### 3.485 $$\int \frac{e^{-3 \coth ^{-1}(a x)}}{\sqrt{c-\frac{c}{a x}}} \, dx$$

Optimal. Leaf size=118 $\frac{x \sqrt{c-\frac{c}{a x}}}{c \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{5 \sqrt{c-\frac{c}{a x}}}{a c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{a \sqrt{c}}$

[Out]

(5*Sqrt[c - c/(a*x)])/(a*c*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[c - c/(a*x)]*x)/(c*Sqrt[1 - 1/(a^2*x^2)]) - (5*ArcTa
nh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(a*Sqrt[c])

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Rubi [A]  time = 0.221027, antiderivative size = 118, normalized size of antiderivative = 1., 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, 879, 869, 875, 208} $\frac{x \sqrt{c-\frac{c}{a x}}}{c \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{5 \sqrt{c-\frac{c}{a x}}}{a c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{a \sqrt{c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*Sqrt[c - c/(a*x)])/(a*c*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[c - c/(a*x)]*x)/(c*Sqrt[1 - 1/(a^2*x^2)]) - (5*ArcTa
nh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(a*Sqrt[c])

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

Mathematica [C]  time = 0.040337, size = 69, normalized size = 0.58 $\frac{\sqrt{1-\frac{1}{a x}} \left (5 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{1}{a x}+1\right )+a x\right )}{a \sqrt{\frac{1}{a x}+1} \sqrt{c-\frac{c}{a x}}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 - 1/(a*x)]*(a*x + 5*Hypergeometric2F1[-1/2, 1, 1/2, 1 + 1/(a*x)]))/(a*Sqrt[1 + 1/(a*x)]*Sqrt[c - c/(a*
x)])

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Maple [A]  time = 0.184, size = 149, normalized size = 1.3 \begin{align*}{\frac{ \left ( ax+1 \right ) x}{2\, \left ( ax-1 \right ) ^{2}c} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 2\,{a}^{3/2}x\sqrt{ \left ( ax+1 \right ) x}-5\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}+2\,ax+1}{\sqrt{a}}} \right ) xa+10\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}-5\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}+2\,ax+1}{\sqrt{a}}} \right ) \right ){\frac{1}{\sqrt{ \left ( ax+1 \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(c-c/a/x)^(1/2),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*(2*a^(3/2)*x*((a*x+1)*x)^(1/2)-5*ln(1/2*
(2*((a*x+1)*x)^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2))*x*a+10*((a*x+1)*x)^(1/2)*a^(1/2)-5*ln(1/2*(2*((a*x+1)*x)^(1/2)*
a^(1/2)+2*a*x+1)/a^(1/2)))/a^(1/2)/c/((a*x+1)*x)^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2879, size = 651, normalized size = 5.52 \begin{align*} \left [\frac{5 \,{\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} + 5 \, a x\right )} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{4 \,{\left (a^{2} c x - a c\right )}}, \frac{5 \,{\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} + 5 \, a x\right )} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{2 \,{\left (a^{2} c x - a c\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(5*(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 + 5*a*x)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*
x - c)/(a*x)))/(a^2*c*x - a*c), 1/2*(5*(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 + 5*a*x)*sqrt((a*x - 1)/(a*x + 1))*sqr
t((a*c*x - c)/(a*x)))/(a^2*c*x - a*c)]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError