### 3.181 $$\int e^{3 \coth ^{-1}(a x)} (c-a c x) \, dx$$

Optimal. Leaf size=65 $-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}}-2 c x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}$

[Out]

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

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Rubi [A]  time = 0.18547, antiderivative size = 65, 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, 852, 1807, 807, 266, 63, 208} $-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}}-2 c x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(3*ArcCoth[a*x])*(c - a*c*x),x]

[Out]

-2*c*Sqrt[1 - 1/(a^2*x^2)]*x - (a*c*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 - (3*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 852

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

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

Mathematica [A]  time = 0.0746661, size = 53, normalized size = 0.82 $-\frac{c \left (a x \sqrt{1-\frac{1}{a^2 x^2}} (a x+4)+3 \log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )\right )}{2 a}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.174, size = 162, normalized size = 2.5 \begin{align*} -{\frac{ \left ( ax-1 \right ) ^{2}c}{ \left ( 2\,ax+2 \right ) a} \left ( \sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa+4\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) a+4\,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{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02008, size = 182, normalized size = 2.8 \begin{align*} -\frac{1}{2} \, a{\left (\frac{2 \,{\left (3 \, c \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} - 5 \, 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{3 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{3 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(2*(3*c*((a*x - 1)/(a*x + 1))^(3/2) - 5*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) + 3*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 3*c*log(sqrt((a*x - 1)/(a*x + 1))
- 1)/a^2)

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Fricas [A]  time = 1.55294, size = 197, normalized size = 3.03 \begin{align*} -\frac{3 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 3 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (a^{2} c x^{2} + 5 \, a c x + 4 \, 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(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c),x, algorithm="fricas")

[Out]

-1/2*(3*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*c*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (a^2*c*x^2 + 5*a*c*x +
4*c)*sqrt((a*x - 1)/(a*x + 1)))/a

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.23544, size = 167, normalized size = 2.57 \begin{align*} -\frac{1}{2} \, a{\left (\frac{3 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{3 \, c \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2}} - \frac{2 \,{\left (\frac{3 \,{\left (a x - 1\right )} c \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1} - 5 \, c \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{a^{2}{\left (\frac{a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(3*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 3*c*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^2 - 2*(3*(a
*x - 1)*c*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - 5*c*sqrt((a*x - 1)/(a*x + 1)))/(a^2*((a*x - 1)/(a*x + 1) - 1)^
2))