3.632 \(\int \frac {e^{\tanh ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {a x+1}{a c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\sin ^{-1}(a x)}{a c} \]

[Out]

arcsin(a*x)/a/c+(-a*x-1)/a/c/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/a/c

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Rubi [A]  time = 0.13, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6157, 6148, 797, 641, 216, 637} \[ -\frac {a x+1}{a c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - c/(a^2*x^2)),x]

[Out]

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx &=-\frac {a^2 \int \frac {e^{\tanh ^{-1}(a x)} x^2}{1-a^2 x^2} \, dx}{c}\\ &=-\frac {a^2 \int \frac {x^2 (1+a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac {\int \frac {1+a x}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}+\frac {\int \frac {1+a x}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {1+a x}{a c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {1+a x}{a c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\sin ^{-1}(a x)}{a c}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.87 \[ \frac {a^2 x^2+\sqrt {1-a^2 x^2} \sin ^{-1}(a x)-a x-2}{a c \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(c - c/(a^2*x^2)),x]

[Out]

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

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fricas [A]  time = 0.77, size = 68, normalized size = 1.11 \[ -\frac {2 \, a x + 2 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x - 2\right )} - 2}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*x + 2*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(a*x - 2) - 2)/(a^2*c*x - a*
c)

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giac [A]  time = 0.20, size = 72, normalized size = 1.18 \[ \frac {\arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{c {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a c} - \frac {2}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(a*x)*sgn(a)/(c*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c) - 2/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*
abs(a))

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maple [A]  time = 0.04, size = 94, normalized size = 1.54 \[ -\frac {\sqrt {-a^{2} x^{2}+1}}{a c}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c \sqrt {a^{2}}}+\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{2} c \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2+1)^(1/2)/a/c+1/c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/a^2/c/(x-1/a)*(-a^2*(x-1/a)
^2-2*a*(x-1/a))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a^{2} x^{2}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.84, size = 90, normalized size = 1.48 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c}-\frac {\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(x*(-a^2)^(1/2))/(c*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(a*c) - (1 - a^2*x^2)^(1/2)/(c*(x*(-a^2)^(1/2) -
(-a^2)^(1/2)/a)*(-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \int \frac {x^{2}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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