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

Optimal. Leaf size=42 \[ \frac {3 \left (a+b e^{2 x}\right )^{4/3}}{8 b^2}-\frac {3 a \sqrt [3]{a+b e^{2 x}}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2248, 43} \[ \frac {3 \left (a+b e^{2 x}\right )^{4/3}}{8 b^2}-\frac {3 a \sqrt [3]{a+b e^{2 x}}}{2 b^2} \]

Antiderivative was successfully verified.

[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 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^{4 x}}{\left (a+b e^{2 x}\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{(a+b x)^{2/3}} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{2} \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.74 \[ \frac {3 \left (b e^{2 x}-3 a\right ) \sqrt [3]{a+b e^{2 x}}}{8 b^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.41, size = 25, normalized size = 0.60 \[ \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} {\left (b e^{\left (2 \, x\right )} - 3 \, a\right )}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.27, size = 32, normalized size = 0.76 \[ \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {4}{3}}}{8 \, b^{2}} - \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} a}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 27, normalized size = 0.64 \[ -\frac {3 \left (b \,{\mathrm e}^{2 x}+a \right )^{\frac {1}{3}} \left (-b \,{\mathrm e}^{2 x}+3 a \right )}{8 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 32, normalized size = 0.76 \[ \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {4}{3}}}{8 \, b^{2}} - \frac {3 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{\frac {1}{3}} a}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.49, size = 26, normalized size = 0.62 \[ -\frac {3\,\left (3\,a-b\,{\mathrm {e}}^{2\,x}\right )\,{\left (a+b\,{\mathrm {e}}^{2\,x}\right )}^{1/3}}{8\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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