3.267 \(\int \frac{e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{a c^2 \sqrt{c-a c x}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a c^{5/2}} \]

[Out]

1/(a*c^2*Sqrt[c - a*c*x]) - ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(5/2))

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Rubi [A]  time = 0.0604231, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6130, 21, 51, 63, 206} \[ \frac{1}{a c^2 \sqrt{c-a c x}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(a*c^2*Sqrt[c - a*c*x]) - ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(5/2))

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

Mathematica [C]  time = 0.0215347, size = 36, normalized size = 0.63 \[ \frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{1}{2} (1-a x)\right )}{a c^2 \sqrt{c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 50, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{ac} \left ( -1/4\,{\frac{\sqrt{2}}{{c}^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-acx+c}\sqrt{2}}{\sqrt{c}}} \right ) }+1/2\,{\frac{1}{c\sqrt{-acx+c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/c/a*(-1/4/c^(3/2)*2^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))+1/2/c/(-a*c*x+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(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54433, size = 360, normalized size = 6.32 \begin{align*} \left [\frac{\sqrt{2}{\left (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 \, \sqrt{-a c x + c}}{4 \,{\left (a^{2} c^{3} x - a c^{3}\right )}}, \frac{\sqrt{2}{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) - 2 \, \sqrt{-a c x + c}}{2 \,{\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*(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*sqrt(-a*
c*x + c))/(a^2*c^3*x - a*c^3), 1/2*(sqrt(2)*(a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c)
- 2*sqrt(-a*c*x + c))/(a^2*c^3*x - a*c^3)]

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Sympy [A]  time = 21.7705, size = 60, normalized size = 1.05 \begin{align*} \frac{1}{a c^{2} \sqrt{- a c x + c}} + \frac{\sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \sqrt{- c}} \right )}}{2 a c^{2} \sqrt{- c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*c**2*sqrt(-a*c*x + c)) + sqrt(2)*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/(2*a*c**2*sqrt(-c))

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Giac [A]  time = 1.25676, size = 72, normalized size = 1.26 \begin{align*} \frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{2 \, a \sqrt{-c} c^{2}} + \frac{1}{\sqrt{-a c x + c} a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^2) + 1/(sqrt(-a*c*x + c)*a*c^2)