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

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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

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fricas [A]  time = 0.58, size = 100, normalized size = 1.47 \[ \frac {3 \, {\left (b x + a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a - 5\right )}}{b^{2} x + {\left (a - 1\right )} b} \]

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)

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giac [A]  time = 0.31, size = 75, normalized size = 1.10 \[ \frac {3 \, \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b} + \frac {8}{{\left (\frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \]

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))

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maple [B]  time = 0.04, size = 388, normalized size = 5.71 \[ \frac {2 \left (1+a \right )^{3} \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {3 a}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 a^{2} x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 a^{3}}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {4 a^{2}}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a^{3} x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a^{4}}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {b \,x^{2}}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {5 a x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {5}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}} \]

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]

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

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maxima [B]  time = 0.42, size = 753, normalized size = 11.07 \[ -\frac {6 \, a^{3} b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {5 \, {\left (a^{2} - 1\right )} a b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {{\left (a^{2} - 1\right )} a^{2} b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {6 \, {\left (a b^{2} + b^{2}\right )} a^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {3 \, {\left (a^{2} b + 2 \, a b + b\right )} a b x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {b x^{2}}{\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {3 \, {\left (a^{2} b + 2 \, a b + b\right )} a^{2}}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} a b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {3 \, {\left (a b^{2} + b^{2}\right )} {\left (a^{2} - 1\right )} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} - \frac {3 \, a \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b} + \frac {3 \, {\left (a b^{2} + b^{2}\right )} {\left (a^{2} - 1\right )} a}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} b} - \frac {2 \, {\left (a^{2} - 1\right )}}{\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} b} + \frac {3 \, {\left (a b^{2} + b^{2}\right )} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{3}} + \frac {3 \, {\left (a^{2} b + 2 \, a b + b\right )}}{\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} b^{2}} \]

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]

-6*a^3*b^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 5*(a^2 - 1)*a*b^2*x/((a^2*b^2 -
(a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - (a^2 - 1)*a^2*b/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2
- 2*a*b*x - a^2 + 1)) + 6*(a*b^2 + b^2)*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) -
3*(a^2*b + 2*a*b + b)*a*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + (a^3 + 3*a^2 + 3
*a + 1)*b^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - b*x^2/sqrt(-b^2*x^2 - 2*a*b*x -
a^2 + 1) - 3*(a^2*b + 2*a*b + b)*a^2/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + (a^3 +
3*a^2 + 3*a + 1)*a*b/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 3*(a*b^2 + b^2)*(a^2 - 1
)*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 3*a*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 -
(a^2 - 1)*b^2))/b + 3*(a*b^2 + b^2)*(a^2 - 1)*a/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*
b) - 2*(a^2 - 1)/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 3*(a*b^2 + b^2)*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 -
(a^2 - 1)*b^2))/b^3 + 3*(a^2*b + 2*a*b + b)/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,x+1\right )}^3}{{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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