3.23 \(\int \frac {x}{(b e^{-p x}+a e^{p x})^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac {\log \left (a e^{2 p x}+b\right )}{4 a b p^2}+\frac {x}{2 a b p}-\frac {x}{2 a p \left (a e^{2 p x}+b\right )} \]

[Out]

1/2*x/a/b/p-1/2*x/a/(b+a*exp(2*p*x))/p-1/4*ln(b+a*exp(2*p*x))/a/b/p^2

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Rubi [A]  time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2283, 2191, 2282, 36, 29, 31} \[ -\frac {\log \left (a e^{2 p x}+b\right )}{4 a b p^2}+\frac {x}{2 a b p}-\frac {x}{2 a p \left (a e^{2 p x}+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(b/E^(p*x) + a*E^(p*x))^2,x]

[Out]

x/(2*a*b*p) - x/(2*a*(b + a*E^(2*p*x))*p) - Log[b + a*E^(2*p*x)]/(4*a*b*p^2)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2283

Int[(u_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n*v)*(a + b*F^ExpandToSum[w - v, x])
^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ[n, 0] && LinearQ[{v, w}, x]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 49, normalized size = 0.79 \[ \frac {\frac {2 p x e^{2 p x}}{a e^{2 p x}+b}-\frac {\log \left (a e^{2 p x}+b\right )}{a}}{4 b p^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*E^(2*p*x)*p*x)/(b + a*E^(2*p*x)) - Log[b + a*E^(2*p*x)]/a)/(4*b*p^2)

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fricas [A]  time = 0.74, size = 58, normalized size = 0.94 \[ \frac {2 \, a p x e^{\left (2 \, p x\right )} - {\left (a e^{\left (2 \, p x\right )} + b\right )} \log \left (a e^{\left (2 \, p x\right )} + b\right )}{4 \, {\left (a^{2} b p^{2} e^{\left (2 \, p x\right )} + a b^{2} p^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*p*x*e^(2*p*x) - (a*e^(2*p*x) + b)*log(a*e^(2*p*x) + b))/(a^2*b*p^2*e^(2*p*x) + a*b^2*p^2)

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giac [A]  time = 1.21, size = 74, normalized size = 1.19 \[ \frac {2 \, a p x e^{\left (2 \, p x\right )} - a e^{\left (2 \, p x\right )} \log \left (-a e^{\left (2 \, p x\right )} - b\right ) - b \log \left (-a e^{\left (2 \, p x\right )} - b\right )}{4 \, {\left (a^{2} b p^{2} e^{\left (2 \, p x\right )} + a b^{2} p^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*p*x*e^(2*p*x) - a*e^(2*p*x)*log(-a*e^(2*p*x) - b) - b*log(-a*e^(2*p*x) - b))/(a^2*b*p^2*e^(2*p*x) + a
*b^2*p^2)

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maple [A]  time = 0.01, size = 51, normalized size = 0.82 \[ \frac {x \,{\mathrm e}^{2 p x}}{2 \left (a \,{\mathrm e}^{2 p x}+b \right ) b p}-\frac {\ln \left (a \,{\mathrm e}^{2 p x}+b \right )}{4 a b \,p^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/p^2/b/a*ln(a*exp(p*x)^2+b)+1/2/p*x*exp(p*x)^2/b/(a*exp(p*x)^2+b)

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maxima [A]  time = 0.59, size = 51, normalized size = 0.82 \[ \frac {x e^{\left (2 \, p x\right )}}{2 \, {\left (a b p e^{\left (2 \, p x\right )} + b^{2} p\right )}} - \frac {\log \left (\frac {a e^{\left (2 \, p x\right )} + b}{a}\right )}{4 \, a b p^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*e^(2*p*x)/(a*b*p*e^(2*p*x) + b^2*p) - 1/4*log((a*e^(2*p*x) + b)/a)/(a*b*p^2)

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mupad [B]  time = 0.41, size = 47, normalized size = 0.76 \[ \frac {x\,{\mathrm {e}}^{2\,p\,x}}{2\,b\,p\,\left (b+a\,{\mathrm {e}}^{2\,p\,x}\right )}-\frac {\ln \left (b+a\,{\mathrm {e}}^{2\,p\,x}\right )}{4\,a\,b\,p^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*exp(p*x) + b*exp(-p*x))^2,x)

[Out]

(x*exp(2*p*x))/(2*b*p*(b + a*exp(2*p*x))) - log(b + a*exp(2*p*x))/(4*a*b*p^2)

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sympy [A]  time = 0.18, size = 51, normalized size = 0.82 \[ \frac {x}{2 a b p + 2 b^{2} p e^{- 2 p x}} - \frac {x}{2 a b p} - \frac {\log {\left (\frac {a}{b} + e^{- 2 p x} \right )}}{4 a b p^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*a*b*p + 2*b**2*p*exp(-2*p*x)) - x/(2*a*b*p) - log(a/b + exp(-2*p*x))/(4*a*b*p**2)

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