3.9.20 \(\int e^{\tanh ^{-1}(a+b x)} x \, dx\) [820]

Optimal. Leaf size=84 \[ -\frac {(1-2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2}}{2 b^2}+\frac {(1-2 a) \text {ArcSin}(a+b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, 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 = {6298, 81, 52, 55, 633, 222} \begin {gather*} \frac {(1-2 a) \text {ArcSin}(a+b x)}{2 b^2}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}-\frac {(1-2 a) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]*x,x]

[Out]

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

Rule 52

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 55

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 222

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 633

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 6298

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 130, normalized size = 1.55 \begin {gather*} \frac {\sqrt {b} (-2+a-b x) \sqrt {1-a^2-2 a b x-b^2 x^2}+2 \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )+4 a \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{2 b^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]*x,x]

[Out]

(Sqrt[b]*(-2 + a - b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] + 2*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(S
qrt[2]*Sqrt[b])] + 4*a*Sqrt[-b]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(2*b^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(245\) vs. \(2(70)=140\).
time = 0.07, size = 246, normalized size = 2.93

method	result	size
risch	\(-\frac {\left (-b x +a -2\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b \sqrt {b^{2}}}\)	\(146\)
default	\(b \left (-\frac {x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 b^{2}}-\frac {3 a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{2}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}\right )}{2 b}+\frac {\left (-a^{2}+1\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{2} \sqrt {b^{2}}}\right )+\left (1+a \right ) \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{2}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}\right )\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)*x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (70) = 140\).
time = 0.46, size = 209, normalized size = 2.49 \begin {gather*} \frac {{\left (a + 1\right )} a \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{2}} - \frac {3 \, a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{2 \, b} + \frac {{\left (a^{2} - 1\right )} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a + 1\right )}}{b^{2}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 92, normalized size = 1.10 \begin {gather*} \frac {{\left (2 \, 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 + 2\right )}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*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 + 2))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (a + b x + 1\right )}{\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 68, normalized size = 0.81 \begin {gather*} -\frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\frac {x}{b} - \frac {a b - 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a - 1\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (a+b\,x+1\right )}{\sqrt {1-{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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