3.275 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 x \sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}}+\frac {6 \sqrt {1-\frac {1}{a x}}}{a \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6176, 6181, 78, 37} \[ \frac {2 x \sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}}+\frac {6 \sqrt {1-\frac {1}{a x}}}{a \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*Sqrt[c - a*c*x]),x]

[Out]

(6*Sqrt[1 - 1/(a*x)])/(a*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x]) + (2*Sqrt[1 - 1/(a*x)]*x)/(Sqrt[1 + 1/(a*x)]*Sqrt[
c - a*c*x])

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 6176

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

Rule 6181

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

Rubi steps

\begin {align*} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx &=\frac {\left (\sqrt {1-\frac {1}{a x}} \sqrt {x}\right ) \int \frac {e^{-3 \coth ^{-1}(a x)}}{\sqrt {1-\frac {1}{a x}} \sqrt {x}} \, dx}{\sqrt {c-a c x}}\\ &=-\frac {\sqrt {1-\frac {1}{a x}} \operatorname {Subst}\left (\int \frac {1-\frac {x}{a}}{x^{3/2} \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {\frac {1}{x}} \sqrt {c-a c x}}\\ &=\frac {2 \sqrt {1-\frac {1}{a x}} x}{\sqrt {1+\frac {1}{a x}} \sqrt {c-a c x}}+\frac {\left (3 \sqrt {1-\frac {1}{a x}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{a \sqrt {\frac {1}{x}} \sqrt {c-a c x}}\\ &=\frac {6 \sqrt {1-\frac {1}{a x}}}{a \sqrt {1+\frac {1}{a x}} \sqrt {c-a c x}}+\frac {2 \sqrt {1-\frac {1}{a x}} x}{\sqrt {1+\frac {1}{a x}} \sqrt {c-a c x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.56 \[ \frac {2 \sqrt {1-\frac {1}{a x}} (a x+3)}{a \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*Sqrt[c - a*c*x]),x]

[Out]

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

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fricas [A]  time = 0.61, size = 44, normalized size = 0.52 \[ -\frac {2 \, \sqrt {-a c x + c} {\left (a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 42, normalized size = 0.49 \[ 2 \, {\left (\frac {\sqrt {-a c x - c}}{a c^{2}} - \frac {2}{\sqrt {-a c x - c} a c}\right )} {\left | c \right |} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(-a*c*x - c)/(a*c^2) - 2/(sqrt(-a*c*x - c)*a*c))*abs(c)

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maple [A]  time = 0.04, size = 47, normalized size = 0.55 \[ \frac {2 \left (a x +1\right ) \left (a x +3\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \left (a x -1\right ) \sqrt {-a c x +c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 48, normalized size = 0.56 \[ \frac {2 \, {\left (a^{2} x^{2} + 4 \, a x + 3\right )} {\left (a x - 1\right )}}{{\left (a^{2} \sqrt {-c} x - a \sqrt {-c}\right )} {\left (a x + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.40, size = 34, normalized size = 0.40 \[ \frac {\left (2\,x+\frac {6}{a}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{\sqrt {c-a\,c\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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