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

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

[Out]

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

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Rubi [A]  time = 0.223204, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6177, 879, 865, 875, 208} \[ \frac{c^3 x \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-\frac{c}{a x}\right )^{3/2}}+\frac{c^2 \sqrt{1-\frac{1}{a^2 x^2}}}{a \sqrt{c-\frac{c}{a x}}}-\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(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 - 1, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 865

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0518382, size = 70, normalized size = 0.6 \[ \frac{c \sqrt{c-\frac{c}{a x}} \left (\sqrt{\frac{1}{a x}+1} (a x+2)-\tanh ^{-1}\left (\sqrt{\frac{1}{a x}+1}\right )\right )}{a \sqrt{1-\frac{1}{a x}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.178, size = 105, normalized size = 0.9 \begin{align*} -{\frac{c}{2}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -2\,{a}^{3/2}x\sqrt{ \left ( ax+1 \right ) x}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}+2\,ax+1 \right ){\frac{1}{\sqrt{a}}}} \right ) xa-4\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a} \right ){\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( ax+1 \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c - \frac{c}{a x}\right )}^{\frac{3}{2}}}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.21834, size = 667, normalized size = 5.7 \begin{align*} \left [\frac{{\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} + 3 \, 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{{\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} + 3 \, 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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((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 + 3*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*((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 + 3*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c - \frac{c}{a x}\right )}^{\frac{3}{2}}}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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