3.483 \(\int e^{-3 \coth ^{-1}(a x)} (c-\frac {c}{a x})^{3/2} \, dx\)

Optimal. Leaf size=158 \[ \frac {x \left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {\frac {1}{a x}+1}}+\frac {\left (21 a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {\frac {1}{a x}+1}}-\frac {9 \left (c-\frac {c}{a x}\right )^{3/2} \tanh ^{-1}\left (\sqrt {\frac {1}{a x}+1}\right )}{a \left (1-\frac {1}{a x}\right )^{3/2}} \]

[Out]

-9*(c-c/a/x)^(3/2)*arctanh((1+1/a/x)^(1/2))/a/(1-1/a/x)^(3/2)+(21*a+1/x)*(c-c/a/x)^(3/2)/a^2/(1-1/a/x)^(3/2)/(
1+1/a/x)^(1/2)+(a-1/x)^2*(c-c/a/x)^(3/2)*x/a^2/(1-1/a/x)^(3/2)/(1+1/a/x)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6182, 6179, 98, 146, 63, 208} \[ \frac {x \left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {\frac {1}{a x}+1}}+\frac {\left (21 a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {\frac {1}{a x}+1}}-\frac {9 \left (c-\frac {c}{a x}\right )^{3/2} \tanh ^{-1}\left (\sqrt {\frac {1}{a x}+1}\right )}{a \left (1-\frac {1}{a x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^(3/2)/E^(3*ArcCoth[a*x]),x]

[Out]

((21*a + x^(-1))*(c - c/(a*x))^(3/2))/(a^2*(1 - 1/(a*x))^(3/2)*Sqrt[1 + 1/(a*x)]) + ((a - x^(-1))^2*(c - c/(a*
x))^(3/2)*x)/(a^2*(1 - 1/(a*x))^(3/2)*Sqrt[1 + 1/(a*x)]) - (9*(c - c/(a*x))^(3/2)*ArcTanh[Sqrt[1 + 1/(a*x)]])/
(a*(1 - 1/(a*x))^(3/2))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 98

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

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 6179

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

Rule 6182

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

Rubi steps

\begin {align*} \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^{3/2} \, dx &=\frac {\left (c-\frac {c}{a x}\right )^{3/2} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{3/2} \, dx}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=-\frac {\left (c-\frac {c}{a x}\right )^{3/2} \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^3}{x^2 \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {\left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2} x}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (c-\frac {c}{a x}\right )^{3/2} \operatorname {Subst}\left (\int \frac {\left (\frac {9}{2 a}-\frac {x}{2 a^2}\right ) \left (1-\frac {x}{a}\right )}{x \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {\left (21 a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2} x}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (9 \left (c-\frac {c}{a x}\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a \left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {\left (21 a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2} x}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (9 \left (c-\frac {c}{a x}\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-a+a x^2} \, dx,x,\sqrt {1+\frac {1}{a x}}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {\left (21 a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}+\frac {\left (a-\frac {1}{x}\right )^2 \left (c-\frac {c}{a x}\right )^{3/2} x}{a^2 \left (1-\frac {1}{a x}\right )^{3/2} \sqrt {1+\frac {1}{a x}}}-\frac {9 \left (c-\frac {c}{a x}\right )^{3/2} \tanh ^{-1}\left (\sqrt {1+\frac {1}{a x}}\right )}{a \left (1-\frac {1}{a x}\right )^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 71, normalized size = 0.45 \[ \frac {c \sqrt {c-\frac {c}{a x}} \left (a^2 x^2+9 a x \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+\frac {1}{a x}\right )+10 a x+2\right )}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^(3/2)/E^(3*ArcCoth[a*x]),x]

[Out]

(c*Sqrt[c - c/(a*x)]*(2 + 10*a*x + a^2*x^2 + 9*a*x*Hypergeometric2F1[-1/2, 1, 1/2, 1 + 1/(a*x)]))/(a^2*Sqrt[1
- 1/(a^2*x^2)]*x)

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fricas [A]  time = 0.80, size = 315, normalized size = 1.99 \[ \left [\frac {9 \, {\left (a c x - c\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (a^{2} c x^{2} + 19 \, a c x + 2 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} x - a\right )}}, \frac {9 \, {\left (a c x - c\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (a^{2} c x^{2} + 19 \, a c x + 2 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} x - a\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(9*(a*c*x - c)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x -
1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(a^2*c*x^2 + 19*a*c*x + 2*c)*sqrt((a*x - 1)/(a*x + 1
))*sqrt((a*c*x - c)/(a*x)))/(a^2*x - a), 1/2*(9*(a*c*x - c)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a
*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(a^2*c*x^2 + 19*a*c*x + 2*c)*sqrt((a
*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a^2*x - a)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x),abs(a*x+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.08, size = 169, normalized size = 1.07 \[ \frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, c \left (2 a^{\frac {5}{2}} x^{2} \sqrt {\left (a x +1\right ) x}+38 a^{\frac {3}{2}} x \sqrt {\left (a x +1\right ) x}-9 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) x^{2} a^{2}-9 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) x a +4 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}\right )}{2 \left (a x -1\right )^{2} a^{\frac {3}{2}} \sqrt {\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x-1)/(a*x+1))^(3/2)*(a*x+1)/(a*x-1)^2*(c*(a*x-1)/a/x)^(1/2)*c*(2*a^(5/2)*x^2*((a*x+1)*x)^(1/2)+38*a^(3
/2)*x*((a*x+1)*x)^(1/2)-9*ln(1/2*(2*((a*x+1)*x)^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2))*x^2*a^2-9*ln(1/2*(2*((a*x+1)*x
)^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2))*x*a+4*((a*x+1)*x)^(1/2)*a^(1/2))/a^(3/2)/((a*x+1)*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c - \frac {c}{a x}\right )}^{\frac {3}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c-\frac {c}{a\,x}\right )}^{3/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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