3.1.21 \(\int \frac {e^{2 x}}{(a+b e^x)^4} \, dx\) [21]

Optimal. Leaf size=34 \[ \frac {a}{3 b^2 \left (a+b e^x\right )^3}-\frac {1}{2 b^2 \left (a+b e^x\right )^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2280, 45} \begin {gather*} \frac {a}{3 b^2 \left (a+b e^x\right )^3}-\frac {1}{2 b^2 \left (a+b e^x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(a + b*E^x)^4,x]

[Out]

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

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Mathematica [A]
time = 0.03, size = 26, normalized size = 0.76 \begin {gather*} \frac {-a-3 b e^x}{6 b^2 \left (a+b e^x\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(a + b*E^x)^4,x]

[Out]

(-a - 3*b*E^x)/(6*b^2*(a + b*E^x)^3)

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Maple [A]
time = 0.02, size = 29, normalized size = 0.85

method	result	size
risch	\(-\frac {3 b \,{\mathrm e}^{x}+a}{6 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{3}}\)	\(21\)
norman	\(\frac {-\frac {a}{6 b^{2}}-\frac {{\mathrm e}^{x}}{2 b}}{\left (a +b \,{\mathrm e}^{x}\right )^{3}}\)	\(24\)
default	\(\frac {a}{3 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{3}}-\frac {1}{2 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (28) = 56\).
time = 0.27, size = 85, normalized size = 2.50 \begin {gather*} -\frac {b e^{x}}{2 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} - \frac {a}{6 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 47, normalized size = 1.38 \begin {gather*} -\frac {3 \, b e^{x} + a}{6 \, {\left (b^{5} e^{\left (3 \, x\right )} + 3 \, a b^{4} e^{\left (2 \, x\right )} + 3 \, a^{2} b^{3} e^{x} + a^{3} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.06, size = 51, normalized size = 1.50 \begin {gather*} \frac {- a - 3 b e^{x}}{6 a^{3} b^{2} + 18 a^{2} b^{3} e^{x} + 18 a b^{4} e^{2 x} + 6 b^{5} e^{3 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a - 3*b*exp(x))/(6*a**3*b**2 + 18*a**2*b**3*exp(x) + 18*a*b**4*exp(2*x) + 6*b**5*exp(3*x))

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Giac [A]
time = 3.56, size = 20, normalized size = 0.59 \begin {gather*} -\frac {3 \, b e^{x} + a}{6 \, {\left (b e^{x} + a\right )}^{3} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.60, size = 53, normalized size = 1.56 \begin {gather*} \frac {\frac {{\mathrm {e}}^{2\,x}}{2\,a}+\frac {b\,{\mathrm {e}}^{3\,x}}{6\,a^2}}{a^3+3\,{\mathrm {e}}^x\,a^2\,b+3\,{\mathrm {e}}^{2\,x}\,a\,b^2+{\mathrm {e}}^{3\,x}\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(a + b*exp(x))^4,x)

[Out]

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

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