3.2.84 \(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx\) [184]

Optimal. Leaf size=67 \[ \frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac {1}{x}\right )^6}-\frac {6 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac {1}{x}\right )^5} \]

[Out]

1/7*a^5*(1-1/a^2/x^2)^(5/2)/c^3/(a-1/x)^6-6/35*a^4*(1-1/a^2/x^2)^(5/2)/c^3/(a-1/x)^5

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Rubi [A]
time = 0.09, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6310, 6313, 807, 665} \begin {gather*} \frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac {1}{x}\right )^6}-\frac {6 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac {1}{x}\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2))/(7*c^3*(a - x^(-1))^6) - (6*a^4*(1 - 1/(a^2*x^2))^(5/2))/(35*c^3*(a - x^(-1))^5)

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rule 6310

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx &=-\frac {\int \frac {e^{3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^3 x^3} \, dx}{a^3 c^3}\\ &=\frac {\text {Subst}\left (\int \frac {x \left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^6} \, dx,x,\frac {1}{x}\right )}{a^3 c^3}\\ &=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac {1}{x}\right )^6}-\frac {6 \text {Subst}\left (\int \frac {\left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^5} \, dx,x,\frac {1}{x}\right )}{7 a^2 c^3}\\ &=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac {1}{x}\right )^6}-\frac {6 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac {1}{x}\right )^5}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 41, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (-6+a x) (1+a x)^2}{35 c^3 (-1+a x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

-1/35*(Sqrt[1 - 1/(a^2*x^2)]*x*(-6 + a*x)*(1 + a*x)^2)/(c^3*(-1 + a*x)^4)

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Maple [A]
time = 0.14, size = 41, normalized size = 0.61

method	result	size
gosper	\(-\frac {\left (a x -6\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{2} c^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(41\)
default	\(-\frac {\left (a x -6\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{2} c^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(41\)
trager	\(-\frac {\left (a x +1\right ) \left (a^{3} x^{3}-4 a^{2} x^{2}-11 a x -6\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{3} \left (a x -1\right )^{4}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/35*(a*x-6)*(a*x+1)/(a*x-1)^2/c^3/((a*x-1)/(a*x+1))^(3/2)/a

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Maxima [A]
time = 0.25, size = 39, normalized size = 0.58 \begin {gather*} -\frac {\frac {7 \, {\left (a x - 1\right )}}{a x + 1} - 5}{70 \, a c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(7*(a*x - 1)/(a*x + 1) - 5)/(a*c^3*((a*x - 1)/(a*x + 1))^(7/2))

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Fricas [A]
time = 0.35, size = 95, normalized size = 1.42 \begin {gather*} -\frac {{\left (a^{4} x^{4} - 3 \, a^{3} x^{3} - 15 \, a^{2} x^{2} - 17 \, a x - 6\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

-1/35*(a^4*x^4 - 3*a^3*x^3 - 15*a^2*x^2 - 17*a*x - 6)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 +
6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{\frac {a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {4 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {6 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {4 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**3,x)

[Out]

-Integral(1/(a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 4*a**3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x +
1))/(a*x + 1) + 6*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 4*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x +
1))/(a*x + 1) + sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1)), x)/c**3

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (59) = 118\).
time = 0.47, size = 125, normalized size = 1.87 \begin {gather*} \frac {2 \, {\left (35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{5} x^{5} + 35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 70 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 14 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 7 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{35 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{7} a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

2/35*(35*(a + sqrt(a^2 - 1/x^2))^5*x^5 + 35*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 70*(a + sqrt(a^2 - 1/x^2))^3*x^3 +
14*(a + sqrt(a^2 - 1/x^2))^2*x^2 + 7*(a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^7*a*c^3)

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Mupad [B]
time = 1.18, size = 39, normalized size = 0.58 \begin {gather*} -\frac {\frac {a\,x-1}{5\,\left (a\,x+1\right )}-\frac {1}{7}}{2\,a\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c - a*c*x)^3*((a*x - 1)/(a*x + 1))^(3/2)),x)

[Out]

-((a*x - 1)/(5*(a*x + 1)) - 1/7)/(2*a*c^3*((a*x - 1)/(a*x + 1))^(7/2))

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