3.631 \(\int e^{\tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\)

Optimal. Leaf size=58 \[ \frac {c \sqrt {1-a^2 x^2} (1-a x)}{a^2 x}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}+\frac {c \sin ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6157, 6148, 813, 844, 216, 266, 63, 208} \[ \frac {c \sqrt {1-a^2 x^2} (1-a x)}{a^2 x}+\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}+\frac {c \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(1 - a*x)*Sqrt[1 - a^2*x^2])/(a^2*x) + (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 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 6148

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

Rule 6157

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx &=-\frac {c \int \frac {e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=-\frac {c \int \frac {(1+a x) \sqrt {1-a^2 x^2}}{x^2} \, dx}{a^2}\\ &=\frac {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {c \int \frac {-2 a+2 a^2 x}{x \sqrt {1-a^2 x^2}} \, dx}{2 a^2}\\ &=\frac {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+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 {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {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 {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {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 {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {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 = 0.04, size = 55, normalized size = 0.95 \[ \frac {c \left (\sqrt {1-a^2 x^2} (1-a x)+a x \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+a x \sin ^{-1}(a x)\right )}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 84, normalized size = 1.45 \[ -\frac {2 \, a c x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + a c x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + a c x + \sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}}{a^{2} x} \]

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^2/x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 128, normalized size = 2.21 \[ -\frac {a^{2} c x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c}{2 \, a^{2} x {\left | a \right |}} \]

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^2/x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 85, normalized size = 1.47 \[ -\frac {c \sqrt {-a^{2} x^{2}+1}}{a}+\frac {c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {c \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{a}+\frac {c \sqrt {-a^{2} x^{2}+1}}{a^{2} x} \]

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^2/x^2),x)

[Out]

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

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maxima [A]  time = 0.40, size = 79, normalized size = 1.36 \[ \frac {c \arcsin \left (a x\right )}{a} + \frac {c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a^{2} x} \]

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^2/x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.83, size = 76, normalized size = 1.31 \[ \frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}-\frac {c\,\sqrt {1-a^2\,x^2}}{a}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c\,\sqrt {1-a^2\,x^2}}{a^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.60, size = 144, normalized size = 2.48 \[ a c \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) - \frac {c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{a^{2}} \]

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**2/x**2),x)

[Out]

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

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