3.631 \(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=97 \[ -\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {2 (a x+1)}{7 a \left (c-a^2 c x^2\right )^{7/2}} \]

[Out]

-2/7*(a*x+1)/a/(-a^2*c*x^2+c)^(7/2)-1/7*x/c/(-a^2*c*x^2+c)^(5/2)-4/21*x/c^2/(-a^2*c*x^2+c)^(3/2)-8/21*x/c^3/(-
a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6167, 6141, 653, 192, 191} \[ -\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {2 (a x+1)}{7 a \left (c-a^2 c x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(7/2),x]

[Out]

(-2*(1 + a*x))/(7*a*(c - a^2*c*x^2)^(7/2)) - x/(7*c*(c - a^2*c*x^2)^(5/2)) - (4*x)/(21*c^2*(c - a^2*c*x^2)^(3/
2)) - (8*x)/(21*c^3*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 6141

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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]) && IGt
Q[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 )^{7/2}} \, dx &=-\int \frac {e^{2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\\ &=-\left (c \int \frac {(1+a x)^2}{\left (c-a^2 c x^2\right )^{9/2}} \, dx\right )\\ &=-\frac {2 (1+a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {5}{7} \int \frac {1}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\\ &=-\frac {2 (1+a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {4 \int \frac {1}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{7 c}\\ &=-\frac {2 (1+a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {8 \int \frac {1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{21 c^2}\\ &=-\frac {2 (1+a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 96, normalized size = 0.99 \[ -\frac {\sqrt {1-a^2 x^2} \left (-8 a^5 x^5+16 a^4 x^4+4 a^3 x^3-24 a^2 x^2+9 a x+6\right )}{21 a c^3 (1-a x)^{7/2} (a x+1)^{3/2} \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(7/2),x]

[Out]

-1/21*(Sqrt[1 - a^2*x^2]*(6 + 9*a*x - 24*a^2*x^2 + 4*a^3*x^3 + 16*a^4*x^4 - 8*a^5*x^5))/(a*c^3*(1 - a*x)^(7/2)
*(1 + a*x)^(3/2)*Sqrt[c - a^2*c*x^2])

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fricas [A]  time = 0.82, size = 124, normalized size = 1.28 \[ \frac {{\left (8 \, a^{5} x^{5} - 16 \, a^{4} x^{4} - 4 \, a^{3} x^{3} + 24 \, a^{2} x^{2} - 9 \, a x - 6\right )} \sqrt {-a^{2} c x^{2} + c}}{21 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

1/21*(8*a^5*x^5 - 16*a^4*x^4 - 4*a^3*x^3 + 24*a^2*x^2 - 9*a*x - 6)*sqrt(-a^2*c*x^2 + c)/(a^7*c^4*x^6 - 2*a^6*c
^4*x^5 - a^5*c^4*x^4 + 4*a^4*c^4*x^3 - a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(7/2)*(a*x - 1)), x)

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maple [A]  time = 0.04, size = 64, normalized size = 0.66 \[ \frac {\left (a x +1\right )^{2} \left (8 x^{5} a^{5}-16 x^{4} a^{4}-4 x^{3} a^{3}+24 a^{2} x^{2}-9 a x -6\right )}{21 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(a*x-1)/(-a^2*c*x^2+c)^(7/2),x)

[Out]

1/21*(a*x+1)^2*(8*a^5*x^5-16*a^4*x^4-4*a^3*x^3+24*a^2*x^2-9*a*x-6)/a/(-a^2*c*x^2+c)^(7/2)

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maxima [A]  time = 0.32, size = 99, normalized size = 1.02 \[ \frac {2}{7 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a^{2} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a c\right )}} - \frac {8 \, x}{21 \, \sqrt {-a^{2} c x^{2} + c} c^{3}} - \frac {4 \, x}{21 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {x}{7 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

2/7/((-a^2*c*x^2 + c)^(5/2)*a^2*c*x - (-a^2*c*x^2 + c)^(5/2)*a*c) - 8/21*x/(sqrt(-a^2*c*x^2 + c)*c^3) - 4/21*x
/((-a^2*c*x^2 + c)^(3/2)*c^2) - 1/7*x/((-a^2*c*x^2 + c)^(5/2)*c)

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mupad [B]  time = 1.45, size = 134, normalized size = 1.38 \[ \frac {\sqrt {c-a^2\,c\,x^2}}{14\,a\,c^4\,{\left (a\,x-1\right )}^3}-\frac {\sqrt {c-a^2\,c\,x^2}}{28\,a\,c^4\,{\left (a\,x-1\right )}^4}-\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {11\,x}{42\,c^4}+\frac {5}{28\,a\,c^4}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}+\frac {8\,x\,\sqrt {c-a^2\,c\,x^2}}{21\,c^4\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/((c - a^2*c*x^2)^(7/2)*(a*x - 1)),x)

[Out]

(c - a^2*c*x^2)^(1/2)/(14*a*c^4*(a*x - 1)^3) - (c - a^2*c*x^2)^(1/2)/(28*a*c^4*(a*x - 1)^4) - ((c - a^2*c*x^2)
^(1/2)*((11*x)/(42*c^4) + 5/(28*a*c^4)))/((a*x - 1)^2*(a*x + 1)^2) + (8*x*(c - a^2*c*x^2)^(1/2))/(21*c^4*(a*x
- 1)*(a*x + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x - 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a**2*c*x**2+c)**(7/2),x)

[Out]

Integral((a*x + 1)/((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a*x - 1)), x)

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