### 3.849 $$\int e^{3 \coth ^{-1}(a x)} (c-\frac{c}{a^2 x^2})^{3/2} \, dx$$

Optimal. Leaf size=148 $\frac{c x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}-\frac{3 c \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{2 a^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 c \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.128052, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {6197, 6193, 43} $\frac{c x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}-\frac{3 c \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{2 a^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 c \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6197

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

Rule 6193

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.044434, size = 59, normalized size = 0.4 $\frac{\left (c-\frac{c}{a^2 x^2}\right )^{3/2} \left (a^3 x+3 a^2 \log (x)-\frac{3 a}{x}-\frac{1}{2 x^2}\right )}{a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.244, size = 69, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,{x}^{3}{a}^{3}+6\,{a}^{2}\ln \left ( x \right ){x}^{2}-6\,ax-1 \right ) x}{2\, \left ( ax+1 \right ) ^{3}} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.61299, size = 103, normalized size = 0.7 \begin{align*} \frac{{\left (2 \, a^{3} c x^{3} + 6 \, a^{2} c x^{2} \log \left (x\right ) - 6 \, a c x - c\right )} \sqrt{a^{2} c}}{2 \, a^{4} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3*c*x^3 + 6*a^2*c*x^2*log(x) - 6*a*c*x - c)*sqrt(a^2*c)/(a^4*x^2)

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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))**(3/2)*(c-c/a**2/x**2)**(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^{2} x^{2}}\right )}^{\frac{3}{2}}}{\left (\frac{a x - 1}{a x + 1}\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))^(3/2)*(c-c/a^2/x^2)^(3/2),x, algorithm="giac")

[Out]

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