3.658 \(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{9/2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac {16 x}{45 c^4 \sqrt {c-a^2 c x^2}}-\frac {8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}+\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}} \]

[Out]

2/9*(-a*x+1)/a/(-a^2*c*x^2+c)^(9/2)-1/9*x/c/(-a^2*c*x^2+c)^(7/2)-2/15*x/c^2/(-a^2*c*x^2+c)^(5/2)-8/45*x/c^3/(-
a^2*c*x^2+c)^(3/2)-16/45*x/c^4/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6167, 6142, 653, 192, 191} \[ -\frac {16 x}{45 c^4 \sqrt {c-a^2 c x^2}}-\frac {8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}+\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^(9/2)),x]

[Out]

(2*(1 - a*x))/(9*a*(c - a^2*c*x^2)^(9/2)) - x/(9*c*(c - a^2*c*x^2)^(7/2)) - (2*x)/(15*c^2*(c - a^2*c*x^2)^(5/2
)) - (8*x)/(45*c^3*(c - a^2*c*x^2)^(3/2)) - (16*x)/(45*c^4*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]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{9/2}} \, dx &=-\int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{9/2}} \, dx\\ &=-\left (c \int \frac {(1-a x)^2}{\left (c-a^2 c x^2\right )^{11/2}} \, dx\right )\\ &=\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac {7}{9} \int \frac {1}{\left (c-a^2 c x^2\right )^{9/2}} \, dx\\ &=\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac {2 \int \frac {1}{\left (c-a^2 c x^2\right )^{7/2}} \, dx}{3 c}\\ &=\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac {2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {8 \int \frac {1}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{15 c^2}\\ &=\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac {2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac {16 \int \frac {1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{45 c^3}\\ &=\frac {2 (1-a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac {x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac {2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac {16 x}{45 c^4 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 112, normalized size = 0.93 \[ \frac {\sqrt {1-a^2 x^2} \left (-16 a^7 x^7-32 a^6 x^6+24 a^5 x^5+80 a^4 x^4+10 a^3 x^3-60 a^2 x^2-25 a x+10\right )}{45 a c^4 (1-a x)^{5/2} (a x+1)^{9/2} \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^(9/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(10 - 25*a*x - 60*a^2*x^2 + 10*a^3*x^3 + 80*a^4*x^4 + 24*a^5*x^5 - 32*a^6*x^6 - 16*a^7*x^7)
)/(45*a*c^4*(1 - a*x)^(5/2)*(1 + a*x)^(9/2)*Sqrt[c - a^2*c*x^2])

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fricas [A]  time = 1.32, size = 152, normalized size = 1.26 \[ \frac {{\left (16 \, a^{7} x^{7} + 32 \, a^{6} x^{6} - 24 \, a^{5} x^{5} - 80 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 60 \, a^{2} x^{2} + 25 \, a x - 10\right )} \sqrt {-a^{2} c x^{2} + c}}{45 \, {\left (a^{9} c^{5} x^{8} + 2 \, a^{8} c^{5} x^{7} - 2 \, a^{7} c^{5} x^{6} - 6 \, a^{6} c^{5} x^{5} + 6 \, a^{4} c^{5} x^{3} + 2 \, a^{3} c^{5} x^{2} - 2 \, a^{2} c^{5} x - a c^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="fricas")

[Out]

1/45*(16*a^7*x^7 + 32*a^6*x^6 - 24*a^5*x^5 - 80*a^4*x^4 - 10*a^3*x^3 + 60*a^2*x^2 + 25*a*x - 10)*sqrt(-a^2*c*x
^2 + c)/(a^9*c^5*x^8 + 2*a^8*c^5*x^7 - 2*a^7*c^5*x^6 - 6*a^6*c^5*x^5 + 6*a^4*c^5*x^3 + 2*a^3*c^5*x^2 - 2*a^2*c
^5*x - a*c^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x - 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{2}} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="giac")

[Out]

integrate((a*x - 1)/((-a^2*c*x^2 + c)^(9/2)*(a*x + 1)), x)

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maple [A]  time = 0.04, size = 80, normalized size = 0.66 \[ -\frac {\left (a x -1\right )^{2} \left (16 a^{7} x^{7}+32 x^{6} a^{6}-24 x^{5} a^{5}-80 x^{4} a^{4}-10 x^{3} a^{3}+60 a^{2} x^{2}+25 a x -10\right )}{45 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(a*x-1)/(-a^2*c*x^2+c)^(9/2),x)

[Out]

-1/45*(a*x-1)^2*(16*a^7*x^7+32*a^6*x^6-24*a^5*x^5-80*a^4*x^4-10*a^3*x^3+60*a^2*x^2+25*a*x-10)/a/(-a^2*c*x^2+c)
^(9/2)

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maxima [A]  time = 0.33, size = 117, normalized size = 0.97 \[ \frac {2}{9 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} a^{2} c x + {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} a c\right )}} - \frac {16 \, x}{45 \, \sqrt {-a^{2} c x^{2} + c} c^{4}} - \frac {8 \, x}{45 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {2 \, x}{15 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c^{2}} - \frac {x}{9 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="maxima")

[Out]

2/9/((-a^2*c*x^2 + c)^(7/2)*a^2*c*x + (-a^2*c*x^2 + c)^(7/2)*a*c) - 16/45*x/(sqrt(-a^2*c*x^2 + c)*c^4) - 8/45*
x/((-a^2*c*x^2 + c)^(3/2)*c^3) - 2/15*x/((-a^2*c*x^2 + c)^(5/2)*c^2) - 1/9*x/((-a^2*c*x^2 + c)^(7/2)*c)

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mupad [B]  time = 1.49, size = 177, normalized size = 1.46 \[ \frac {5\,\sqrt {c-a^2\,c\,x^2}}{144\,a\,c^5\,{\left (a\,x+1\right )}^4}+\frac {\sqrt {c-a^2\,c\,x^2}}{72\,a\,c^5\,{\left (a\,x+1\right )}^5}+\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {31\,x}{120\,c^5}-\frac {5}{24\,a\,c^5}\right )}{{\left (a\,x-1\right )}^3\,{\left (a\,x+1\right )}^3}-\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {8\,x}{45\,c^5}+\frac {5}{144\,a\,c^5}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}+\frac {16\,x\,\sqrt {c-a^2\,c\,x^2}}{45\,c^5\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x - 1)/((c - a^2*c*x^2)^(9/2)*(a*x + 1)),x)

[Out]

(5*(c - a^2*c*x^2)^(1/2))/(144*a*c^5*(a*x + 1)^4) + (c - a^2*c*x^2)^(1/2)/(72*a*c^5*(a*x + 1)^5) + ((c - a^2*c
*x^2)^(1/2)*((31*x)/(120*c^5) - 5/(24*a*c^5)))/((a*x - 1)^3*(a*x + 1)^3) - ((c - a^2*c*x^2)^(1/2)*((8*x)/(45*c
^5) + 5/(144*a*c^5)))/((a*x - 1)^2*(a*x + 1)^2) + (16*x*(c - a^2*c*x^2)^(1/2))/(45*c^5*(a*x - 1)*(a*x + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x - 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {9}{2}} \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a**2*c*x**2+c)**(9/2),x)

[Out]

Integral((a*x - 1)/((-c*(a*x - 1)*(a*x + 1))**(9/2)*(a*x + 1)), x)

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