\(\int e^{-3 \coth ^{-1}(a x)} (c-a^2 c x^2)^{7/2} \, dx\) [660]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {2 (1-a x)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}-\frac {4 (1-a x)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}+\frac {(1-a x)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \]

[Out]

2/3*(-a*x+1)^6*(-a^2*c*x^2+c)^(7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7-4/7*(-a*x+1)^7*(-a^2*c*x^2+c)^(7/2)/a^8/(1-1/a
^2/x^2)^(7/2)/x^7+1/8*(-a*x+1)^8*(-a^2*c*x^2+c)^(7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6327, 6328, 45} \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {(1-a x)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {4 (1-a x)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}+\frac {2 (1-a x)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}} \]

[In]

Int[(c - a^2*c*x^2)^(7/2)/E^(3*ArcCoth[a*x]),x]

[Out]

(2*(1 - a*x)^6*(c - a^2*c*x^2)^(7/2))/(3*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) - (4*(1 - a*x)^7*(c - a^2*c*x^2)^(7/
2))/(7*a^8*(1 - 1/(a^2*x^2))^(7/2)*x^7) + ((1 - a*x)^8*(c - a^2*c*x^2)^(7/2))/(8*a^8*(1 - 1/(a^2*x^2))^(7/2)*x
^7)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6327

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

Rule 6328

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c-a^2 c x^2\right )^{7/2} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \, dx}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \\ & = \frac {\left (c-a^2 c x^2\right )^{7/2} \int (-1+a x)^5 (1+a x)^2 \, dx}{a^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \\ & = \frac {\left (c-a^2 c x^2\right )^{7/2} \int \left (4 (-1+a x)^5+4 (-1+a x)^6+(-1+a x)^7\right ) \, dx}{a^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \\ & = \frac {2 (1-a x)^6 \left (c-a^2 c x^2\right )^{7/2}}{3 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}-\frac {4 (1-a x)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}+\frac {(1-a x)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.44 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {c^3 (-1+a x)^6 \left (37+54 a x+21 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{168 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \]

[In]

Integrate[(c - a^2*c*x^2)^(7/2)/E^(3*ArcCoth[a*x]),x]

[Out]

-1/168*(c^3*(-1 + a*x)^6*(37 + 54*a*x + 21*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.70

method	result	size
gosper	\(\frac {x \left (21 a^{7} x^{7}-72 a^{6} x^{6}+28 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-56 a^{2} x^{2}+252 a x -168\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{168 \left (a x +1\right )^{2} \left (a x -1\right )^{5}}\)	\(100\)
default	\(-\frac {\left (21 a^{7} x^{7}-72 a^{6} x^{6}+28 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-56 a^{2} x^{2}+252 a x -168\right ) x \,c^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{168 \left (a x -1\right )^{2}}\)	\(102\)

[In]

int((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/168*x*(21*a^7*x^7-72*a^6*x^6+28*a^5*x^5+168*a^4*x^4-210*a^3*x^3-56*a^2*x^2+252*a*x-168)*(-a^2*c*x^2+c)^(7/2)
*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)^2/(a*x-1)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {{\left (21 \, a^{7} c^{3} x^{8} - 72 \, a^{6} c^{3} x^{7} + 28 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} - 210 \, a^{3} c^{3} x^{4} - 56 \, a^{2} c^{3} x^{3} + 252 \, a c^{3} x^{2} - 168 \, c^{3} x\right )} \sqrt {-a^{2} c}}{168 \, a} \]

[In]

integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

-1/168*(21*a^7*c^3*x^8 - 72*a^6*c^3*x^7 + 28*a^5*c^3*x^6 + 168*a^4*c^3*x^5 - 210*a^3*c^3*x^4 - 56*a^2*c^3*x^3
+ 252*a*c^3*x^2 - 168*c^3*x)*sqrt(-a^2*c)/a

Sympy [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\text {Timed out} \]

[In]

integrate((-a**2*c*x**2+c)**(7/2)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.21 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {{\left (21 \, a^{9} \sqrt {-c} c^{3} x^{9} - 51 \, a^{8} \sqrt {-c} c^{3} x^{8} - 44 \, a^{7} \sqrt {-c} c^{3} x^{7} + 196 \, a^{6} \sqrt {-c} c^{3} x^{6} - 42 \, a^{5} \sqrt {-c} c^{3} x^{5} - 266 \, a^{4} \sqrt {-c} c^{3} x^{4} + 196 \, a^{3} \sqrt {-c} c^{3} x^{3} + 84 \, a^{2} \sqrt {-c} c^{3} x^{2} + 168 \, \sqrt {-c} c^{3}\right )} {\left (a x - 1\right )}^{2}}{168 \, {\left (a^{3} x^{2} - 2 \, a^{2} x + a\right )} {\left (a x + 1\right )}} \]

[In]

integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

-1/168*(21*a^9*sqrt(-c)*c^3*x^9 - 51*a^8*sqrt(-c)*c^3*x^8 - 44*a^7*sqrt(-c)*c^3*x^7 + 196*a^6*sqrt(-c)*c^3*x^6
- 42*a^5*sqrt(-c)*c^3*x^5 - 266*a^4*sqrt(-c)*c^3*x^4 + 196*a^3*sqrt(-c)*c^3*x^3 + 84*a^2*sqrt(-c)*c^3*x^2 + 1
68*sqrt(-c)*c^3)*(a*x - 1)^2/((a^3*x^2 - 2*a^2*x + a)*(a*x + 1))

Giac [F]

\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(7/2)*((a*x - 1)/(a*x + 1))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int {\left (c-a^2\,c\,x^2\right )}^{7/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

[In]

int((c - a^2*c*x^2)^(7/2)*((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

int((c - a^2*c*x^2)^(7/2)*((a*x - 1)/(a*x + 1))^(3/2), x)