Integrand size = 22, antiderivative size = 69 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {x}{a b d \log (F)}-\frac {x}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {\log \left (a+b F^{c+d x}\right )}{a b d^2 \log ^2(F)} \] Output:
x/a/b/d/ln(F)-x/b/d/(a+b*F^(d*x+c))/ln(F)-ln(a+b*F^(d*x+c))/a/b/d^2/ln(F)^ 2
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {\frac {d F^{c+d x} x \log (F)}{a+b F^{c+d x}}-\frac {\log \left (a+b F^{c+d x}\right )}{b}}{a d^2 \log ^2(F)} \] Input:
Integrate[(F^(c + d*x)*x)/(a + b*F^(c + d*x))^2,x]
Output:
((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)
Time = 0.42 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2621, 2720, 47, 14, 16}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x F^{c+d x}}{\left (a+b F^{c+d x}\right )^2} \, dx\) |
\(\Big \downarrow \) 2621 |
\(\displaystyle \frac {\int \frac {1}{b F^{c+d x}+a}dx}{b d \log (F)}-\frac {x}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {F^{-c-d x}}{b F^{c+d x}+a}dF^{c+d x}}{b d^2 \log ^2(F)}-\frac {x}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {\int F^{-c-d x}dF^{c+d x}}{a}-\frac {b \int \frac {1}{b F^{c+d x}+a}dF^{c+d x}}{a}}{b d^2 \log ^2(F)}-\frac {x}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {\log \left (F^{c+d x}\right )}{a}-\frac {b \int \frac {1}{b F^{c+d x}+a}dF^{c+d x}}{a}}{b d^2 \log ^2(F)}-\frac {x}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {\log \left (F^{c+d x}\right )}{a}-\frac {\log \left (a+b F^{c+d x}\right )}{a}}{b d^2 \log ^2(F)}-\frac {x}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
Input:
Int[(F^(c + d*x)*x)/(a + b*F^(c + d*x))^2,x]
Output:
-(x/(b*d*(a + b*F^(c + d*x))*Log[F])) + (Log[F^(c + d*x)]/a - Log[a + b*F^ (c + d*x)]/a)/(b*d^2*Log[F]^2)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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] - Simp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {x \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}}{\ln \left (F \right ) a d \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}-\frac {\ln \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}{\ln \left (F \right )^{2} b \,d^{2} a}\) | \(67\) |
parallelrisch | \(\frac {b d x \,F^{d x +c} \ln \left (F \right )-\ln \left (a +b \,F^{d x +c}\right ) F^{d x +c} b -\ln \left (a +b \,F^{d x +c}\right ) a}{\ln \left (F \right )^{2} a b \,d^{2} \left (a +b \,F^{d x +c}\right )}\) | \(79\) |
risch | \(\frac {x}{a b d \ln \left (F \right )}+\frac {c}{\ln \left (F \right ) b \,d^{2} a}-\frac {x}{b d \left (a +b \,F^{d x +c}\right ) \ln \left (F \right )}-\frac {\ln \left (F^{d x +c}+\frac {a}{b}\right )}{\ln \left (F \right )^{2} b \,d^{2} a}\) | \(87\) |
Input:
int(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {F^{d x + c} b d x \log \left (F\right ) - {\left (F^{d x + c} b + a\right )} \log \left (F^{d x + c} b + a\right )}{F^{d x + c} a b^{2} d^{2} \log \left (F\right )^{2} + a^{2} b d^{2} \log \left (F\right )^{2}} \] Input:
integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="fricas")
Output:
(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)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=- \frac {x}{F^{c + d x} b^{2} d \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {x}{a b d \log {\left (F \right )}} - \frac {\log {\left (F^{c + d x} + \frac {a}{b} \right )}}{a b d^{2} \log {\left (F \right )}^{2}} \] Input:
integrate(F**(d*x+c)*x/(a+b*F**(d*x+c))**2,x)
Output:
-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)
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {F^{d x} F^{c} x}{F^{d x} F^{c} a b d \log \left (F\right ) + a^{2} d \log \left (F\right )} - \frac {\log \left (\frac {F^{d x} F^{c} b + a}{F^{c} b}\right )}{a b d^{2} \log \left (F\right )^{2}} \] Input:
integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="maxima")
Output:
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)
\[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\int { \frac {F^{d x + c} x}{{\left (F^{d x + c} b + a\right )}^{2}} \,d x } \] Input:
integrate(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x, algorithm="giac")
Output:
integrate(F^(d*x + c)*x/(F^(d*x + c)*b + a)^2, x)
Time = 22.87 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {F^c\,F^{d\,x}\,x}{a\,d\,\ln \left (F\right )\,\left (a+F^c\,F^{d\,x}\,b\right )}-\frac {\ln \left (a+F^c\,F^{d\,x}\,b\right )}{a\,b\,d^2\,{\ln \left (F\right )}^2} \] Input:
int((F^(c + d*x)*x)/(a + F^(c + d*x)*b)^2,x)
Output:
(F^c*F^(d*x)*x)/(a*d*log(F)*(a + F^c*F^(d*x)*b)) - log(a + F^c*F^(d*x)*b)/ (a*b*d^2*log(F)^2)
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {-f^{d x +c} \mathrm {log}\left (f^{d x +c} b +a \right ) b +f^{d x +c} \mathrm {log}\left (f \right ) b d x -\mathrm {log}\left (f^{d x +c} b +a \right ) a}{\mathrm {log}\left (f \right )^{2} a b \,d^{2} \left (f^{d x +c} b +a \right )} \] Input:
int(F^(d*x+c)*x/(a+b*F^(d*x+c))^2,x)
Output:
( - f**(c + d*x)*log(f**(c + d*x)*b + a)*b + f**(c + d*x)*log(f)*b*d*x - l og(f**(c + d*x)*b + a)*a)/(log(f)**2*a*b*d**2*(f**(c + d*x)*b + a))