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

Optimal. Leaf size=275 \[ \frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{32 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{32 (a x+1)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{12 (a x+1)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{16 (a x+1)^4 \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \tanh ^{-1}(a x)}{32 \left (c-a^2 c x^2\right )^{7/2}} \]

[Out]

1/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)-1/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+1)^4/(-a^
2*c*x^2+c)^(7/2)-1/12*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+1)^3/(-a^2*c*x^2+c)^(7/2)-3/32*a^6*(1-1/a^2/x^2)^(7/2)*
x^7/(a*x+1)^2/(-a^2*c*x^2+c)^(7/2)-1/8*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+1)/(-a^2*c*x^2+c)^(7/2)+5/32*a^6*(1-1/
a^2/x^2)^(7/2)*x^7*arctanh(a*x)/(-a^2*c*x^2+c)^(7/2)

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Rubi [A]  time = 0.22, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6192, 6193, 44, 207} \[ \frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{32 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{32 (a x+1)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{12 (a x+1)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{16 (a x+1)^4 \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \tanh ^{-1}(a x)}{32 \left (c-a^2 c x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(7/2)),x]

[Out]

(a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7)/(32*(1 - a*x)*(c - a^2*c*x^2)^(7/2)) - (a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7)/(16
*(1 + a*x)^4*(c - a^2*c*x^2)^(7/2)) - (a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7)/(12*(1 + a*x)^3*(c - a^2*c*x^2)^(7/2))
- (3*a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7)/(32*(1 + a*x)^2*(c - a^2*c*x^2)^(7/2)) - (a^6*(1 - 1/(a^2*x^2))^(7/2)*x
^7)/(8*(1 + a*x)*(c - a^2*c*x^2)^(7/2)) + (5*a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7*ArcTanh[a*x])/(32*(c - a^2*c*x^2)
^(7/2))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6192

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 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), 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*} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx &=\frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7\right ) \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \, dx}{\left (c-a^2 c x^2\right )^{7/2}}\\ &=\frac {\left (a^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7\right ) \int \frac {1}{(-1+a x)^2 (1+a x)^5} \, dx}{\left (c-a^2 c x^2\right )^{7/2}}\\ &=\frac {\left (a^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7\right ) \int \left (\frac {1}{32 (-1+a x)^2}+\frac {1}{4 (1+a x)^5}+\frac {1}{4 (1+a x)^4}+\frac {3}{16 (1+a x)^3}+\frac {1}{8 (1+a x)^2}-\frac {5}{32 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{7/2}}\\ &=\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1+a x)^4 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{12 (1+a x)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {\left (5 a^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{32 \left (c-a^2 c x^2\right )^{7/2}}\\ &=\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1+a x)^4 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{12 (1+a x)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \tanh ^{-1}(a x)}{32 \left (c-a^2 c x^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 99, normalized size = 0.36 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (-15 a^4 x^4-45 a^3 x^3-35 a^2 x^2+15 a x+15 (a x-1) (a x+1)^4 \tanh ^{-1}(a x)+32\right )}{96 c^3 (a x-1) (a x+1)^4 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(7/2)),x]

[Out]

-1/96*(Sqrt[1 - 1/(a^2*x^2)]*x*(32 + 15*a*x - 35*a^2*x^2 - 45*a^3*x^3 - 15*a^4*x^4 + 15*(-1 + a*x)*(1 + a*x)^4
*ArcTanh[a*x]))/(c^3*(-1 + a*x)*(1 + a*x)^4*Sqrt[c - a^2*c*x^2])

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fricas [A]  time = 1.30, size = 193, normalized size = 0.70 \[ -\frac {15 \, {\left (a^{6} x^{5} + 3 \, a^{5} x^{4} + 2 \, a^{4} x^{3} - 2 \, a^{3} x^{2} - 3 \, a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (15 \, a^{4} x^{4} + 45 \, a^{3} x^{3} + 35 \, a^{2} x^{2} - 15 \, a x - 32\right )} \sqrt {-a^{2} c}}{192 \, {\left (a^{7} c^{4} x^{5} + 3 \, a^{6} c^{4} x^{4} + 2 \, a^{5} c^{4} x^{3} - 2 \, a^{4} c^{4} x^{2} - 3 \, a^{3} c^{4} x - a^{2} c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(15*(a^6*x^5 + 3*a^5*x^4 + 2*a^4*x^3 - 2*a^3*x^2 - 3*a^2*x - a)*sqrt(-c)*log((a^2*c*x^2 + 2*sqrt(-a^2*c
)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) + 2*(15*a^4*x^4 + 45*a^3*x^3 + 35*a^2*x^2 - 15*a*x - 32)*sqrt(-a^2*c))/(a^7*c
^4*x^5 + 3*a^6*c^4*x^4 + 2*a^5*c^4*x^3 - 2*a^4*c^4*x^2 - 3*a^3*c^4*x - a^2*c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 241, normalized size = 0.88 \[ -\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x -1\right ) x^{5} a^{5}-15 \ln \left (a x +1\right ) x^{5} a^{5}+45 \ln \left (a x -1\right ) x^{4} a^{4}-45 \ln \left (a x +1\right ) x^{4} a^{4}+30 x^{4} a^{4}+30 \ln \left (a x -1\right ) x^{3} a^{3}-30 a^{3} x^{3} \ln \left (a x +1\right )+90 x^{3} a^{3}-30 \ln \left (a x -1\right ) x^{2} a^{2}+30 \ln \left (a x +1\right ) x^{2} a^{2}+70 a^{2} x^{2}-45 \ln \left (a x -1\right ) x a +45 a x \ln \left (a x +1\right )-30 a x -15 \ln \left (a x -1\right )+15 \ln \left (a x +1\right )-64\right )}{192 \left (a x +1\right )^{2} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)^2/(a*x-1)^2*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x-1)*x^5*a^5-15*ln(a*x+1)*x
^5*a^5+45*ln(a*x-1)*x^4*a^4-45*ln(a*x+1)*x^4*a^4+30*x^4*a^4+30*ln(a*x-1)*x^3*a^3-30*a^3*x^3*ln(a*x+1)+90*x^3*a
^3-30*ln(a*x-1)*x^2*a^2+30*ln(a*x+1)*x^2*a^2+70*a^2*x^2-45*ln(a*x-1)*x*a+45*a*x*ln(a*x+1)-30*a*x-15*ln(a*x-1)+
15*ln(a*x+1)-64)/(a^2*x^2-1)/c^4/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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