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

Optimal. Leaf size=38 \[ \frac {a}{6 b^2 \left (a+b e^{2 x}\right )^3}-\frac {1}{4 b^2 \left (a+b e^{2 x}\right )^2} \]

[Out]

1/6*a/b^2/(a+b*exp(2*x))^3-1/4/b^2/(a+b*exp(2*x))^2

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2248, 43} \[ \frac {a}{6 b^2 \left (a+b e^{2 x}\right )^3}-\frac {1}{4 b^2 \left (a+b e^{2 x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*x)/(a + b*E^(2*x))^4,x]

[Out]

a/(6*b^2*(a + b*E^(2*x))^3) - 1/(4*b^2*(a + b*E^(2*x))^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 )^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{(a+b x)^4} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a}{b (a+b x)^4}+\frac {1}{b (a+b x)^3}\right ) \, dx,x,e^{2 x}\right )\\ &=\frac {a}{6 b^2 \left (a+b e^{2 x}\right )^3}-\frac {1}{4 b^2 \left (a+b e^{2 x}\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 0.74 \[ -\frac {a+3 b e^{2 x}}{12 b^2 \left (a+b e^{2 x}\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(a + 3*b*E^(2*x))/(b^2*(a + b*E^(2*x))^3)

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fricas [A]  time = 0.43, size = 51, normalized size = 1.34 \[ -\frac {3 \, b e^{\left (2 \, x\right )} + a}{12 \, {\left (b^{5} e^{\left (6 \, x\right )} + 3 \, a b^{4} e^{\left (4 \, x\right )} + 3 \, a^{2} b^{3} e^{\left (2 \, x\right )} + a^{3} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b*e^(2*x) + a)/(b^5*e^(6*x) + 3*a*b^4*e^(4*x) + 3*a^2*b^3*e^(2*x) + a^3*b^2)

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giac [A]  time = 0.31, size = 24, normalized size = 0.63 \[ -\frac {3 \, b e^{\left (2 \, x\right )} + a}{12 \, {\left (b e^{\left (2 \, x\right )} + a\right )}^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b*e^(2*x) + a)/((b*e^(2*x) + a)^3*b^2)

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maple [A]  time = 0.02, size = 33, normalized size = 0.87 \[ \frac {a}{6 \left (b \,{\mathrm e}^{2 x}+a \right )^{3} b^{2}}-\frac {1}{4 \left (b \,{\mathrm e}^{2 x}+a \right )^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a/b^2/(a+b*exp(x)^2)^3-1/4/b^2/(a+b*exp(x)^2)^2

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maxima [B]  time = 0.68, size = 91, normalized size = 2.39 \[ -\frac {b e^{\left (2 \, x\right )}}{4 \, {\left (b^{5} e^{\left (6 \, x\right )} + 3 \, a b^{4} e^{\left (4 \, x\right )} + 3 \, a^{2} b^{3} e^{\left (2 \, x\right )} + a^{3} b^{2}\right )}} - \frac {a}{12 \, {\left (b^{5} e^{\left (6 \, x\right )} + 3 \, a b^{4} e^{\left (4 \, x\right )} + 3 \, a^{2} b^{3} e^{\left (2 \, x\right )} + a^{3} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*e^(2*x)/(b^5*e^(6*x) + 3*a*b^4*e^(4*x) + 3*a^2*b^3*e^(2*x) + a^3*b^2) - 1/12*a/(b^5*e^(6*x) + 3*a*b^4*e
^(4*x) + 3*a^2*b^3*e^(2*x) + a^3*b^2)

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mupad [B]  time = 3.60, size = 55, normalized size = 1.45 \[ \frac {\frac {{\mathrm {e}}^{4\,x}}{4\,a}+\frac {b\,{\mathrm {e}}^{6\,x}}{12\,a^2}}{a^3+3\,{\mathrm {e}}^{2\,x}\,a^2\,b+3\,{\mathrm {e}}^{4\,x}\,a\,b^2+{\mathrm {e}}^{6\,x}\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*x)/(a + b*exp(2*x))^4,x)

[Out]

(exp(4*x)/(4*a) + (b*exp(6*x))/(12*a^2))/(b^3*exp(6*x) + a^3 + 3*a^2*b*exp(2*x) + 3*a*b^2*exp(4*x))

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sympy [A]  time = 0.17, size = 54, normalized size = 1.42 \[ \frac {- a - 3 b e^{2 x}}{12 a^{3} b^{2} + 36 a^{2} b^{3} e^{2 x} + 36 a b^{4} e^{4 x} + 12 b^{5} e^{6 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a - 3*b*exp(2*x))/(12*a**3*b**2 + 36*a**2*b**3*exp(2*x) + 36*a*b**4*exp(4*x) + 12*b**5*exp(6*x))

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