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

Optimal. Leaf size=222 \[ -\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2} a}+\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2} a}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2} a}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2} a} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6125, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2} a}+\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2} a}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2} a}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2} a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - a*x)^(3/4)*(1 + a*x)^(1/4))/a) + ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(Sqrt[2]*a) - A
rcTan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(Sqrt[2]*a) - Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] - (Sqrt
[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(2*Sqrt[2]*a) + Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt[2]*(1 - a*x)
^(1/4))/(1 + a*x)^(1/4)]/(2*Sqrt[2]*a)

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 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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 6125

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 149, normalized size = 0.67 \[ \frac {-\frac {8 e^{\frac {1}{2} \tanh ^{-1}(a x)}}{e^{2 \tanh ^{-1}(a x)}+1}-\sqrt {2} \log \left (-\sqrt {2} e^{\frac {1}{2} \tanh ^{-1}(a x)}+e^{\tanh ^{-1}(a x)}+1\right )+\sqrt {2} \log \left (\sqrt {2} e^{\frac {1}{2} \tanh ^{-1}(a x)}+e^{\tanh ^{-1}(a x)}+1\right )-2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} e^{\frac {1}{2} \tanh ^{-1}(a x)}\right )+2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} e^{\frac {1}{2} \tanh ^{-1}(a x)}+1\right )}{4 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-8*E^(ArcTanh[a*x]/2))/(1 + E^(2*ArcTanh[a*x])) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^(ArcTanh[a*x]/2)] + 2*Sqrt[
2]*ArcTan[1 + Sqrt[2]*E^(ArcTanh[a*x]/2)] - Sqrt[2]*Log[1 - Sqrt[2]*E^(ArcTanh[a*x]/2) + E^ArcTanh[a*x]] + Sqr
t[2]*Log[1 + Sqrt[2]*E^(ArcTanh[a*x]/2) + E^ArcTanh[a*x]])/(4*a)

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fricas [B]  time = 0.63, size = 511, normalized size = 2.30 \[ -\frac {4 \, \sqrt {2} a \frac {1}{a^{4}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{3} \sqrt {\frac {\sqrt {2} {\left (a^{2} x - a\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {1}{4}} + {\left (a^{3} x - a^{2}\right )} \sqrt {\frac {1}{a^{4}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {3}{4}} - \sqrt {2} a^{3} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {3}{4}} - 1\right ) + 4 \, \sqrt {2} a \frac {1}{a^{4}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{3} \sqrt {-\frac {\sqrt {2} {\left (a^{2} x - a\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {1}{4}} - {\left (a^{3} x - a^{2}\right )} \sqrt {\frac {1}{a^{4}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {3}{4}} - \sqrt {2} a^{3} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {3}{4}} + 1\right ) - \sqrt {2} a \frac {1}{a^{4}}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} {\left (a^{2} x - a\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {1}{4}} + {\left (a^{3} x - a^{2}\right )} \sqrt {\frac {1}{a^{4}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) + \sqrt {2} a \frac {1}{a^{4}}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} {\left (a^{2} x - a\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{4}}^{\frac {1}{4}} - {\left (a^{3} x - a^{2}\right )} \sqrt {\frac {1}{a^{4}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) - 4 \, {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*a*(a^(-4))^(1/4)*arctan(sqrt(2)*a^3*sqrt((sqrt(2)*(a^2*x - a)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x -
1))*(a^(-4))^(1/4) + (a^3*x - a^2)*sqrt(a^(-4)) - sqrt(-a^2*x^2 + 1))/(a*x - 1))*(a^(-4))^(3/4) - sqrt(2)*a^3*
sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-4))^(3/4) - 1) + 4*sqrt(2)*a*(a^(-4))^(1/4)*arctan(sqrt(2)*a^3*sqrt(-
(sqrt(2)*(a^2*x - a)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-4))^(1/4) - (a^3*x - a^2)*sqrt(a^(-4)) + sqrt(-a
^2*x^2 + 1))/(a*x - 1))*(a^(-4))^(3/4) - sqrt(2)*a^3*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-4))^(3/4) + 1) -
sqrt(2)*a*(a^(-4))^(1/4)*log((sqrt(2)*(a^2*x - a)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-4))^(1/4) + (a^3*x
- a^2)*sqrt(a^(-4)) - sqrt(-a^2*x^2 + 1))/(a*x - 1)) + sqrt(2)*a*(a^(-4))^(1/4)*log(-(sqrt(2)*(a^2*x - a)*sqr
t(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-4))^(1/4) - (a^3*x - a^2)*sqrt(a^(-4)) + sqrt(-a^2*x^2 + 1))/(a*x - 1))
- 4*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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