3.528 \(\int \sqrt [4]{1-2 e^{x/3}} \, dx\)

Optimal. Leaf size=54 \[ 12 \sqrt [4]{1-2 e^{x/3}}-6 \tan ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right )-6 \tanh ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right ) \]

[Out]

12*(1 - 2*E^(x/3))^(1/4) - 6*ArcTan[(1 - 2*E^(x/3))^(1/4)] - 6*ArcTanh[(1 - 2*E^(x/3))^(1/4)]

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Rubi [A]  time = 0.021602, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2282, 50, 63, 212, 206, 203} \[ 12 \sqrt [4]{1-2 e^{x/3}}-6 \tan ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right )-6 \tanh ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*E^(x/3))^(1/4),x]

[Out]

12*(1 - 2*E^(x/3))^(1/4) - 6*ArcTan[(1 - 2*E^(x/3))^(1/4)] - 6*ArcTanh[(1 - 2*E^(x/3))^(1/4)]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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 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 212

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

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \sqrt [4]{1-2 e^{x/3}} \, dx &=3 \operatorname{Subst}\left (\int \frac{\sqrt [4]{1-2 x}}{x} \, dx,x,e^{x/3}\right )\\ &=12 \sqrt [4]{1-2 e^{x/3}}+3 \operatorname{Subst}\left (\int \frac{1}{(1-2 x)^{3/4} x} \, dx,x,e^{x/3}\right )\\ &=12 \sqrt [4]{1-2 e^{x/3}}-6 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}-\frac{x^4}{2}} \, dx,x,\sqrt [4]{1-2 e^{x/3}}\right )\\ &=12 \sqrt [4]{1-2 e^{x/3}}-6 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [4]{1-2 e^{x/3}}\right )-6 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [4]{1-2 e^{x/3}}\right )\\ &=12 \sqrt [4]{1-2 e^{x/3}}-6 \tan ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right )-6 \tanh ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0141379, size = 54, normalized size = 1. \[ 12 \sqrt [4]{1-2 e^{x/3}}-6 \tan ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right )-6 \tanh ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*E^(x/3))^(1/4),x]

[Out]

12*(1 - 2*E^(x/3))^(1/4) - 6*ArcTan[(1 - 2*E^(x/3))^(1/4)] - 6*ArcTanh[(1 - 2*E^(x/3))^(1/4)]

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Maple [A]  time = 0.006, size = 57, normalized size = 1.1 \begin{align*} 12\,\sqrt [4]{1-2\,{{\rm e}^{x/3}}}+3\,\ln \left ( -1+\sqrt [4]{1-2\,{{\rm e}^{x/3}}} \right ) -3\,\ln \left ( 1+\sqrt [4]{1-2\,{{\rm e}^{x/3}}} \right ) -6\,\arctan \left ( \sqrt [4]{1-2\,{{\rm e}^{x/3}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*exp(1/3*x))^(1/4),x)

[Out]

12*(1-2*exp(1/3*x))^(1/4)+3*ln(-1+(1-2*exp(1/3*x))^(1/4))-3*ln(1+(1-2*exp(1/3*x))^(1/4))-6*arctan((1-2*exp(1/3
*x))^(1/4))

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Maxima [A]  time = 1.42446, size = 76, normalized size = 1.41 \begin{align*} 12 \,{\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 6 \, \arctan \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}}\right ) - 3 \, \log \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} + 1\right ) + 3 \, \log \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*exp(1/3*x))^(1/4),x, algorithm="maxima")

[Out]

12*(-2*e^(1/3*x) + 1)^(1/4) - 6*arctan((-2*e^(1/3*x) + 1)^(1/4)) - 3*log((-2*e^(1/3*x) + 1)^(1/4) + 1) + 3*log
((-2*e^(1/3*x) + 1)^(1/4) - 1)

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Fricas [A]  time = 1.69538, size = 192, normalized size = 3.56 \begin{align*} 12 \,{\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 6 \, \arctan \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}}\right ) - 3 \, \log \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} + 1\right ) + 3 \, \log \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*exp(1/3*x))^(1/4),x, algorithm="fricas")

[Out]

12*(-2*e^(1/3*x) + 1)^(1/4) - 6*arctan((-2*e^(1/3*x) + 1)^(1/4)) - 3*log((-2*e^(1/3*x) + 1)^(1/4) + 1) + 3*log
((-2*e^(1/3*x) + 1)^(1/4) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [4]{1 - 2 e^{\frac{x}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*exp(1/3*x))**(1/4),x)

[Out]

Integral((1 - 2*exp(x/3))**(1/4), x)

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Giac [A]  time = 1.14576, size = 77, normalized size = 1.43 \begin{align*} 12 \,{\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 6 \, \arctan \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}}\right ) - 3 \, \log \left ({\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} + 1\right ) + 3 \, \log \left ({\left |{\left (-2 \, e^{\left (\frac{1}{3} \, x\right )} + 1\right )}^{\frac{1}{4}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*exp(1/3*x))^(1/4),x, algorithm="giac")

[Out]

12*(-2*e^(1/3*x) + 1)^(1/4) - 6*arctan((-2*e^(1/3*x) + 1)^(1/4)) - 3*log((-2*e^(1/3*x) + 1)^(1/4) + 1) + 3*log
(abs((-2*e^(1/3*x) + 1)^(1/4) - 1))