3.432 \(\int e^{-3 \coth ^{-1}(a x)} (c-\frac {c}{a x}) \, dx\)

Optimal. Leaf size=75 \[ \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {4 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}+\frac {c \csc ^{-1}(a x)}{a} \]

[Out]

c*arccsc(a*x)/a-4*c*arctanh((1-1/a^2/x^2)^(1/2))/a+8*c*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+c*x*(1-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6177, 1805, 1807, 844, 216, 266, 63, 208} \[ \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {4 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}+\frac {c \csc ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.54, size = 234, normalized size = 3.12 \[ \frac {5 a^2 c x^2 \left ((a x+1) \left (\sqrt {\frac {1}{a x}+1} \left (a^2 x^2-3 a x+2\right )+6 a x \sqrt {1-\frac {1}{a x}} \sin ^{-1}\left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-2 a x \sqrt {1-\frac {1}{a x}} \sin ^{-1}\left (\frac {1}{a x}\right )\right )-4 a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {\frac {1}{a x}+1} \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )+\sqrt {2} c (a x+1) (a x-1)^3 \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )}{5 a^4 x^3 \sqrt {1-\frac {1}{a x}} (a x+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(5*a^2*c*x^2*((1 + a*x)*(Sqrt[1 + 1/(a*x)]*(2 - 3*a*x + a^2*x^2) + 6*a*Sqrt[1 - 1/(a*x)]*x*ArcSin[Sqrt[1 - 1/(
a*x)]/Sqrt[2]] - 2*a*Sqrt[1 - 1/(a*x)]*x*ArcSin[1/(a*x)]) - 4*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 + 1/(a*x)]*x^2*
ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]) + Sqrt[2]*c*(-1 + a*x)^3*(1 + a*x)*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - 1/(a*
x))/2])/(5*a^4*Sqrt[1 - 1/(a*x)]*x^3*(1 + a*x))

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fricas [A]  time = 0.65, size = 92, normalized size = 1.23 \[ -\frac {2 \, c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + 4 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 4 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a c x + 9 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*c*arctan(sqrt((a*x - 1)/(a*x + 1))) + 4*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 4*c*log(sqrt((a*x - 1)/(a*x
+ 1)) - 1) - (a*c*x + 9*c)*sqrt((a*x - 1)/(a*x + 1)))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 0.05, size = 376, normalized size = 5.01 \[ -\frac {\left (-4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{2} a^{2}-a^{2} x^{2} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+4 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-8 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x a -2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a -2 a x \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+8 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}-4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}-\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}+4 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right ) c \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-4*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x^2*a^2-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a^2-a^2*x^2*(a^2)^(1/2)*arc
tan(1/(a^2*x^2-1)^(1/2))+4*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+4*((a*x-1)*(a*x
+1))^(3/2)*(a^2)^(1/2)-8*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x*a-2*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x*a-2*a*x*(a^
2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+8*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-4*((a
*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)-arctan(1/(a^2*x^2-1)^(1/2))*(a^2)^(1/2)+4*a*ln(
(a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(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 = 0.42, size = 135, normalized size = 1.80 \[ -2 \, a {\left (\frac {c \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} + \frac {2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {4 \, c \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 107, normalized size = 1.43 \[ \frac {2\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}-\frac {2\,c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {8\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}+\frac {c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*((a*x - 1)/(a*x + 1))^(1/2))/(a - (a*(a*x - 1))/(a*x + 1)) - (2*c*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a +
(c*atan(((a*x - 1)/(a*x + 1))^(1/2)*1i)*8i)/a + (8*c*((a*x - 1)/(a*x + 1))^(1/2))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c \left (\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x^{2} + x}\, dx + \int \left (- \frac {2 a \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {a^{2} x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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