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]

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)

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Rubi [A]  time = 0.0907172, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0648487, 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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Maple [A]  time = 0.012, size = 51, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ( a \left ({{\rm e}^{px}} \right ) ^{2}+b \right ) }{4\,{p}^{2}ba}}+{\frac{x \left ({{\rm e}^{px}} \right ) ^{2}}{2\,bp \left ( a \left ({{\rm e}^{px}} \right ) ^{2}+b \right ) }} \end{align*}

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 = 0.955416, size = 69, normalized size = 1.11 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 2.1886, size = 135, normalized size = 2.18 \begin{align*} \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 )}} \end{align*}

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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Sympy [A]  time = 0.158276, size = 51, normalized size = 0.82 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.07893, size = 100, normalized size = 1.61 \begin{align*} \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 )}} \end{align*}

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)