### 3.268 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx$$

Optimal. Leaf size=83 $-\frac{1}{2 a c^3 \sqrt{c-a c x}}-\frac{1}{3 a c^2 (c-a c x)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a c^{7/2}}$

[Out]

-1/(3*a*c^2*(c - a*c*x)^(3/2)) - 1/(2*a*c^3*Sqrt[c - a*c*x]) + ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(2*S
qrt[2]*a*c^(7/2))

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Rubi [A]  time = 0.115963, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, 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} $-\frac{1}{2 a c^3 \sqrt{c-a c x}}-\frac{1}{3 a c^2 (c-a c x)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a c^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(3*a*c^2*(c - a*c*x)^(3/2)) - 1/(2*a*c^3*Sqrt[c - a*c*x]) + ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(2*S
qrt[2]*a*c^(7/2))

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

Mathematica [C]  time = 0.0276531, size = 39, normalized size = 0.47 $-\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{1}{2} (1-a x)\right )}{3 a c^2 (c-a c x)^{3/2}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 64, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{ac} \left ( 1/4\,{\frac{1}{\sqrt{-acx+c}{c}^{2}}}+1/6\,{\frac{1}{c \left ( -acx+c \right ) ^{3/2}}}-1/8\,{\frac{\sqrt{2}}{{c}^{5/2}}{\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)^(7/2),x)

[Out]

-2/c/a*(1/4/c^2/(-a*c*x+c)^(1/2)+1/6/c/(-a*c*x+c)^(3/2)-1/8/c^(5/2)*2^(1/2)*arctanh(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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.90501, size = 478, normalized size = 5.76 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left (a^{2} x^{2} - 2 \, 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}{\left (3 \, a x - 5\right )}}{24 \,{\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}}, -\frac{3 \, \sqrt{2}{\left (a^{2} x^{2} - 2 \, 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}{\left (3 \, a x - 5\right )}}{12 \,{\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\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)^(7/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(2)*(a^2*x^2 - 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)*(3*a*x - 5))/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4), -1/12*(3*sqrt(2)*(a^2*x^2 - 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)*(3*a*x - 5))/(a^3*c^4*x^2 - 2
*a^2*c^4*x + a*c^4)]

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Sympy [A]  time = 48.9461, size = 82, normalized size = 0.99 \begin{align*} - \frac{1}{3 a c^{2} \left (- a c x + c\right )^{\frac{3}{2}}} - \frac{1}{2 a c^{3} \sqrt{- a c x + c}} - \frac{\sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \sqrt{- c}} \right )}}{4 a c^{3} \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)**(7/2),x)

[Out]

-1/(3*a*c**2*(-a*c*x + c)**(3/2)) - 1/(2*a*c**3*sqrt(-a*c*x + c)) - sqrt(2)*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*s
qrt(-c)))/(4*a*c**3*sqrt(-c))

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Giac [A]  time = 1.12433, size = 99, normalized size = 1.19 \begin{align*} -\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{4 \, a \sqrt{-c} c^{3}} - \frac{3 \, a c x - 5 \, c}{6 \,{\left (a c x - c\right )} \sqrt{-a c x + c} a c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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