3.5.13 \(\int \frac {\sqrt {1+\frac {1}{c^2 x^2}}}{\sqrt {1-c^4 x^4}} \, dx\) [413]

Optimal. Leaf size=40 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {1-c^4 x^4}}{c \sqrt {1+\frac {1}{c^2 x^2}} x}\right )}{c} \]

[Out]

-arctanh((-c^4*x^4+1)^(1/2)/c/x/(1+1/c^2/x^2)^(1/2))/c

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1462, 1266, 862, 65, 214} \begin {gather*} -\frac {x \sqrt {\frac {1}{c^2 x^2}+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + 1/(c^2*x^2)]/Sqrt[1 - c^4*x^4],x]

[Out]

-((Sqrt[1 + 1/(c^2*x^2)]*x*ArcTanh[Sqrt[1 - c^2*x^2]])/Sqrt[1 + c^2*x^2])

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 862

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

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1462

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Dist[(e^IntPart[q]*((d + e*
x^mn)^FracPart[q]/(1 + d*(1/(x^mn*e)))^FracPart[q]))/x^(mn*FracPart[q]), Int[x^(mn*q)*(1 + d*(1/(x^mn*e)))^q*(
a + c*x^n2)^p, x], x] /; FreeQ[{a, c, d, e, mn, p, q}, x] && EqQ[n2, -2*mn] &&  !IntegerQ[p] &&  !IntegerQ[q]
&& PosQ[n2]

Rubi steps

\begin {align*} \int \frac {\sqrt {1+\frac {1}{c^2 x^2}}}{\sqrt {1-c^4 x^4}} \, dx &=\frac {\left (\sqrt {1+\frac {1}{c^2 x^2}} x\right ) \int \frac {\sqrt {1+c^2 x^2}}{x \sqrt {1-c^4 x^4}} \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (\sqrt {1+\frac {1}{c^2 x^2}} x\right ) \text {Subst}\left (\int \frac {\sqrt {1+c^2 x}}{x \sqrt {1-c^4 x^2}} \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}\\ &=\frac {\left (\sqrt {1+\frac {1}{c^2 x^2}} x\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {\left (\sqrt {1+\frac {1}{c^2 x^2}} x\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{c^2 \sqrt {1+c^2 x^2}}\\ &=-\frac {\sqrt {1+\frac {1}{c^2 x^2}} x \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {1+c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.47, size = 58, normalized size = 1.45 \begin {gather*} -\frac {\sqrt {1+\frac {1}{c^2 x^2}} x \tanh ^{-1}\left (\frac {\sqrt {1-c^4 x^4}}{\sqrt {1+c^2 x^2}}\right )}{\sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + 1/(c^2*x^2)]/Sqrt[1 - c^4*x^4],x]

[Out]

-((Sqrt[1 + 1/(c^2*x^2)]*x*ArcTanh[Sqrt[1 - c^4*x^4]/Sqrt[1 + c^2*x^2]])/Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.13, size = 101, normalized size = 2.52

method	result	size
default	\(-\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \sqrt {-c^{4} x^{4}+1}\, \mathrm {csgn}\left (\frac {1}{c}\right ) \ln \left (\frac {2 \,\mathrm {csgn}\left (\frac {1}{c}\right ) c \sqrt {-\frac {c^{2} x^{2}-1}{c^{2}}}+2}{c^{2} x}\right )}{\left (c^{2} x^{2}+1\right ) \sqrt {-\frac {c^{2} x^{2}-1}{c^{2}}}\, c}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+1/c^2/x^2)^(1/2)/(-c^4*x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-((c^2*x^2+1)/c^2/x^2)^(1/2)*x*(-c^4*x^4+1)^(1/2)*csgn(1/c)*ln(2*(csgn(1/c)*c*(-(c^2*x^2-1)/c^2)^(1/2)+1)/c^2/
x)/(c^2*x^2+1)/(-(c^2*x^2-1)/c^2)^(1/2)/c

________________________________________________________________________________________

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+1/c^2/x^2)^(1/2)/(-c^4*x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(1/(c^2*x^2) + 1)/sqrt(-c^4*x^4 + 1), x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (36) = 72\).
time = 0.37, size = 120, normalized size = 3.00 \begin {gather*} -\frac {\log \left (\frac {c^{2} x^{2} + \sqrt {-c^{4} x^{4} + 1} c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c^{2} x^{2} + 1}\right ) - \log \left (-\frac {c^{2} x^{2} - \sqrt {-c^{4} x^{4} + 1} c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c^{2} x^{2} + 1}\right )}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+1/c^2/x^2)^(1/2)/(-c^4*x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(log((c^2*x^2 + sqrt(-c^4*x^4 + 1)*c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c^2*x^2 + 1)) - log(-(c^2*x^2
- sqrt(-c^4*x^4 + 1)*c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c^2*x^2 + 1)))/c

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+1/c**2/x**2)**(1/2)/(-c**4*x**4+1)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 5.06, size = 42, normalized size = 1.05 \begin {gather*} -\frac {{\left (\log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \log \left (-\sqrt {-c^{2} x^{2} + 1} + 1\right )\right )} {\left | c \right |}}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+1/c^2/x^2)^(1/2)/(-c^4*x^4+1)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {\frac {1}{c^2\,x^2}+1}}{\sqrt {1-c^4\,x^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/(c^2*x^2) + 1)^(1/2)/(1 - c^4*x^4)^(1/2),x)

[Out]

int((1/(c^2*x^2) + 1)^(1/2)/(1 - c^4*x^4)^(1/2), x)

________________________________________________________________________________________