3.6.28 \(\int \sqrt [4]{1-2 e^{x/3}} \, dx\) [528]

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*exp(1/3*x))^(1/4)-6*arctan((1-2*exp(1/3*x))^(1/4))-6*arctanh((1-2*exp(1/3*x))^(1/4))

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Rubi [A]
time = 0.01, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2320, 52, 65, 218, 212, 209} \begin {gather*} -6 \text {ArcTan}\left (\sqrt [4]{1-2 e^{x/3}}\right )+12 \sqrt [4]{1-2 e^{x/3}}-6 \tanh ^{-1}\left (\sqrt [4]{1-2 e^{x/3}}\right ) \end {gather*}

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 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 65

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 2320

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]]]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 54, normalized size = 1.00 \begin {gather*} 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 {gather*}

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 = 0.05, size = 57, normalized size = 1.06

method	result	size
derivativedivides	\(12 \left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}+3 \ln \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}-1\right )-3 \ln \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}+1\right )-6 \arctan \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}\right )\)	\(57\)
default	\(12 \left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}+3 \ln \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}-1\right )-3 \ln \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}+1\right )-6 \arctan \left (\left (1-2 \,{\mathrm e}^{\frac {x}{3}}\right )^{\frac {1}{4}}\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.59, size = 56, normalized size = 1.04 \begin {gather*} 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 {gather*}

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 = 0.66, size = 56, normalized size = 1.04 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [4]{1 - 2 e^{\frac {x}{3}}}\, dx \end {gather*}

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 = 1.13, size = 57, normalized size = 1.06 \begin {gather*} 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 {gather*}

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))

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Mupad [B]
time = 0.35, size = 33, normalized size = 0.61 \begin {gather*} \frac {12\,{\left (2-4\,{\mathrm {e}}^{x/3}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ \frac {{\mathrm {e}}^{-\frac {x}{3}}}{2}\right )}{{\left (2-{\mathrm {e}}^{-\frac {x}{3}}\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*exp(x/3))^(1/4),x)

[Out]

(12*(2 - 4*exp(x/3))^(1/4)*hypergeom([-1/4, -1/4], 3/4, exp(-x/3)/2))/(2 - exp(-x/3))^(1/4)

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