimal. Leaf size=81 \[ -\frac {2 \sqrt {d^2-e^2 x^2}}{d^3 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x (d+e x)}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3} \]

[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)

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Rubi [A]
time = 0.04, antiderivative size = 81, 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 = {871, 821, 272, 65, 214} \begin {gather*} \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}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.23, size = 77, normalized size = 0.95 \begin {gather*} -\frac {\frac {(d+2 e x) \sqrt {d^2-e^2 x^2}}{x (d+e x)}+2 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 108, normalized size = 1.33

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 1.43, size = 88, normalized size = 1.09 \begin {gather*} -\frac {x^{2} e^{2} + d x e + {\left (x^{2} e^{2} + d x e\right )} \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 )}}{d^{3} x^{2} e + d^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (74) = 148\).
time = 0.79, size = 164, normalized size = 2.02 \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^{3}} + \frac {x {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{x} + e\right )} e^{2}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{3} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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