### 3.241 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx$$

Optimal. Leaf size=40 $\frac{2}{3 a c (c-a c x)^{3/2}}-\frac{4}{5 a (c-a c x)^{5/2}}$

[Out]

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

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Rubi [A]  time = 0.0886908, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {6167, 6130, 21, 43} $\frac{2}{3 a c (c-a c x)^{3/2}}-\frac{4}{5 a (c-a c x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0585227, size = 34, normalized size = 0.85 $\frac{2 (5 a x+1) \sqrt{c-a c x}}{15 a c^3 (a x-1)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(1 + 5*a*x)*Sqrt[c - a*c*x])/(15*a*c^3*(-1 + a*x)^3)

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Maple [A]  time = 0.045, size = 21, normalized size = 0.5 \begin{align*} -{\frac{10\,ax+2}{15\,a} \left ( -acx+c \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*a*x+1)/a/(-a*c*x+c)^(5/2)

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Maxima [A]  time = 1.0351, size = 32, normalized size = 0.8 \begin{align*} -\frac{2 \,{\left (5 \, a c x + c\right )}}{15 \,{\left (-a c x + c\right )}^{\frac{5}{2}} a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*a*c*x + c)/((-a*c*x + c)^(5/2)*a*c)

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Fricas [A]  time = 1.48353, size = 117, normalized size = 2.92 \begin{align*} \frac{2 \, \sqrt{-a c x + c}{\left (5 \, a x + 1\right )}}{15 \,{\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*sqrt(-a*c*x + c)*(5*a*x + 1)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)

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Sympy [A]  time = 25.0109, size = 31, normalized size = 0.78 \begin{align*} - \frac{4}{5 a \left (- a c x + c\right )^{\frac{5}{2}}} + \frac{2}{3 a c \left (- a c x + c\right )^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(5*a*(-a*c*x + c)**(5/2)) + 2/(3*a*c*(-a*c*x + c)**(3/2))

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Giac [A]  time = 1.12408, size = 46, normalized size = 1.15 \begin{align*} -\frac{2 \,{\left (5 \, a c x + c\right )}}{15 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*a*c*x + c)/((a*c*x - c)^2*sqrt(-a*c*x + c)*a*c)