\(\int \frac {e^{4 x}}{(a+b e^{2 x})^{2/3}} \, dx\) [26]

Optimal result

Integrand size = 19, antiderivative size = 42 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=-\frac {3 a \sqrt [3]{a+b e^{2 x}}}{2 b^2}+\frac {3 \left (a+b e^{2 x}\right )^{4/3}}{8 b^2} \]

[Out]

-3/2*a*(a+b*exp(2*x))^(1/3)/b^2+3/8*(a+b*exp(2*x))^(4/3)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2280, 45} \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {3 \left (a+b e^{2 x}\right )^{4/3}}{8 b^2}-\frac {3 a \sqrt [3]{a+b e^{2 x}}}{2 b^2} \]

[In]

Int[E^(4*x)/(a + b*E^(2*x))^(2/3),x]

[Out]

(-3*a*(a + b*E^(2*x))^(1/3))/(2*b^2) + (3*(a + b*E^(2*x))^(4/3))/(8*b^2)

Rule 45

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 2280

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(a+b x)^{2/3}} \, dx,x,e^{2 x}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^{2/3}}+\frac {\sqrt [3]{a+b x}}{b}\right ) \, dx,x,e^{2 x}\right ) \\ & = -\frac {3 a \sqrt [3]{a+b e^{2 x}}}{2 b^2}+\frac {3 \left (a+b e^{2 x}\right )^{4/3}}{8 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {3 \left (-3 a+b e^{2 x}\right ) \sqrt [3]{a+b e^{2 x}}}{8 b^2} \]

[In]

Integrate[E^(4*x)/(a + b*E^(2*x))^(2/3),x]

[Out]

(3*(-3*a + b*E^(2*x))*(a + b*E^(2*x))^(1/3))/(8*b^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {3 \left (a +b \,{\mathrm e}^{2 x}\right )^{\frac {1}{3}} \left (-b \,{\mathrm e}^{2 x}+3 a \right )}{8 b^{2}}\)	\(27\)

[In]

int(exp(4*x)/(a+b*exp(2*x))^(2/3),x,method=_RETURNVERBOSE)

[Out]

-3/8*(a+b*exp(2*x))^(1/3)*(-b*exp(2*x)+3*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.60 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} {\left (b e^{\left (2 \, x\right )} - 3 \, a\right )}}{8 \, b^{2}} \]

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^(2/3),x, algorithm="fricas")

[Out]

3/8*(b*e^(2*x) + a)^(1/3)*(b*e^(2*x) - 3*a)/b^2

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {\begin {cases} \frac {3 \left (- a \sqrt [3]{a + b e^{2 x}} + \frac {\left (a + b e^{2 x}\right )^{\frac {4}{3}}}{4}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {e^{4 x}}{2 a^{\frac {2}{3}}} & \text {otherwise} \end {cases}}{2} \]

[In]

integrate(exp(4*x)/(a+b*exp(2*x))**(2/3),x)

[Out]

Piecewise((3*(-a*(a + b*exp(2*x))**(1/3) + (a + b*exp(2*x))**(4/3)/4)/b**2, Ne(b, 0)), (exp(4*x)/(2*a**(2/3)),
True))/2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {4}{3}}}{8 \, b^{2}} - \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} a}{2 \, b^{2}} \]

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^(2/3),x, algorithm="maxima")

[Out]

3/8*(b*e^(2*x) + a)^(4/3)/b^2 - 3/2*(b*e^(2*x) + a)^(1/3)*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=\frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {4}{3}}}{8 \, b^{2}} - \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} a}{2 \, b^{2}} \]

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^(2/3),x, algorithm="giac")

[Out]

3/8*(b*e^(2*x) + a)^(4/3)/b^2 - 3/2*(b*e^(2*x) + a)^(1/3)*a/b^2

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.62 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx=-\frac {3\,\left (3\,a-b\,{\mathrm {e}}^{2\,x}\right )\,{\left (a+b\,{\mathrm {e}}^{2\,x}\right )}^{1/3}}{8\,b^2} \]

[In]

int(exp(4*x)/(a + b*exp(2*x))^(2/3),x)

[Out]

-(3*(3*a - b*exp(2*x))*(a + b*exp(2*x))^(1/3))/(8*b^2)