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

Optimal. Leaf size=77 \[ -\frac {c \sqrt {1-a^2 x^2}}{a}-\frac {8 c (1-a x)}{a \sqrt {1-a^2 x^2}}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}-\frac {4 c \sin ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6131, 6128, 1805, 1809, 844, 216, 266, 63, 208} \[ -\frac {c \sqrt {1-a^2 x^2}}{a}-\frac {8 c (1-a x)}{a \sqrt {1-a^2 x^2}}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}-\frac {4 c \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c*(1 - a*x))/(a*Sqrt[1 - a^2*x^2]) - (c*Sqrt[1 - a^2*x^2])/a - (4*c*ArcSin[a*x])/a + (c*ArcTanh[Sqrt[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 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 6128

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

Rule 6131

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 61, normalized size = 0.79 \[ \frac {c \left (-\frac {\sqrt {1-a^2 x^2} (a x+9)}{a x+1}+\log \left (\sqrt {1-a^2 x^2}+1\right )-4 \sin ^{-1}(a x)-\log (x)\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*(-(((9 + a*x)*Sqrt[1 - a^2*x^2])/(1 + a*x)) - 4*ArcSin[a*x] - Log[x] + Log[1 + Sqrt[1 - a^2*x^2]]))/a

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fricas [A]  time = 0.43, size = 97, normalized size = 1.26 \[ -\frac {9 \, a c x - 8 \, {\left (a c x + c\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a c x + c\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a c x + 9 \, c\right )} + 9 \, c}{a^{2} x + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(9*a*c*x - 8*(a*c*x + c)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a*c*x + c)*log((sqrt(-a^2*x^2 + 1) - 1)/x)
+ sqrt(-a^2*x^2 + 1)*(a*c*x + 9*c) + 9*c)/(a^2*x + a)

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giac [A]  time = 4.04, size = 104, normalized size = 1.35 \[ -\frac {4 \, c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {16 \, c}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*arcsin(a*x)*sgn(a)/abs(a) + c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt
(-a^2*x^2 + 1)*c/a + 16*c/(((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))

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maple [B]  time = 0.05, size = 223, normalized size = 2.90 \[ -\frac {c \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a}-\frac {c \sqrt {-a^{2} x^{2}+1}}{a}+\frac {c \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{a}-\frac {2 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{4} \left (x +\frac {1}{a}\right )^{3}}-\frac {3 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{3} \left (x +\frac {1}{a}\right )^{2}}-\frac {8 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3 a}-4 c \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x -\frac {4 c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*(-a^2*x^2+1)^(3/2)/a-c*(-a^2*x^2+1)^(1/2)/a+c/a*arctanh(1/(-a^2*x^2+1)^(1/2))-2*c/a^4/(x+1/a)^3*(-a^2*(
x+1/a)^2+2*a*(x+1/a))^(5/2)-3*c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-8/3*c/a*(-a^2*(x+1/a)^2+2*a*(
x+1/a))^(3/2)-4*c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-4*c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*
a*(x+1/a))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a x}\right )}}{{\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.84, size = 102, normalized size = 1.32 \[ \frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}-\frac {c\,\sqrt {1-a^2\,x^2}}{a}-\frac {4\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {8\,c\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*atanh((1 - a^2*x^2)^(1/2)))/a - (c*(1 - a^2*x^2)^(1/2))/a - (4*c*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2) + (8*c
*(1 - a^2*x^2)^(1/2))/((x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{4} + 3 a^{2} x^{3} + 3 a x^{2} + x}\right )\, dx + \int \frac {a x \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{4} + 3 a^{2} x^{3} + 3 a x^{2} + x}\, dx + \int \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{4} + 3 a^{2} x^{3} + 3 a x^{2} + x}\, dx + \int \left (- \frac {a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{4} + 3 a^{2} x^{3} + 3 a x^{2} + x}\right )\, dx\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(Integral(-sqrt(-a**2*x**2 + 1)/(a**3*x**4 + 3*a**2*x**3 + 3*a*x**2 + x), x) + Integral(a*x*sqrt(-a**2*x**2
+ 1)/(a**3*x**4 + 3*a**2*x**3 + 3*a*x**2 + x), x) + Integral(a**2*x**2*sqrt(-a**2*x**2 + 1)/(a**3*x**4 + 3*a**
2*x**3 + 3*a*x**2 + x), x) + Integral(-a**3*x**3*sqrt(-a**2*x**2 + 1)/(a**3*x**4 + 3*a**2*x**3 + 3*a*x**2 + x)
, x))/a

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