### 3.159 $$\int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx$$

Optimal. Leaf size=105 $-\frac{1}{4} a^3 c^3 x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}+\frac{2}{3} a^2 c^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}-\frac{5}{8} a c^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{5 c^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{8 a}$

[Out]

(-5*a*c^3*Sqrt[1 - 1/(a^2*x^2)]*x^2)/8 + (2*a^2*c^3*(1 - 1/(a^2*x^2))^(3/2)*x^3)/3 - (a^3*c^3*(1 - 1/(a^2*x^2)
)^(3/2)*x^4)/4 + (5*c^3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(8*a)

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Rubi [A]  time = 0.225517, antiderivative size = 105, 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, 1807, 807, 266, 47, 63, 208} $-\frac{1}{4} a^3 c^3 x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}+\frac{2}{3} a^2 c^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}-\frac{5}{8} a c^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{5 c^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{8 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[a*x]*(c - a*c*x)^3,x]

[Out]

(-5*a*c^3*Sqrt[1 - 1/(a^2*x^2)]*x^2)/8 + (2*a^2*c^3*(1 - 1/(a^2*x^2))^(3/2)*x^3)/3 - (a^3*c^3*(1 - 1/(a^2*x^2)
)^(3/2)*x^4)/4 + (5*c^3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(8*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 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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx &=-\left (\left (a^3 c^3\right ) \int e^{\coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^3 x^3 \, dx\right )\\ &=\left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^2 \sqrt{1-\frac{x^2}{a^2}}}{x^5} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4-\frac{1}{4} \left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{8}{a}-\frac{5 x}{a^2}\right ) \sqrt{1-\frac{x^2}{a^2}}}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{3} a^2 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4+\frac{1}{4} \left (5 a c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{3} a^2 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4+\frac{1}{8} \left (5 a c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x}{a^2}}}{x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{5}{8} a c^3 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{2}{3} a^2 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4-\frac{\left (5 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{16 a}\\ &=-\frac{5}{8} a c^3 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{2}{3} a^2 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4+\frac{1}{8} \left (5 a c^3\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )\\ &=-\frac{5}{8} a c^3 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{2}{3} a^2 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3-\frac{1}{4} a^3 c^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^4+\frac{5 c^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{8 a}\\ \end{align*}

Mathematica [A]  time = 0.153045, size = 73, normalized size = 0.7 $\frac{c^3 \left (15 \log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )-a x \sqrt{1-\frac{1}{a^2 x^2}} \left (6 a^3 x^3-16 a^2 x^2+9 a x+16\right )\right )}{24 a}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(-(a*Sqrt[1 - 1/(a^2*x^2)]*x*(16 + 9*a*x - 16*a^2*x^2 + 6*a^3*x^3)) + 15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)]
)*x]))/(24*a)

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Maple [A]  time = 0.132, size = 141, normalized size = 1.3 \begin{align*} -{\frac{ \left ( ax-1 \right ){c}^{3}}{24\,a} \left ( 6\,\sqrt{{a}^{2}} \left ({a}^{2}{x}^{2}-1 \right ) ^{3/2}xa+15\,\sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa-16\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}-15\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}}{\sqrt{{a}^{2}}}} \right ) a \right ){\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{\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))^(1/2)*(-a*c*x+c)^3,x)

[Out]

-1/24*(a*x-1)*c^3/a*(6*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*x*a+15*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*x*a-16*((a*x-1)*(a*x
+1))^(3/2)*(a^2)^(1/2)-15*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a)/((a*x-1)/(a*x+1))^(1/2)/((a
*x-1)*(a*x+1))^(1/2)/(a^2)^(1/2)

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Maxima [B]  time = 0.991732, size = 298, normalized size = 2.84 \begin{align*} \frac{1}{24} \,{\left (\frac{15 \, c^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{15 \, c^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac{2 \,{\left (15 \, c^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}} + 73 \, c^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} - 55 \, c^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + 15 \, c^{3} \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{\frac{4 \,{\left (a x - 1\right )} a^{2}}{a x + 1} - \frac{6 \,{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac{4 \,{\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac{{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} - a^{2}}\right )} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + 2*(15*c^
3*((a*x - 1)/(a*x + 1))^(7/2) + 73*c^3*((a*x - 1)/(a*x + 1))^(5/2) - 55*c^3*((a*x - 1)/(a*x + 1))^(3/2) + 15*c
^3*sqrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)*a^2/(a*x + 1) - 6*(a*x - 1)^2*a^2/(a*x + 1)^2 + 4*(a*x - 1)^3*a^2/(
a*x + 1)^3 - (a*x - 1)^4*a^2/(a*x + 1)^4 - a^2))*a

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Fricas [A]  time = 1.5496, size = 263, normalized size = 2.5 \begin{align*} \frac{15 \, c^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 15 \, c^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) -{\left (6 \, a^{4} c^{3} x^{4} - 10 \, a^{3} c^{3} x^{3} - 7 \, a^{2} c^{3} x^{2} + 25 \, a c^{3} x + 16 \, c^{3}\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{24 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (6*a^4*c^3*x^4 -
10*a^3*c^3*x^3 - 7*a^2*c^3*x^2 + 25*a*c^3*x + 16*c^3)*sqrt((a*x - 1)/(a*x + 1)))/a

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

[Out]

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

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Giac [B]  time = 1.21364, size = 270, normalized size = 2.57 \begin{align*} \frac{1}{24} \,{\left (\frac{15 \, c^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{15 \, c^{3} \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (\frac{55 \,{\left (a x - 1\right )} c^{3} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1} - \frac{73 \,{\left (a x - 1\right )}^{2} c^{3} \sqrt{\frac{a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{2}} - \frac{15 \,{\left (a x - 1\right )}^{3} c^{3} \sqrt{\frac{a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{3}} - 15 \, c^{3} \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{a^{2}{\left (\frac{a x - 1}{a x + 1} - 1\right )}^{4}}\right )} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^3*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^2 + 2*(
55*(a*x - 1)*c^3*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - 73*(a*x - 1)^2*c^3*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^
2 - 15*(a*x - 1)^3*c^3*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^3 - 15*c^3*sqrt((a*x - 1)/(a*x + 1)))/(a^2*((a*x -
1)/(a*x + 1) - 1)^4))*a