3.84 \(\int \frac{F^{c+d x} x}{(a+b F^{c+d x})^2} \, dx\)

Optimal. Leaf size=69 \[ -\frac{\log \left (a+b F^{c+d x}\right )}{a b d^2 \log ^2(F)}-\frac{x}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac{x}{a b d \log (F)} \]

[Out]

x/(a*b*d*Log[F]) - x/(b*d*(a + b*F^(c + d*x))*Log[F]) - Log[a + b*F^(c + d*x)]/(a*b*d^2*Log[F]^2)

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Rubi [A]  time = 0.0738341, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {2191, 2282, 36, 29, 31} \[ -\frac{\log \left (a+b F^{c+d x}\right )}{a b d^2 \log ^2(F)}-\frac{x}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac{x}{a b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c + d*x)*x)/(a + b*F^(c + d*x))^2,x]

[Out]

x/(a*b*d*Log[F]) - x/(b*d*(a + b*F^(c + d*x))*Log[F]) - Log[a + b*F^(c + d*x)]/(a*b*d^2*Log[F]^2)

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{F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx &=-\frac{x}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac{\int \frac{1}{a+b F^{c+d x}} \, dx}{b d \log (F)}\\ &=-\frac{x}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)} \, dx,x,F^{c+d x}\right )}{b d^2 \log ^2(F)}\\ &=-\frac{x}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac{\operatorname{Subst}\left (\int \frac{1}{a+b x} \, dx,x,F^{c+d x}\right )}{a d^2 \log ^2(F)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,F^{c+d x}\right )}{a b d^2 \log ^2(F)}\\ &=\frac{x}{a b d \log (F)}-\frac{x}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac{\log \left (a+b F^{c+d x}\right )}{a b d^2 \log ^2(F)}\\ \end{align*}

Mathematica [A]  time = 0.0701456, size = 54, normalized size = 0.78 \[ \frac{\frac{d x \log (F) F^{c+d x}}{a+b F^{c+d x}}-\frac{\log \left (a+b F^{c+d x}\right )}{b}}{a d^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c + d*x)*x)/(a + b*F^(c + d*x))^2,x]

[Out]

((d*F^(c + d*x)*x*Log[F])/(a + b*F^(c + d*x)) - Log[a + b*F^(c + d*x)]/b)/(a*d^2*Log[F]^2)

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Maple [A]  time = 0.01, size = 67, normalized size = 1. \begin{align*}{\frac{x{{\rm e}^{ \left ( dx+c \right ) \ln \left ( F \right ) }}}{\ln \left ( F \right ) ad \left ( a+b{{\rm e}^{ \left ( dx+c \right ) \ln \left ( F \right ) }} \right ) }}-{\frac{\ln \left ( a+b{{\rm e}^{ \left ( dx+c \right ) \ln \left ( F \right ) }} \right ) }{ \left ( \ln \left ( F \right ) \right ) ^{2}ab{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x)

[Out]

1/ln(F)/a/d*x*exp((d*x+c)*ln(F))/(a+b*exp((d*x+c)*ln(F)))-1/ln(F)^2/b/d^2/a*ln(a+b*exp((d*x+c)*ln(F)))

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Maxima [A]  time = 1.0704, size = 97, normalized size = 1.41 \begin{align*} \frac{F^{d x} F^{c} x}{F^{d x} F^{c} a b d \log \left (F\right ) + a^{2} d \log \left (F\right )} - \frac{\log \left (\frac{F^{d x} F^{c} b + a}{F^{c} b}\right )}{a b d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="maxima")

[Out]

F^(d*x)*F^c*x/(F^(d*x)*F^c*a*b*d*log(F) + a^2*d*log(F)) - log((F^(d*x)*F^c*b + a)/(F^c*b))/(a*b*d^2*log(F)^2)

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Fricas [A]  time = 1.51882, size = 171, normalized size = 2.48 \begin{align*} \frac{F^{d x + c} b d x \log \left (F\right ) -{\left (F^{d x + c} b + a\right )} \log \left (F^{d x + c} b + a\right )}{F^{d x + c} a b^{2} d^{2} \log \left (F\right )^{2} + a^{2} b d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="fricas")

[Out]

(F^(d*x + c)*b*d*x*log(F) - (F^(d*x + c)*b + a)*log(F^(d*x + c)*b + a))/(F^(d*x + c)*a*b^2*d^2*log(F)^2 + a^2*
b*d^2*log(F)^2)

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Sympy [A]  time = 0.169933, size = 58, normalized size = 0.84 \begin{align*} - \frac{x}{F^{c + d x} b^{2} d \log{\left (F \right )} + a b d \log{\left (F \right )}} + \frac{x}{a b d \log{\left (F \right )}} - \frac{\log{\left (F^{c + d x} + \frac{a}{b} \right )}}{a b d^{2} \log{\left (F \right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(d*x+c)*x/(a+b*F**(d*x+c))**2,x)

[Out]

-x/(F**(c + d*x)*b**2*d*log(F) + a*b*d*log(F)) + x/(a*b*d*log(F)) - log(F**(c + d*x) + a/b)/(a*b*d**2*log(F)**
2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{d x + c} x}{{\left (F^{d x + c} b + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(F^(d*x + c)*x/(F^(d*x + c)*b + a)^2, x)