3.1212 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 a c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 a c (a x+1) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6143, 6140, 44, 207} \[ \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 a c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 a c (a x+1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]

[Out]

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

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

Rule 6140

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

Rule 6143

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 59, normalized size = 0.66 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {\tanh ^{-1}(a x)}{2 a}-\frac {1}{2 a (a x+1)}\right )}{c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.71, size = 343, normalized size = 3.81 \[ \left [-\frac {4 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x - {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right )}{8 \, {\left (a^{4} c^{2} x^{3} + a^{3} c^{2} x^{2} - a^{2} c^{2} x - a c^{2}\right )}}, -\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x - {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right )}{4 \, {\left (a^{4} c^{2} x^{3} + a^{3} c^{2} x^{2} - a^{2} c^{2} x - a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(4*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x - (a^3*x^3 + a^2*x^2 - a*x - 1)*sqrt(c)*log(-(a^6*c*x^6 +
5*a^4*c*x^4 - 5*a^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 -
3*a^4*x^4 + 3*a^2*x^2 - 1)))/(a^4*c^2*x^3 + a^3*c^2*x^2 - a^2*c^2*x - a*c^2), -1/4*(2*sqrt(-a^2*c*x^2 + c)*sq
rt(-a^2*x^2 + 1)*a*x - (a^3*x^3 + a^2*x^2 - a*x - 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)
*a*sqrt(-c)*x/(a^4*c*x^4 - c)))/(a^4*c^2*x^3 + a^3*c^2*x^2 - a^2*c^2*x - a*c^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((-a^2*c*x^2 + c)^(3/2)*(a*x + 1)), x)

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maple [A]  time = 0.05, size = 88, normalized size = 0.98 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x -1\right ) x a -a x \ln \left (a x +1\right )+\ln \left (a x -1\right )-\ln \left (a x +1\right )+2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.34, size = 50, normalized size = 0.56 \[ -\frac {\sqrt {c}}{2 \, {\left (a^{2} c^{2} x + a c^{2}\right )}} + \frac {\log \left (a x + 1\right )}{4 \, a c^{\frac {3}{2}}} - \frac {\log \left (a x - 1\right )}{4 \, a c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(c)/(a^2*c^2*x + a*c^2) + 1/4*log(a*x + 1)/(a*c^(3/2)) - 1/4*log(a*x - 1)/(a*c^(3/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^(3/2)*(a*x + 1)),x)

[Out]

int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^(3/2)*(a*x + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1)*(a*x + 1))**(3/2)*(a*x + 1)), x)

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