### 3.843 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^{3/2}} \, dx$$

Optimal. Leaf size=123 $\frac{2 (5-2 a x) (1-a x) (a x+1)^2}{3 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{(a x+1)^2}{3 a^2 x \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{2 (1-a x)^{3/2} (a x+1)^{3/2} \sin ^{-1}(a x)}{a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}$

[Out]

-(1 + a*x)^2/(3*a^2*(c - c/(a^2*x^2))^(3/2)*x) + (2*(5 - 2*a*x)*(1 - a*x)*(1 + a*x)^2)/(3*a^4*(c - c/(a^2*x^2)
)^(3/2)*x^3) - (2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2)*ArcSin[a*x])/(a^4*(c - c/(a^2*x^2))^(3/2)*x^3)

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Rubi [A]  time = 0.435983, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.292, Rules used = {6167, 6159, 6129, 98, 143, 41, 216} $\frac{2 (5-2 a x) (1-a x) (a x+1)^2}{3 a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{(a x+1)^2}{3 a^2 x \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}-\frac{2 (1-a x)^{3/2} (a x+1)^{3/2} \sin ^{-1}(a x)}{a^4 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(1 + a*x)^2/(3*a^2*(c - c/(a^2*x^2))^(3/2)*x) + (2*(5 - 2*a*x)*(1 - a*x)*(1 + a*x)^2)/(3*a^4*(c - c/(a^2*x^2)
)^(3/2)*x^3) - (2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2)*ArcSin[a*x])/(a^4*(c - c/(a^2*x^2))^(3/2)*x^3)

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6159

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

Rule 6129

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

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 143

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{2 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \, dx\\ &=-\frac{\left ((1-a x)^{3/2} (1+a x)^{3/2}\right ) \int \frac{e^{2 \tanh ^{-1}(a x)} x^3}{(1-a x)^{3/2} (1+a x)^{3/2}} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac{\left ((1-a x)^{3/2} (1+a x)^{3/2}\right ) \int \frac{x^3}{(1-a x)^{5/2} \sqrt{1+a x}} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac{(1+a x)^2}{3 a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x}+\frac{\left ((1-a x)^{3/2} (1+a x)^{3/2}\right ) \int \frac{x (2+4 a x)}{(1-a x)^{3/2} \sqrt{1+a x}} \, dx}{3 a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac{(1+a x)^2}{3 a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x}+\frac{2 (5-2 a x) (1-a x) (1+a x)^2}{3 a^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}-\frac{\left (2 (1-a x)^{3/2} (1+a x)^{3/2}\right ) \int \frac{1}{\sqrt{1-a x} \sqrt{1+a x}} \, dx}{a^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac{(1+a x)^2}{3 a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x}+\frac{2 (5-2 a x) (1-a x) (1+a x)^2}{3 a^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}-\frac{\left (2 (1-a x)^{3/2} (1+a x)^{3/2}\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac{(1+a x)^2}{3 a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x}+\frac{2 (5-2 a x) (1-a x) (1+a x)^2}{3 a^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}-\frac{2 (1-a x)^{3/2} (1+a x)^{3/2} \sin ^{-1}(a x)}{a^4 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3}\\ \end{align*}

Mathematica [A]  time = 0.0836096, size = 95, normalized size = 0.77 $\frac{3 a^3 x^3-11 a^2 x^2+6 (a x-1) \sqrt{a^2 x^2-1} \log \left (\sqrt{a^2 x^2-1}+a x\right )-4 a x+10}{3 a^2 c x (a x-1) \sqrt{c-\frac{c}{a^2 x^2}}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10 - 4*a*x - 11*a^2*x^2 + 3*a^3*x^3 + 6*(-1 + a*x)*Sqrt[-1 + a^2*x^2]*Log[a*x + Sqrt[-1 + a^2*x^2]])/(3*a^2*c
*Sqrt[c - c/(a^2*x^2)]*x*(-1 + a*x))

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Maple [B]  time = 0.184, size = 326, normalized size = 2.7 \begin{align*}{\frac{ax+1}{3\,{a}^{4}{x}^{3}} \left ( 3\,{c}^{3/2}\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}{x}^{3}{a}^{3}+4\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}{c}^{3/2}{x}^{2}{a}^{2}-15\,{x}^{2}{a}^{2}{c}^{3/2}\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}+6\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) \sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}x{a}^{2}c-4\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}{c}^{3/2}xa-6\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) \sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}ac-2\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}{c}^{3/2}+12\,{c}^{3/2}\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}} \right ){\frac{1}{\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}}} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*c^(3/2)*((a*x-1)*(a*x+1)*c/a^2)^(1/2)*x^3*a^3+4*(c*(a^2*x^2-1)/a^2)^(1/2)*c^(3/2)*x^2*a^2-15*x^2*a^2*c^
(3/2)*((a*x-1)*(a*x+1)*c/a^2)^(1/2)+6*(c*(a^2*x^2-1)/a^2)^(1/2)*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2)^(1/2))*((a*x-
1)*(a*x+1)*c/a^2)^(1/2)*x*a^2*c-4*(c*(a^2*x^2-1)/a^2)^(1/2)*c^(3/2)*x*a-6*(c*(a^2*x^2-1)/a^2)^(1/2)*ln(x*c^(1/
2)+(c*(a^2*x^2-1)/a^2)^(1/2))*((a*x-1)*(a*x+1)*c/a^2)^(1/2)*a*c-2*(c*(a^2*x^2-1)/a^2)^(1/2)*c^(3/2)+12*c^(3/2)
*((a*x-1)*(a*x+1)*c/a^2)^(1/2))*(a*x+1)/((a*x-1)*(a*x+1)*c/a^2)^(1/2)/x^3/(c*(a^2*x^2-1)/a^2/x^2)^(3/2)/a^4/c^
(3/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.79856, size = 591, normalized size = 4.8 \begin{align*} \left [\frac{3 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt{c} \log \left (2 \, a^{2} c x^{2} + 2 \, a^{2} \sqrt{c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right ) +{\left (3 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 10 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{3 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}}, -\frac{6 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt{-c} \arctan \left (\frac{a^{2} \sqrt{-c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) -{\left (3 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 10 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{3 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*log(2*a^2*c*x^2 + 2*a^2*sqrt(c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c)
+ (3*a^3*x^3 - 14*a^2*x^2 + 10*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2), -1/
3*(6*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*arctan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^2 - c)) -
(3*a^3*x^3 - 14*a^2*x^2 + 10*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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