3.534 \(\int \frac {e^{2 x}}{(3-e^{x/2})^{3/4}} \, dx\)

Optimal. Leaf size=73 \[ \frac {8}{13} \left (3-e^{x/2}\right )^{13/4}-8 \left (3-e^{x/2}\right )^{9/4}+\frac {216}{5} \left (3-e^{x/2}\right )^{5/4}-216 \sqrt [4]{3-e^{x/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2248, 43} \[ \frac {8}{13} \left (3-e^{x/2}\right )^{13/4}-8 \left (3-e^{x/2}\right )^{9/4}+\frac {216}{5} \left (3-e^{x/2}\right )^{5/4}-216 \sqrt [4]{3-e^{x/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(3 - E^(x/2))^(3/4),x]

[Out]

-216*(3 - E^(x/2))^(1/4) + (216*(3 - E^(x/2))^(5/4))/5 - 8*(3 - E^(x/2))^(9/4) + (8*(3 - E^(x/2))^(13/4))/13

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {align*} \int \frac {e^{2 x}}{\left (3-e^{x/2}\right )^{3/4}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3}{(3-x)^{3/4}} \, dx,x,e^{x/2}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {27}{(3-x)^{3/4}}-27 \sqrt [4]{3-x}+9 (3-x)^{5/4}-(3-x)^{9/4}\right ) \, dx,x,e^{x/2}\right )\\ &=-216 \sqrt [4]{3-e^{x/2}}+\frac {216}{5} \left (3-e^{x/2}\right )^{5/4}-8 \left (3-e^{x/2}\right )^{9/4}+\frac {8}{13} \left (3-e^{x/2}\right )^{13/4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.60 \[ -\frac {8}{65} \sqrt [4]{3-e^{x/2}} \left (96 e^{x/2}+20 e^x+5 e^{3 x/2}+1152\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(3 - E^(x/2))^(3/4),x]

[Out]

(-8*(3 - E^(x/2))^(1/4)*(1152 + 96*E^(x/2) + 20*E^x + 5*E^((3*x)/2)))/65

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{2 x}}{\left (3-e^{x/2}\right )^{3/4}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[E^(2*x)/(3 - E^(x/2))^(3/4),x]

[Out]

Could not integrate

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fricas [A]  time = 1.25, size = 30, normalized size = 0.41 \[ -\frac {8}{65} \, {\left (5 \, e^{\left (\frac {3}{2} \, x\right )} + 96 \, e^{\left (\frac {1}{2} \, x\right )} + 20 \, e^{x} + 1152\right )} {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/65*(5*e^(3/2*x) + 96*e^(1/2*x) + 20*e^x + 1152)*(-e^(1/2*x) + 3)^(1/4)

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giac [A]  time = 0.66, size = 65, normalized size = 0.89 \[ -\frac {8}{13} \, {\left (e^{\left (\frac {1}{2} \, x\right )} - 3\right )}^{3} {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {1}{4}} - 8 \, {\left (e^{\left (\frac {1}{2} \, x\right )} - 3\right )}^{2} {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {1}{4}} + \frac {216}{5} \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {5}{4}} - 216 \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/13*(e^(1/2*x) - 3)^3*(-e^(1/2*x) + 3)^(1/4) - 8*(e^(1/2*x) - 3)^2*(-e^(1/2*x) + 3)^(1/4) + 216/5*(-e^(1/2*x
) + 3)^(5/4) - 216*(-e^(1/2*x) + 3)^(1/4)

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maple [A]  time = 0.03, size = 37, normalized size = 0.51




method result size



risch \(\frac {8 \left (5 \,{\mathrm e}^{\frac {3 x}{2}}+20 \,{\mathrm e}^{x}+96 \,{\mathrm e}^{\frac {x}{2}}+1152\right ) \left (-3+{\mathrm e}^{\frac {x}{2}}\right )}{65 \left (3-{\mathrm e}^{\frac {x}{2}}\right )^{\frac {3}{4}}}\) \(37\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/65/(3-exp(1/2*x))^(3/4)*(5*exp(3/2*x)+20*exp(x)+96*exp(1/2*x)+1152)*(-3+exp(1/2*x))

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maxima [A]  time = 0.57, size = 49, normalized size = 0.67 \[ \frac {8}{13} \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {13}{4}} - 8 \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {9}{4}} + \frac {216}{5} \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {5}{4}} - 216 \, {\left (-e^{\left (\frac {1}{2} \, x\right )} + 3\right )}^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/13*(-e^(1/2*x) + 3)^(13/4) - 8*(-e^(1/2*x) + 3)^(9/4) + 216/5*(-e^(1/2*x) + 3)^(5/4) - 216*(-e^(1/2*x) + 3)^
(1/4)

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mupad [B]  time = 0.11, size = 30, normalized size = 0.41 \[ -{\left (3-{\mathrm {e}}^{x/2}\right )}^{1/4}\,\left (\frac {768\,{\mathrm {e}}^{x/2}}{65}+\frac {8\,{\mathrm {e}}^{\frac {3\,x}{2}}}{13}+\frac {32\,{\mathrm {e}}^x}{13}+\frac {9216}{65}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(3 - exp(x/2))^(3/4),x)

[Out]

-(3 - exp(x/2))^(1/4)*((768*exp(x/2))/65 + (8*exp((3*x)/2))/13 + (32*exp(x))/13 + 9216/65)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{2 x}}{\left (3 - e^{\frac {x}{2}}\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(2*x)/(3 - exp(x/2))**(3/4), x)

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