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

Optimal. Leaf size=97 \[ \frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{2 a^3 x^2}-\frac{c^3 (a x+4) \sqrt{1-a^2 x^2}}{2 a^2 x}+\frac{c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 a}-\frac{2 c^3 \sin ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.191305, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {6131, 6128, 1807, 813, 844, 216, 266, 63, 208} \[ \frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{2 a^3 x^2}-\frac{c^3 (a x+4) \sqrt{1-a^2 x^2}}{2 a^2 x}+\frac{c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 a}-\frac{2 c^3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - c/(a*x))^3,x]

[Out]

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

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]

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 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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.185543, size = 75, normalized size = 0.77 \[ \frac{c^3 \left (\frac{\sqrt{1-a^2 x^2} \left (-2 a^2 x^2-4 a x+1\right )}{a^2 x^2}+\log \left (\sqrt{1-a^2 x^2}+1\right )-\log (a x)-4 \sin ^{-1}(a x)\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 119, normalized size = 1.2 \begin{align*} -{\frac{{c}^{3}}{a}\sqrt{-{a}^{2}{x}^{2}+1}}-2\,{\frac{{c}^{3}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-2\,{\frac{{c}^{3}\sqrt{-{a}^{2}{x}^{2}+1}}{{a}^{2}x}}+{\frac{{c}^{3}}{2\,{x}^{2}{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{{c}^{3}}{2\,a}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.46918, size = 162, normalized size = 1.67 \begin{align*} -\frac{2 \, c^{3} \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{3}}{a} + \frac{{\left (a^{2} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-a^{2} x^{2} + 1}}{x^{2}}\right )} c^{3}}{2 \, a^{3}} - \frac{2 \, \sqrt{-a^{2} x^{2} + 1} c^{3}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.25605, size = 252, normalized size = 2.6 \begin{align*} \frac{8 \, a^{2} c^{3} x^{2} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) - a^{2} c^{3} x^{2} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - 2 \, a^{2} c^{3} x^{2} -{\left (2 \, a^{2} c^{3} x^{2} + 4 \, a c^{3} x - c^{3}\right )} \sqrt{-a^{2} x^{2} + 1}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 6.93348, size = 228, normalized size = 2.35 \begin{align*} a c^{3} \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: a^{2} = 0 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{a^{2}} & \text{otherwise} \end{cases}\right ) - 2 c^{3} \left (\begin{cases} \sqrt{\frac{1}{a^{2}}} \operatorname{asin}{\left (x \sqrt{a^{2}} \right )} & \text{for}\: a^{2} > 0 \\\sqrt{- \frac{1}{a^{2}}} \operatorname{asinh}{\left (x \sqrt{- a^{2}} \right )} & \text{for}\: a^{2} < 0 \end{cases}\right ) + \frac{2 c^{3} \left (\begin{cases} - \frac{i \sqrt{a^{2} x^{2} - 1}}{x} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{x} & \text{otherwise} \end{cases}\right )}{a^{2}} - \frac{c^{3} \left (\begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{2} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a}{2 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{2 a x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**3*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) - 2*c**3*Piecewise((sqrt(a**(-2))*
asin(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) + 2*c**3*Piecewise((-I*sqrt(a**
2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))/a**2 - c**3*Piecewise((-a**2*acosh(1/(a*x
))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a/(2*x*sqrt(1 - 1/
(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True))/a**3

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Giac [B]  time = 1.20748, size = 279, normalized size = 2.88 \begin{align*} -\frac{{\left (c^{3} - \frac{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{3}}{a^{2} x}\right )} a^{4} x^{2}}{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}{\left | a \right |}} - \frac{2 \, c^{3} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{c^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \,{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{3}}{a} - \frac{\frac{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{3}{\left | a \right |}}{a^{2} x} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{3}{\left | a \right |}}{a^{4} x^{2}}}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c^3 - 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3/(a^2*x))*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*abs(a))
- 2*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/2*c^3*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^3/a - 1/8*(8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3*abs(a)/(a^2*x) - (sqrt(-a^2*x^2 + 1
)*abs(a) + a)^2*c^3*abs(a)/(a^4*x^2))/a^2