\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\) [657]

Optimal result

Integrand size = 24, antiderivative size = 98 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {2 (1-a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6302, 6277, 667, 198, 197} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}+\frac {2 (1-a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}} \]

[In]

Int[1/(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 197

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 198

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 667

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 6277

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 6302

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*} \text {integral}& = -\int \frac {e^{-2 \text {arctanh}(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*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.98 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {\sqrt {1-a^2 x^2} \left (-6+9 a x+24 a^2 x^2+4 a^3 x^3-16 a^4 x^4-8 a^5 x^5\right )}{21 a c^3 (1-a x)^{3/2} (1+a x)^{7/2} \sqrt {c-a^2 c x^2}} \]

[In]

Integrate[1/(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)^(3/2
)*(1 + a*x)^(7/2)*Sqrt[c - a^2*c*x^2])

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {\left (a x -1\right )^{2} \left (8 a^{5} x^{5}+16 a^{4} x^{4}-4 a^{3} x^{3}-24 a^{2} x^{2}-9 a x +6\right )}{21 \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} a}\)	\(64\)
trager	\(\frac {\left (8 a^{5} x^{5}+16 a^{4} x^{4}-4 a^{3} x^{3}-24 a^{2} x^{2}-9 a x +6\right ) \sqrt {-a^{2} c \,x^{2}+c}}{21 c^{4} \left (a x +1\right )^{4} \left (a x -1\right )^{2} a}\)	\(74\)
default	\(\frac {x}{5 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {-a^{2} c \,x^{2}+c}}}{c}-\frac {2 \left (-\frac {1}{7 a c \left (x +\frac {1}{a}\right ) \left (-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}+\frac {6 a \left (-\frac {-2 a^{2} c \left (x +\frac {1}{a}\right )+2 a c}{10 a^{2} c^{2} \left (-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 a^{2} c \left (x +\frac {1}{a}\right )+2 a c \right )}{15 a^{2} c^{2} \left (-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {4 \left (-2 a^{2} c \left (x +\frac {1}{a}\right )+2 a c \right )}{15 a^{2} c^{3} \sqrt {-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c}}}{c}\right )}{7}\right )}{a}\)	\(268\)

[In]

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

[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^2*c*x^2+c)^(7/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {{\left (8 \, a^{5} x^{5} + 16 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 24 \, a^{2} x^{2} - 9 \, a x + 6\right )} \sqrt {-a^{2} c x^{2} + c}}{21 \, {\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

integrate((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)

Sympy [F]

\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {a x - 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x + 1\right )}\, dx \]

[In]

integrate((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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {2}{7 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a^{2} c x + {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a c\right )}} - \frac {8 \, x}{21 \, \sqrt {-a^{2} c x^{2} + c} c^{3}} - \frac {4 \, x}{21 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {x}{7 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c} \]

[In]

integrate((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)

Giac [F]

\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {a x - 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x + 1\right )}} \,d x } \]

[In]

integrate((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)

Mupad [B] (verification not implemented)

Time = 4.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}}{14\,a\,c^4\,{\left (a\,x+1\right )}^3}+\frac {\sqrt {c-a^2\,c\,x^2}}{28\,a\,c^4\,{\left (a\,x+1\right )}^4}-\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {11\,x}{42\,c^4}-\frac {5}{28\,a\,c^4}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}+\frac {8\,x\,\sqrt {c-a^2\,c\,x^2}}{21\,c^4\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \]

[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))