### 3.431 $$\int e^{-3 \coth ^{-1}(a x)} (c-\frac{c}{a x})^2 \, dx$$

Optimal. Leaf size=105 $\frac{16 c^2 \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+c^2 x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{c^2 \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}+\frac{5 c^2 \csc ^{-1}(a x)}{a}$

[Out]

(c^2*Sqrt[1 - 1/(a^2*x^2)])/a + (16*c^2*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + c^2*Sqrt[1 - 1/(a^2*x^2)]*
x + (5*c^2*ArcCsc[a*x])/a - (5*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/a

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Rubi [A]  time = 0.325673, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.409, Rules used = {6177, 1805, 1807, 1809, 844, 216, 266, 63, 208} $\frac{16 c^2 \left (a-\frac{1}{x}\right )}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+c^2 x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{c^2 \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}+\frac{5 c^2 \csc ^{-1}(a x)}{a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c^2*Sqrt[1 - 1/(a^2*x^2)])/a + (16*c^2*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + c^2*Sqrt[1 - 1/(a^2*x^2)]*
x + (5*c^2*ArcCsc[a*x])/a - (5*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/a

Rule 6177

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c + d*x)^(p -
n)*(1 - x^2/a^2)^(n/2))/x^2, x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && IntegerQ[(n - 1)
/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1]) && IntegerQ[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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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]

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

Mathematica [C]  time = 0.388522, size = 424, normalized size = 4.04 $\frac{c^2 \left (7 \sqrt{2} a x (a x-1)^3 (a x+1) \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{1}{2} \left (1-\frac{1}{a x}\right )\right )+5 \sqrt{2} (a x-1)^4 (a x+1) \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{7}{2},\frac{9}{2},\frac{1}{2} \left (1-\frac{1}{a x}\right )\right )+35 a^6 x^6 \sqrt{\frac{1}{a x}+1}-595 a^5 x^5 \sqrt{\frac{1}{a x}+1}+280 a^4 x^4 \sqrt{\frac{1}{a x}+1}+315 a^3 x^3 \sqrt{\frac{1}{a x}+1}-35 a^2 x^2 \sqrt{\frac{1}{a x}+1}+910 a^5 x^5 \sqrt{1-\frac{1}{a x}} \sin ^{-1}\left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{2}}\right )-105 a^5 x^5 \sqrt{1-\frac{1}{a x}} \sin ^{-1}\left (\frac{1}{a x}\right )+910 a^4 x^4 \sqrt{1-\frac{1}{a x}} \sin ^{-1}\left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{2}}\right )-105 a^4 x^4 \sqrt{1-\frac{1}{a x}} \sin ^{-1}\left (\frac{1}{a x}\right )-175 a^5 x^5 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{\frac{1}{a x}+1} \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )\right )}{35 a^5 x^4 \sqrt{1-\frac{1}{a x}} (a x+1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(-35*a^2*Sqrt[1 + 1/(a*x)]*x^2 + 315*a^3*Sqrt[1 + 1/(a*x)]*x^3 + 280*a^4*Sqrt[1 + 1/(a*x)]*x^4 - 595*a^5*
Sqrt[1 + 1/(a*x)]*x^5 + 35*a^6*Sqrt[1 + 1/(a*x)]*x^6 + 910*a^4*Sqrt[1 - 1/(a*x)]*x^4*ArcSin[Sqrt[1 - 1/(a*x)]/
Sqrt[2]] + 910*a^5*Sqrt[1 - 1/(a*x)]*x^5*ArcSin[Sqrt[1 - 1/(a*x)]/Sqrt[2]] - 105*a^4*Sqrt[1 - 1/(a*x)]*x^4*Arc
Sin[1/(a*x)] - 105*a^5*Sqrt[1 - 1/(a*x)]*x^5*ArcSin[1/(a*x)] - 175*a^5*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 + 1/(a*x)]
*x^5*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]] + 7*Sqrt[2]*a*x*(-1 + a*x)^3*(1 + a*x)*Hypergeometric2F1[3/2, 5/2, 7/2, (1
- 1/(a*x))/2] + 5*Sqrt[2]*(-1 + a*x)^4*(1 + a*x)*Hypergeometric2F1[3/2, 7/2, 9/2, (1 - 1/(a*x))/2]))/(35*a^5*
Sqrt[1 - 1/(a*x)]*x^4*(1 + a*x))

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Maple [B]  time = 0.137, size = 600, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^4*a^4+(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^2*a^2-7*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)
*x^3*a^3+ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4-5*a^3*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2
-1)^(1/2))-4*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3+4*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2
)^(1/2))*x^3*a^4+2*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*x*a-11*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a^2+2*ln((a^2*x+(a^2
*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3-10*a^2*x^2*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+8*(a^2)^(1/
2)*((a*x-1)*(a*x+1))^(3/2)*x*a-8*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2+8*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*x^2-1)^(3/2)*(a^2)^(1/2)-5*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*x*a+ln((a^2
*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-5*a*x*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))-4*(a^2)^(1/
2)*((a*x-1)*(a*x+1))^(1/2)*x*a+4*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x*a^2)/a^2*c^2*((
a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/x/((a*x-1)*(a*x+1))^(1/2)/(a*x-1)

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Maxima [A]  time = 1.54951, size = 201, normalized size = 1.91 \begin{align*} -{\left (\frac{4 \, c^{2} \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} + \frac{10 \, c^{2} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a^{2}} + \frac{5 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{5 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac{16 \, c^{2} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2}}\right )} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*c^2*sqrt((a*x - 1)/(a*x + 1))/((a*x - 1)^2*a^2/(a*x + 1)^2 - a^2) + 10*c^2*arctan(sqrt((a*x - 1)/(a*x + 1)
))/a^2 + 5*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 5*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - 16*c^2*
sqrt((a*x - 1)/(a*x + 1))/a^2)*a

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Fricas [A]  time = 1.86976, size = 286, normalized size = 2.72 \begin{align*} -\frac{10 \, a c^{2} x \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) + 5 \, a c^{2} x \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 5 \, a c^{2} x \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) -{\left (a^{2} c^{2} x^{2} + 18 \, a c^{2} x + c^{2}\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(10*a*c^2*x*arctan(sqrt((a*x - 1)/(a*x + 1))) + 5*a*c^2*x*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 5*a*c^2*x*log(
sqrt((a*x - 1)/(a*x + 1)) - 1) - (a^2*c^2*x^2 + 18*a*c^2*x + c^2)*sqrt((a*x - 1)/(a*x + 1)))/(a^2*x)

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

[Out]

undef