3.7.22 \(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\) [622]

Optimal. Leaf size=277 \[ \frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{24 (1-a x)^3 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x)^2 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \tanh ^{-1}(a x)}{16 \left (c-a^2 c x^2\right )^{7/2}} \]

[Out]

1/24*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)+3/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)-1/32*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)+5/16*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 = 0.16, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6327, 6328, 46, 213} \begin {gather*} \frac {3 a^6 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}{16 (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 (1-a x)^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}}{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}}{24 (1-a x)^3 \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)}{16 \left (c-a^2 c x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^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)/(24*(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)) + (3*a^6*(1 - 1/(a^2*x^2))^(7/2)*x^7)/(16*(1 - a*x)*(c - a^2*c*x^2)^(7
/2)) - (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])/(16*(c - a^2*c*x^
2)^(7/2))

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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*} \int \frac {e^{\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^{\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)^4 (1+a x)^3} \, 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}{8 (-1+a x)^4}-\frac {3}{16 (-1+a x)^3}+\frac {3}{16 (-1+a x)^2}+\frac {1}{16 (1+a x)^3}+\frac {1}{8 (1+a x)^2}-\frac {5}{16 \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}{24 (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 {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (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}{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}{16 \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}{24 (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 {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (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}{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)}{16 \left (c-a^2 c x^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 101, normalized size = 0.36 \begin {gather*} -\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-8-25 a x+25 a^2 x^2+15 a^3 x^3-15 a^4 x^4+15 (-1+a x)^3 (1+a x)^2 \tanh ^{-1}(a x)\right )}{48 c^3 (-1+a x)^3 (1+a x)^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(Sqrt[1 - 1/(a^2*x^2)]*x*(-8 - 25*a*x + 25*a^2*x^2 + 15*a^3*x^3 - 15*a^4*x^4 + 15*(-1 + a*x)^3*(1 + a*x)
^2*ArcTanh[a*x]))/(c^3*(-1 + a*x)^3*(1 + a*x)^2*Sqrt[c - a^2*c*x^2])

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Maple [A]
time = 0.10, size = 241, normalized size = 0.87

method	result	size
default	\(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x +1\right ) a^{5} x^{5}-15 x^{5} \ln \left (a x -1\right ) a^{5}-15 \ln \left (a x +1\right ) a^{4} x^{4}+15 x^{4} \ln \left (a x -1\right ) a^{4}-30 a^{4} x^{4}-30 \ln \left (a x +1\right ) a^{3} x^{3}+30 x^{3} \ln \left (a x -1\right ) a^{3}+30 a^{3} x^{3}+30 \ln \left (a x +1\right ) a^{2} x^{2}-30 x^{2} \ln \left (a x -1\right ) a^{2}+50 a^{2} x^{2}+15 \ln \left (a x +1\right ) a x -15 x \ln \left (a x -1\right ) a -50 a x -15 \ln \left (a x +1\right )+15 \ln \left (a x -1\right )-16\right )}{96 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a \left (a x +1\right )^{2}}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)^2*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x+1)*a^5*x^5-15*x^5*ln(a*x-1)*a^5-15*ln
(a*x+1)*a^4*x^4+15*x^4*ln(a*x-1)*a^4-30*a^4*x^4-30*ln(a*x+1)*a^3*x^3+30*x^3*ln(a*x-1)*a^3+30*a^3*x^3+30*ln(a*x
+1)*a^2*x^2-30*x^2*ln(a*x-1)*a^2+50*a^2*x^2+15*ln(a*x+1)*a*x-15*x*ln(a*x-1)*a-50*a*x-15*ln(a*x+1)+15*ln(a*x-1)
-16)/(a^2*x^2-1)/c^4/a/(a*x+1)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 191, normalized size = 0.69 \begin {gather*} -\frac {15 \, {\left (a^{6} x^{5} - a^{5} x^{4} - 2 \, a^{4} x^{3} + 2 \, a^{3} x^{2} + 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} - 15 \, a^{3} x^{3} - 25 \, a^{2} x^{2} + 25 \, a x + 8\right )} \sqrt {-a^{2} c}}{96 \, {\left (a^{7} c^{4} x^{5} - a^{6} c^{4} x^{4} - 2 \, a^{5} c^{4} x^{3} + 2 \, a^{4} c^{4} x^{2} + a^{3} c^{4} x - a^{2} c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(15*(a^6*x^5 - a^5*x^4 - 2*a^4*x^3 + 2*a^3*x^2 + a^2*x - a)*sqrt(-c)*log((a^2*c*x^2 + 2*sqrt(-a^2*c)*sqr
t(-c)*x + c)/(a^2*x^2 - 1)) + 2*(15*a^4*x^4 - 15*a^3*x^3 - 25*a^2*x^2 + 25*a*x + 8)*sqrt(-a^2*c))/(a^7*c^4*x^5
- a^6*c^4*x^4 - 2*a^5*c^4*x^3 + 2*a^4*c^4*x^2 + a^3*c^4*x - a^2*c^4)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3880 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (c-a^2\,c\,x^2\right )}^{7/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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