3.204 \(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {8 a \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )} \]

[Out]

-1/5*a^2*(1-1/a^2/x^2)^(1/2)/c^4/(a-1/x)^3+8/15*a*(1-1/a^2/x^2)^(1/2)/c^4/(a-1/x)^2-7/15*(1-1/a^2/x^2)^(1/2)/c
^4/(a-1/x)

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Rubi [A]  time = 0.23, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6175, 6178, 1639, 793, 659, 651} \[ -\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {8 a \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcCoth[a*x]*(c - a*c*x)^4),x]

[Out]

-(a^2*Sqrt[1 - 1/(a^2*x^2)])/(5*c^4*(a - x^(-1))^3) + (8*a*Sqrt[1 - 1/(a^2*x^2)])/(15*c^4*(a - x^(-1))^2) - (7
*Sqrt[1 - 1/(a^2*x^2)])/(15*c^4*(a - x^(-1)))

Rule 651

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 659

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*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 793

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 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 6175

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 6178

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^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx &=\frac {\int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^4 x^4} \, dx}{a^4 c^4}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1-\frac {x}{a}\right )^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^4 c^4}\\ &=\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{c^4 \left (a-\frac {1}{x}\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {2}{a^2}-\frac {x}{a^3}}{\left (1-\frac {x}{a}\right )^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^4}\\ &=-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {x}{a}\right )^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 a^2 c^4}\\ &=-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {8 a \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{15 a^2 c^4}\\ &=-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {8 a \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 43, normalized size = 0.45 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a^2 x^2-6 a x+7\right )}{15 c^4 (a x-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)^4),x]

[Out]

-1/15*(Sqrt[1 - 1/(a^2*x^2)]*x*(7 - 6*a*x + 2*a^2*x^2))/(c^4*(-1 + a*x)^3)

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fricas [A]  time = 0.58, size = 77, normalized size = 0.81 \[ -\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 7\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

-1/15*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 7)*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x -
a*c^4)

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giac [A]  time = 0.21, size = 65, normalized size = 0.68 \[ -\frac {4 \, {\left (10 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} - 5 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x + 1\right )}}{15 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{5} a c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

-4/15*(10*(a + sqrt(a^2 - 1/x^2))^2*x^2 - 5*(a + sqrt(a^2 - 1/x^2))*x + 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^5*
a*c^4)

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maple [A]  time = 0.04, size = 50, normalized size = 0.53 \[ -\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x +1\right )}{15 \left (a x -1\right )^{3} c^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x)

[Out]

-1/15*((a*x-1)/(a*x+1))^(1/2)*(2*a^2*x^2-6*a*x+7)*(a*x+1)/(a*x-1)^3/c^4/a

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maxima [A]  time = 0.31, size = 55, normalized size = 0.58 \[ \frac {\frac {10 \, {\left (a x - 1\right )}}{a x + 1} - \frac {15 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 3}{60 \, a c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*(a*x - 1)/(a*x + 1) - 15*(a*x - 1)^2/(a*x + 1)^2 - 3)/(a*c^4*((a*x - 1)/(a*x + 1))^(5/2))

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mupad [B]  time = 0.04, size = 55, normalized size = 0.58 \[ -\frac {\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{3\,\left (a\,x+1\right )}+\frac {1}{5}}{4\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a*c*x)^4,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**4,x)

[Out]

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

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