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

Optimal. Leaf size=117 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a c^{3/2}}-\frac {2 x \sqrt {c-\frac {c}{a x}}}{c^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x \sqrt {1-\frac {1}{a^2 x^2}}}{c \sqrt {c-\frac {c}{a x}}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6177, 869, 873, 875, 208} \[ -\frac {2 x \sqrt {c-\frac {c}{a x}}}{c^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a c^{3/2}}+\frac {3 x \sqrt {1-\frac {1}{a^2 x^2}}}{c \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - c/(a*x))^(3/2)),x]

[Out]

(3*Sqrt[1 - 1/(a^2*x^2)]*x)/(c*Sqrt[c - c/(a*x)]) - (2*Sqrt[c - c/(a*x)]*x)/(c^2*Sqrt[1 - 1/(a^2*x^2)]) - (3*A
rcTanh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(a*c^(3/2))

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 869

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

Rule 873

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

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
 NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 6177

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

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 64, normalized size = 0.55 \[ \frac {2 \left (1-\frac {1}{a x}\right )^{3/2} \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};1+\frac {1}{a x}\right )}{a \sqrt {\frac {1}{a x}+1} \left (c-\frac {c}{a x}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - c/(a*x))^(3/2)),x]

[Out]

(2*(1 - 1/(a*x))^(3/2)*Hypergeometric2F1[-1/2, 2, 1/2, 1 + 1/(a*x)])/(a*Sqrt[1 + 1/(a*x)]*(c - c/(a*x))^(3/2))

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fricas [A]  time = 0.67, size = 311, normalized size = 2.66 \[ \left [\frac {3 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + 3 \, a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} c^{2} x - a c^{2}\right )}}, \frac {3 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (a^{2} x^{2} + 3 \, a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} c^{2} x - a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(((a*x-1)/(a*x+1))^(3/2)/(c-c/a/x)^(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(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 = 0.06, size = 149, normalized size = 1.27 \[ -\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (-2 a^{\frac {3}{2}} x \sqrt {\left (a x +1\right ) x}+3 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) x a -6 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+3 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right )\right )}{2 \left (a x -1\right )^{2} \sqrt {a}\, c^{2} \sqrt {\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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