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

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Rubi [A]  time = 0.09, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.51, size = 56, normalized size = 1.40 \[ \frac {2 \, \sqrt {-a c x + c} {\left (5 \, a x + 1\right )}}{15 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

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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giac [A]  time = 0.14, size = 34, normalized size = 0.85 \[ -\frac {2 \, {\left (5 \, a c x + c\right )}}{15 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a c} \]

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)

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maple [A]  time = 0.04, size = 21, normalized size = 0.52 \[ -\frac {2 \left (5 a x +1\right )}{15 a \left (-a c x +c \right )^{\frac {5}{2}}} \]

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 = 0.30, size = 24, normalized size = 0.60 \[ -\frac {2 \, {\left (5 \, a c x + c\right )}}{15 \, {\left (-a c x + c\right )}^{\frac {5}{2}} a c} \]

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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mupad [B]  time = 1.21, size = 20, normalized size = 0.50 \[ -\frac {10\,a\,x+2}{15\,a\,{\left (c-a\,c\,x\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 21.49, size = 31, normalized size = 0.78 \[ - \frac {4}{5 a \left (- a c x + c\right )^{\frac {5}{2}}} + \frac {2}{3 a c \left (- a c x + c\right )^{\frac {3}{2}}} \]

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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