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

Optimal. Leaf size=68 \[ \frac{2 (a+b x+1)^{3/2}}{b \sqrt{-a-b x+1}}+\frac{3 \sqrt{-a-b x+1} \sqrt{a+b x+1}}{b}-\frac{3 \sin ^{-1}(a+b x)}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0343045, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6161, 47, 50, 53, 619, 216} \[ \frac{2 (a+b x+1)^{3/2}}{b \sqrt{-a-b x+1}}+\frac{3 \sqrt{-a-b x+1} \sqrt{a+b x+1}}{b}-\frac{3 \sin ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x]),x]

[Out]

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

Rule 6161

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^(
n/2), x] /; FreeQ[{a, b, c, n}, x]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 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]

Rubi steps

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

Mathematica [A]  time = 0.0385456, size = 43, normalized size = 0.63 \[ \frac{\left (1-\frac{4}{a+b x-1}\right ) \sqrt{1-(a+b x)^2}}{b}-\frac{3 \sin ^{-1}(a+b x)}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a + b*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.037, size = 388, normalized size = 5.7 \begin{align*} 3\,{\frac{x}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}+3\,{\frac{a}{b\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}-3\,{\frac{{a}^{2}x}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}-3\,{\frac{{a}^{3}}{b\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}+2\,{\frac{ \left ( 1+a \right ) ^{3} \left ( -2\,{b}^{2}x-2\,ab \right ) }{ \left ( -4\,{b}^{2} \left ( -{a}^{2}+1 \right ) -4\,{a}^{2}{b}^{2} \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}-{\frac{{a}^{4}}{b}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-5\,{\frac{ax}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}-4\,{\frac{{a}^{2}}{b\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}-{x{a}^{3}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{b{x}^{2}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-3\,{\frac{1}{\sqrt{{b}^{2}}}\arctan \left ({\frac{\sqrt{{b}^{2}}}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}} \left ( x+{\frac{a}{b}} \right ) } \right ) }+5\,{\frac{1}{b\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.99537, size = 234, normalized size = 3.44 \begin{align*} \frac{3 \,{\left (b x + a - 1\right )} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a - 5\right )}}{b^{2} x +{\left (a - 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + 1\right )^{3}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.24711, size = 101, normalized size = 1.49 \begin{align*} \frac{3 \, \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{{\left | b \right |}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{b} + \frac{8}{{\left (\frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left | b \right |} + b}{b^{2} x + a b} - 1\right )}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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