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

Optimal. Leaf size=55 \[ \frac {2 (a x+1)^2}{a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {1-a^2 x^2}}{a}-\frac {3 \sin ^{-1}(a x)}{a} \]

[Out]

-3*arcsin(a*x)/a+2*(a*x+1)^2/a/(-a^2*x^2+1)^(1/2)+3*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6123, 853, 669, 641, 216} \[ \frac {2 (a x+1)^2}{a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {1-a^2 x^2}}{a}-\frac {3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 669

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

Rule 853

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

Rule 6123

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*
x^2]), x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 39, normalized size = 0.71 \[ \frac {\sqrt {1-a^2 x^2} \left (1-\frac {4}{a x-1}\right )}{a}-\frac {3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 65, normalized size = 1.18 \[ \frac {5 \, a x + 6 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x - 5\right )} - 5}{a^{2} x - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*x + 6*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(a*x - 5) - 5)/(a^2*x - a)

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giac [A]  time = 0.23, size = 63, normalized size = 1.15 \[ -\frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a} + \frac {8}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 79, normalized size = 1.44 \[ -\frac {a \,x^{2}}{\sqrt {-a^{2} x^{2}+1}}+\frac {5}{a \sqrt {-a^{2} x^{2}+1}}+\frac {4 x}{\sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 60, normalized size = 1.09 \[ -\frac {a x^{2}}{\sqrt {-a^{2} x^{2} + 1}} + \frac {4 \, x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {3 \, \arcsin \left (a x\right )}{a} + \frac {5}{\sqrt {-a^{2} x^{2} + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*x^2/sqrt(-a^2*x^2 + 1) + 4*x/sqrt(-a^2*x^2 + 1) - 3*arcsin(a*x)/a + 5/(sqrt(-a^2*x^2 + 1)*a)

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mupad [B]  time = 0.81, size = 81, normalized size = 1.47 \[ \frac {\sqrt {1-a^2\,x^2}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {4\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - a^2*x^2)^(1/2)/a - (3*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2) + (4*(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 \[ \int \frac {\left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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