### 3.219 $$\int e^{-3 \coth ^{-1}(a x)} (c-a c x) \, dx$$

Optimal. Leaf size=92 $-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}}+4 c x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{15 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}$

[Out]

(8*c*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + 4*c*Sqrt[1 - 1/(a^2*x^2)]*x - (a*c*Sqrt[1 - 1/(a^2*x^2)]*x^2)
/2 - (15*c*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a)

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Rubi [A]  time = 0.243798, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {6175, 6178, 1805, 1807, 807, 266, 63, 208} $-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}}+4 c x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{15 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c - a*c*x)/E^(3*ArcCoth[a*x]),x]

[Out]

(8*c*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + 4*c*Sqrt[1 - 1/(a^2*x^2)]*x - (a*c*Sqrt[1 - 1/(a^2*x^2)]*x^2)
/2 - (15*c*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a)

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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^{-3 \coth ^{-1}(a x)} (c-a c x) \, dx &=-\left ((a c) \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right ) x \, dx\right )\\ &=(a c) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^4}{x^3 \left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-(a c) \operatorname{Subst}\left (\int \frac{-1+\frac{4 x}{a}-\frac{7 x^2}{a^2}}{x^3 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{2} (a c) \operatorname{Subst}\left (\int \frac{-\frac{8}{a}+\frac{15 x}{a^2}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+4 c \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{(15 c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+4 c \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{(15 c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 a}\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+4 c \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2-\frac{1}{2} (15 a c) \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )\\ &=\frac{8 c \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+4 c \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2-\frac{15 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.133689, size = 68, normalized size = 0.74 $\frac{1}{2} c \left (\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (-a^2 x^2+7 a x+24\right )}{a x+1}-\frac{15 \log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a}\right )$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*((Sqrt[1 - 1/(a^2*x^2)]*x*(24 + 7*a*x - a^2*x^2))/(1 + a*x) - (15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/a))
/2

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Maple [B]  time = 0.16, size = 422, normalized size = 4.6 \begin{align*} -{\frac{c}{2\, \left ( ax-1 \right ) a} \left ( \sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{3}{a}^{3}+2\,\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{2}{a}^{2}-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ){x}^{2}{a}^{3}-16\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+16\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}+\sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa-2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}+8\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}-32\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+32\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) a-16\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }+16\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) \right ) \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3+2*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a^2-ln((a^2*x+(a^2*x^2-1)^(1/2
)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3-16*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2+16*ln((a^2*x+(a^2)^(1/2)*((
a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^2*a^3+(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*x*a-2*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a
^2)^(1/2))/(a^2)^(1/2))*x*a^2+8*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-32*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x*a
+32*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x*a^2-ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))
/(a^2)^(1/2))*a-16*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)+16*a*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^
2)^(1/2)))/a*c*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/((a*x-1)*(a*x+1))^(1/2)/(a*x-1)

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Maxima [A]  time = 1.06275, size = 211, normalized size = 2.29 \begin{align*} \frac{1}{2} \, a{\left (\frac{2 \,{\left (9 \, c \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} - 7 \, c \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{\frac{2 \,{\left (a x - 1\right )} a^{2}}{a x + 1} - \frac{{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} - \frac{15 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac{15 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac{16 \, c \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(2*(9*c*((a*x - 1)/(a*x + 1))^(3/2) - 7*c*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - (a*x -
1)^2*a^2/(a*x + 1)^2 - a^2) - 15*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 15*c*log(sqrt((a*x - 1)/(a*x + 1)
) - 1)/a^2 + 16*c*sqrt((a*x - 1)/(a*x + 1))/a^2)

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Fricas [A]  time = 1.66708, size = 201, normalized size = 2.18 \begin{align*} -\frac{15 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 15 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (a^{2} c x^{2} - 7 \, a c x - 24 \, c\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{2 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(15*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (a^2*c*x^2 - 7*a*c*x
- 24*c)*sqrt((a*x - 1)/(a*x + 1)))/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((-a*c*x+c)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef