3.1250 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 (1-a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6142, 653, 192, 191} \[ \frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 (1-a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - a*x))/(5*a*(c - a^2*c*x^2)^(5/2)) + x/(5*c*(c - a^2*c*x^2)^(3/2)) + (2*x)/(5*c^2*Sqrt[c - a^2*c*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 6142

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

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=c \int \frac {(1-a x)^2}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\\ &=-\frac {2 (1-a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}+\frac {3}{5} \int \frac {1}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\\ &=-\frac {2 (1-a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 \int \frac {1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{5 c}\\ &=-\frac {2 (1-a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 79, normalized size = 1.05 \[ \frac {\sqrt {1-a^2 x^2} \left (2 a^3 x^3+4 a^2 x^2+a x-2\right )}{5 a c^2 \sqrt {1-a x} (a x+1)^{5/2} \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-2 + a*x + 4*a^2*x^2 + 2*a^3*x^3))/(5*a*c^2*Sqrt[1 - a*x]*(1 + a*x)^(5/2)*Sqrt[c - a^2*c*x
^2])

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fricas [A]  time = 0.83, size = 75, normalized size = 1.00 \[ -\frac {{\left (2 \, a^{3} x^{3} + 4 \, a^{2} x^{2} + a x - 2\right )} \sqrt {-a^{2} c x^{2} + c}}{5 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(2*a^3*x^3 + 4*a^2*x^2 + a*x - 2)*sqrt(-a^2*c*x^2 + c)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3
)

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giac [B]  time = 0.28, size = 220, normalized size = 2.93 \[ \frac {a^{3} {\left (\frac {5}{a^{3} c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)} - \frac {a^{12} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{2} c^{20} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right )^{4} \mathrm {sgn}\relax (a)^{4} + 15 \, a^{12} c^{22} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right )^{4} \mathrm {sgn}\relax (a)^{4} + 5 \, a^{12} c^{21} {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right )^{4} \mathrm {sgn}\relax (a)^{4}}{a^{15} c^{25} \mathrm {sgn}\left (\frac {1}{a x + 1}\right )^{5} \mathrm {sgn}\relax (a)^{5}}\right )} - \frac {16 \, \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)}{\sqrt {-c} c^{2}}}{40 \, {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(a^3*(5/(a^3*c^2*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a)) - (a^12*(c - 2*c/(a*x + 1))^2*c^20*sqr
t(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))^4*sgn(a)^4 + 15*a^12*c^22*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))^4*s
gn(a)^4 + 5*a^12*c^21*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x + 1))^4*sgn(a)^4)/(a^15*c^25*sgn(1/(a*x + 1))^5*sg
n(a)^5)) - 16*sgn(1/(a*x + 1))*sgn(a)/(sqrt(-c)*c^2))/abs(a)

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maple [A]  time = 0.03, size = 47, normalized size = 0.63 \[ \frac {\left (a x -1\right )^{2} \left (2 x^{3} a^{3}+4 a^{2} x^{2}+a x -2\right )}{5 \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 79, normalized size = 1.05 \[ -\frac {2}{5 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x + {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a c\right )}} + \frac {2 \, x}{5 \, \sqrt {-a^{2} c x^{2} + c} c^{2}} + \frac {x}{5 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.09, size = 56, normalized size = 0.75 \[ -\frac {\sqrt {c-a^2\,c\,x^2}\,\left (2\,a^3\,x^3+4\,a^2\,x^2+a\,x-2\right )}{5\,a\,c^3\,\left (a\,x-1\right )\,{\left (a\,x+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a x}{a^{5} c^{2} x^{5} \sqrt {- a^{2} c x^{2} + c} + a^{4} c^{2} x^{4} \sqrt {- a^{2} c x^{2} + c} - 2 a^{3} c^{2} x^{3} \sqrt {- a^{2} c x^{2} + c} - 2 a^{2} c^{2} x^{2} \sqrt {- a^{2} c x^{2} + c} + a c^{2} x \sqrt {- a^{2} c x^{2} + c} + c^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \left (- \frac {1}{a^{5} c^{2} x^{5} \sqrt {- a^{2} c x^{2} + c} + a^{4} c^{2} x^{4} \sqrt {- a^{2} c x^{2} + c} - 2 a^{3} c^{2} x^{3} \sqrt {- a^{2} c x^{2} + c} - 2 a^{2} c^{2} x^{2} \sqrt {- a^{2} c x^{2} + c} + a c^{2} x \sqrt {- a^{2} c x^{2} + c} + c^{2} \sqrt {- a^{2} c x^{2} + c}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x/(a**5*c**2*x**5*sqrt(-a**2*c*x**2 + c) + a**4*c**2*x**4*sqrt(-a**2*c*x**2 + c) - 2*a**3*c**2*x**
3*sqrt(-a**2*c*x**2 + c) - 2*a**2*c**2*x**2*sqrt(-a**2*c*x**2 + c) + a*c**2*x*sqrt(-a**2*c*x**2 + c) + c**2*sq
rt(-a**2*c*x**2 + c)), x) - Integral(-1/(a**5*c**2*x**5*sqrt(-a**2*c*x**2 + c) + a**4*c**2*x**4*sqrt(-a**2*c*x
**2 + c) - 2*a**3*c**2*x**3*sqrt(-a**2*c*x**2 + c) - 2*a**2*c**2*x**2*sqrt(-a**2*c*x**2 + c) + a*c**2*x*sqrt(-
a**2*c*x**2 + c) + c**2*sqrt(-a**2*c*x**2 + c)), x)

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