3.1277 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=182 \[ -\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (a x+1)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

-1/6*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)^3/(-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)-1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+1/8*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 = 0.11, antiderivative size = 182, 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} \[ -\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (a x+1)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2)),x]

[Out]

-Sqrt[1 - a^2*x^2]/(6*a*c^2*(1 + a*x)^3*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]/(8*a*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (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^{-3 \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^{-3 \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) (1+a x)^4} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{2 (1+a x)^4}+\frac {1}{4 (1+a x)^3}+\frac {1}{8 (1+a x)^2}-\frac {1}{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}}{6 a c^2 (1+a x)^3 \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}}{8 a c^2 (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}{8 c^2 \sqrt {c-a^2 c x^2}}\\ &=-\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1+a x)^3 \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}}{8 a c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {\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 = 0.06, size = 73, normalized size = 0.40 \[ \frac {\sqrt {1-a^2 x^2} \left (-3 a^2 x^2-9 a x+3 (a x+1)^3 \tanh ^{-1}(a x)-10\right )}{24 a c^2 (a x+1)^3 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-10 - 9*a*x - 3*a^2*x^2 + 3*(1 + a*x)^3*ArcTanh[a*x]))/(24*a*c^2*(1 + a*x)^3*Sqrt[c - a^2*
c*x^2])

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fricas [A]  time = 0.65, size = 461, normalized size = 2.53 \[ \left [\frac {3 \, {\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (10 \, a^{3} x^{3} + 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{96 \, {\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) - 2 \, {\left (10 \, a^{3} x^{3} + 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{48 \, {\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/96*(3*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3 - 2*a^2*x^2 - 3*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*(10*a^3*x^3 + 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + 3*a^
5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x - a*c^3), 1/48*(3*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3 - 2
*a^2*x^2 - 3*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*(10*a^3*x^3 + 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + 3*a^5*c^3*x^4 +
2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x - a*c^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^3), x)

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maple [A]  time = 0.05, size = 159, normalized size = 0.87 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )+9 \ln \left (a x -1\right ) x^{2} a^{2}-9 \ln \left (a x +1\right ) x^{2} a^{2}+6 a^{2} x^{2}+9 \ln \left (a x -1\right ) x a -9 a x \ln \left (a x +1\right )+18 a x +3 \ln \left (a x -1\right )-3 \ln \left (a x +1\right )+20\right )}{48 \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x +1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x)

[Out]

1/48*(-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)+9*ln(a*x-1)*x^2*a^2-9*
ln(a*x+1)*x^2*a^2+6*a^2*x^2+9*ln(a*x-1)*x*a-9*a*x*ln(a*x+1)+18*a*x+3*ln(a*x-1)-3*ln(a*x+1)+20)/(a^2*x^2-1)/c^3
/a/(a*x+1)^3

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maxima [A]  time = 0.35, size = 93, normalized size = 0.51 \[ -\frac {3 \, a^{2} \sqrt {c} x^{2} + 9 \, a \sqrt {c} x + 10 \, \sqrt {c}}{24 \, {\left (a^{4} c^{3} x^{3} + 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{16 \, a c^{\frac {5}{2}}} - \frac {\log \left (a x - 1\right )}{16 \, a c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(3*a^2*sqrt(c)*x^2 + 9*a*sqrt(c)*x + 10*sqrt(c))/(a^4*c^3*x^3 + 3*a^3*c^3*x^2 + 3*a^2*c^3*x + a*c^3) + 1
/16*log(a*x + 1)/(a*c^(5/2)) - 1/16*log(a*x - 1)/(a*c^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^(5/2)*(a*x + 1)^3),x)

[Out]

int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^(5/2)*(a*x + 1)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x + 1)**3), x)

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