3.269 \(\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{9/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{1}{4 a c^4 \sqrt{c-a c x}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{1}{5 a c^2 (c-a c x)^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a c^{9/2}} \]

[Out]

-1/(5*a*c^2*(c - a*c*x)^(5/2)) - 1/(6*a*c^3*(c - a*c*x)^(3/2)) - 1/(4*a*c^4*Sqrt[c - a*c*x]) + ArcTanh[Sqrt[c
- a*c*x]/(Sqrt[2]*Sqrt[c])]/(4*Sqrt[2]*a*c^(9/2))

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Rubi [A]  time = 0.126211, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6167, 6130, 21, 51, 63, 206} \[ -\frac{1}{4 a c^4 \sqrt{c-a c x}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{1}{5 a c^2 (c-a c x)^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

-1/(5*a*c^2*(c - a*c*x)^(5/2)) - 1/(6*a*c^3*(c - a*c*x)^(3/2)) - 1/(4*a*c^4*Sqrt[c - a*c*x]) + ArcTanh[Sqrt[c
- a*c*x]/(Sqrt[2]*Sqrt[c])]/(4*Sqrt[2]*a*c^(9/2))

Rule 6167

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]

Rule 6130

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{9/2}} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^{9/2}} \, dx\\ &=-\int \frac{1-a x}{(1+a x) (c-a c x)^{9/2}} \, dx\\ &=-\frac{\int \frac{1}{(1+a x) (c-a c x)^{7/2}} \, dx}{c}\\ &=-\frac{1}{5 a c^2 (c-a c x)^{5/2}}-\frac{\int \frac{1}{(1+a x) (c-a c x)^{5/2}} \, dx}{2 c^2}\\ &=-\frac{1}{5 a c^2 (c-a c x)^{5/2}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{\int \frac{1}{(1+a x) (c-a c x)^{3/2}} \, dx}{4 c^3}\\ &=-\frac{1}{5 a c^2 (c-a c x)^{5/2}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{1}{4 a c^4 \sqrt{c-a c x}}-\frac{\int \frac{1}{(1+a x) \sqrt{c-a c x}} \, dx}{8 c^4}\\ &=-\frac{1}{5 a c^2 (c-a c x)^{5/2}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{1}{4 a c^4 \sqrt{c-a c x}}+\frac{\operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{c}} \, dx,x,\sqrt{c-a c x}\right )}{4 a c^5}\\ &=-\frac{1}{5 a c^2 (c-a c x)^{5/2}}-\frac{1}{6 a c^3 (c-a c x)^{3/2}}-\frac{1}{4 a c^4 \sqrt{c-a c x}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a c^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0341427, size = 39, normalized size = 0.38 \[ -\frac{\text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\frac{1}{2} (1-a x)\right )}{5 a c^2 (c-a c x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

-Hypergeometric2F1[-5/2, 1, -3/2, (1 - a*x)/2]/(5*a*c^2*(c - a*c*x)^(5/2))

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Maple [A]  time = 0.051, size = 78, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{ac} \left ( 1/8\,{\frac{1}{\sqrt{-acx+c}{c}^{3}}}+1/12\,{\frac{1}{ \left ( -acx+c \right ) ^{3/2}{c}^{2}}}+1/10\,{\frac{1}{c \left ( -acx+c \right ) ^{5/2}}}-1/16\,{\frac{\sqrt{2}}{{c}^{7/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-acx+c}\sqrt{2}}{\sqrt{c}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(a*x-1)/(-a*c*x+c)^(9/2),x)

[Out]

-2/c/a*(1/8/c^3/(-a*c*x+c)^(1/2)+1/12/c^2/(-a*c*x+c)^(3/2)+1/10/c/(-a*c*x+c)^(5/2)-1/16/c^(7/2)*2^(1/2)*arctan
h(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69163, size = 599, normalized size = 5.76 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt{c} \log \left (\frac{a c x - 2 \, \sqrt{2} \sqrt{-a c x + c} \sqrt{c} - 3 \, c}{a x + 1}\right ) + 4 \,{\left (15 \, a^{2} x^{2} - 40 \, a x + 37\right )} \sqrt{-a c x + c}}{240 \,{\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}}, -\frac{15 \, \sqrt{2}{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) - 2 \,{\left (15 \, a^{2} x^{2} - 40 \, a x + 37\right )} \sqrt{-a c x + c}}{120 \,{\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/240*(15*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) -
3*c)/(a*x + 1)) + 4*(15*a^2*x^2 - 40*a*x + 37)*sqrt(-a*c*x + c))/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x -
a*c^5), -1/120*(15*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqr
t(-c)/c) - 2*(15*a^2*x^2 - 40*a*x + 37)*sqrt(-a*c*x + c))/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5)]

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Sympy [A]  time = 28.0085, size = 100, normalized size = 0.96 \begin{align*} - \frac{1}{5 a c^{2} \left (- a c x + c\right )^{\frac{5}{2}}} - \frac{1}{6 a c^{3} \left (- a c x + c\right )^{\frac{3}{2}}} - \frac{1}{4 a c^{4} \sqrt{- a c x + c}} - \frac{\sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \sqrt{- c}} \right )}}{8 a c^{4} \sqrt{- c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)**(9/2),x)

[Out]

-1/(5*a*c**2*(-a*c*x + c)**(5/2)) - 1/(6*a*c**3*(-a*c*x + c)**(3/2)) - 1/(4*a*c**4*sqrt(-a*c*x + c)) - sqrt(2)
*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/(8*a*c**4*sqrt(-c))

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Giac [A]  time = 1.14351, size = 126, normalized size = 1.21 \begin{align*} -\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{8 \, a \sqrt{-c} c^{4}} - \frac{15 \,{\left (a c x - c\right )}^{2} - 10 \,{\left (a c x - c\right )} c + 12 \, c^{2}}{60 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} a c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(9/2),x, algorithm="giac")

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^4) - 1/60*(15*(a*c*x - c)^2 - 10*(a*c
*x - c)*c + 12*c^2)/((a*c*x - c)^2*sqrt(-a*c*x + c)*a*c^4)