3.1213 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a c^2 (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)^2/(-a^2*c*x^2+
c)^(1/2)-1/4*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+3/8*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a/c^2/(
-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 183, 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} \[ \frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a c^2 (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(5/2)),x]

[Out]

Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 - a*x)*Sqrt[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 + a*x)^2*Sqrt[c - a^
2*c*x^2]) - Sqrt[1 - a^2*x^2]/(4*a*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*
a*c^2*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 )^{5/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{-\tanh ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x)^2 (1+a x)^3} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{8 (-1+a x)^2}+\frac {1}{4 (1+a x)^3}+\frac {1}{4 (1+a x)^2}-\frac {3}{8 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {\left (3 \sqrt {1-a^2 x^2}\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{8 c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 83, normalized size = 0.45 \[ \frac {\sqrt {1-a^2 x^2} \left (-3 a^2 x^2-3 a x+3 (a x-1) (a x+1)^2 \tanh ^{-1}(a x)+2\right )}{8 a (a x-1) (a c x+c)^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(2 - 3*a*x - 3*a^2*x^2 + 3*(-1 + a*x)*(1 + a*x)^2*ArcTanh[a*x]))/(8*a*(-1 + a*x)*(c + a*c*x
)^2*Sqrt[c - a^2*c*x^2])

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fricas [A]  time = 0.71, size = 451, normalized size = 2.46 \[ \left [\frac {3 \, {\left (a^{5} x^{5} + a^{4} x^{4} - 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} - 5 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{32 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} + a^{4} x^{4} - 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) - 2 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} - 5 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{16 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}}\right ] \]

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)^(5/2),x, algorithm="fricas")

[Out]

[1/32*(3*(a^5*x^5 + a^4*x^4 - 2*a^3*x^3 - 2*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)) - 4*(2*a^3*x^3 - a^2*x^2 - 5*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + a^5*c^3*x^4 -
2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 + a^2*c^3*x + a*c^3), 1/16*(3*(a^5*x^5 + a^4*x^4 - 2*a^3*x^3 - 2*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)) - 2*(2*a^3*x^3 -
a^2*x^2 - 5*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + a^5*c^3*x^4 - 2*a^4*c^3*x^3 - 2*a^3*c
^3*x^2 + a^2*c^3*x + a*c^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)), x)

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maple [A]  time = 0.05, size = 166, normalized size = 0.91 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )+3 \ln \left (a x -1\right ) x^{2} a^{2}-3 \ln \left (a x +1\right ) x^{2} a^{2}+6 a^{2} x^{2}-3 \ln \left (a x -1\right ) x a +3 a x \ln \left (a x +1\right )+6 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )-4\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x -1\right ) \left (a x +1\right )^{2}} \]

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)^(5/2),x)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(3*ln(a*x-1)*x^3*a^3-3*a^3*x^3*ln(a*x+1)+3*ln(a*x-1)*x^2*a^2-3*
ln(a*x+1)*x^2*a^2+6*a^2*x^2-3*ln(a*x-1)*x*a+3*a*x*ln(a*x+1)+6*a*x-3*ln(a*x-1)+3*ln(a*x+1)-4)/(a^2*x^2-1)/c^3/a
/(a*x-1)/(a*x+1)^2

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maxima [A]  time = 0.34, size = 83, normalized size = 0.45 \[ -\frac {3 \, a^{2} x^{2} + 3 \, a x - 2}{8 \, {\left (a^{4} c^{\frac {5}{2}} x^{3} + a^{3} c^{\frac {5}{2}} x^{2} - a^{2} c^{\frac {5}{2}} x - a c^{\frac {5}{2}}\right )}} + \frac {3 \, \log \left (a x + 1\right )}{16 \, a c^{\frac {5}{2}}} - \frac {3 \, \log \left (a x - 1\right )}{16 \, a c^{\frac {5}{2}}} \]

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)^(5/2),x, algorithm="maxima")

[Out]

-1/8*(3*a^2*x^2 + 3*a*x - 2)/(a^4*c^(5/2)*x^3 + a^3*c^(5/2)*x^2 - a^2*c^(5/2)*x - a*c^(5/2)) + 3/16*log(a*x +
1)/(a*c^(5/2)) - 3/16*log(a*x - 1)/(a*c^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \]

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)^(5/2)*(a*x + 1)),x)

[Out]

int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^(5/2)*(a*x + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \]

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)**(5/2),x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x + 1)), x)

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