3.7.49 \(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [649]

Optimal. Leaf size=183 \[ \frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {3 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 183, 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 = {6327, 6328, 46, 213} \begin {gather*} \frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{4 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {3 a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(
1 + a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(4*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (3
*a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2))

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 )^{5/2}} \, dx &=\frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{(-1+a x)^2 (1+a x)^3} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (\frac {1}{8 (-1+a x)^2}+\frac {1}{4 (1+a x)^3}+\frac {1}{4 (1+a x)^2}-\frac {3}{8 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {\left (3 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{8 \left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {3 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 81, normalized size = 0.44 \begin {gather*} \frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (2-3 a x-3 a^2 x^2+3 (-1+a x) (1+a x)^2 \tanh ^{-1}(a x)\right )}{8 (-1+a x) (c+a c x)^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2)),x]

[Out]

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

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Maple [A]
time = 0.10, size = 169, normalized size = 0.92

method	result	size
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x +1\right ) a^{3} x^{3}-3 x^{3} \ln \left (a x -1\right ) a^{3}+3 \ln \left (a x +1\right ) a^{2} x^{2}-3 x^{2} \ln \left (a x -1\right ) a^{2}-6 a^{2} x^{2}-3 \ln \left (a x +1\right ) a x +3 x \ln \left (a x -1\right ) a -6 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )+4\right )}{16 \left (a x +1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x -1\right )}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/16*((a*x-1)/(a*x+1))^(1/2)/(a*x+1)*(-c*(a^2*x^2-1))^(1/2)*(3*ln(a*x+1)*a^3*x^3-3*x^3*ln(a*x-1)*a^3+3*ln(a*x
+1)*a^2*x^2-3*x^2*ln(a*x-1)*a^2-6*a^2*x^2-3*ln(a*x+1)*a*x+3*x*ln(a*x-1)*a-6*a*x-3*ln(a*x+1)+3*ln(a*x-1)+4)/(a^
2*x^2-1)/c^3/a/(a*x-1)

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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(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(-a^2*c*x^2 + c)^(5/2), x)

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Fricas [A]
time = 0.36, size = 137, normalized size = 0.75 \begin {gather*} -\frac {3 \, {\left (a^{4} x^{3} + 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 (3 \, a^{2} x^{2} + 3 \, a x - 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{5} c^{3} x^{3} + a^{4} c^{3} x^{2} - a^{3} c^{3} x - a^{2} c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(3*(a^4*x^3 + a^3*x^2 - 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*(3*a^2*x^2 + 3*a*x - 2)*sqrt(-a^2*c))/(a^5*c^3*x^3 + a^4*c^3*x^2 - a^3*c^3*x - a^2*c^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(1/2)/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

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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(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(-a^2*c*x^2 + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(5/2),x)

[Out]

int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(5/2), x)

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