Integrand size = 22, antiderivative size = 106 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\frac {1}{2 a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x}{2 a^2 b d \log (F)}-\frac {x}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {\log \left (a+b F^{c+d x}\right )}{2 a^2 b d^2 \log ^2(F)} \]
1/2/a/b/d^2/(a+b*F^(d*x+c))/ln(F)^2+1/2*x/a^2/b/d/ln(F)-1/2*x/b/d/(a+b*F^( d*x+c))^2/ln(F)-1/2*ln(a+b*F^(d*x+c))/a^2/b/d^2/ln(F)^2
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\frac {b d F^{c+d x} \left (2 a+b F^{c+d x}\right ) x \log (F)-\left (a+b F^{c+d x}\right ) \left (-a+\left (a+b F^{c+d x}\right ) \log \left (a+b F^{c+d x}\right )\right )}{2 a^2 b d^2 \left (a+b F^{c+d x}\right )^2 \log ^2(F)} \]
(b*d*F^(c + d*x)*(2*a + b*F^(c + d*x))*x*Log[F] - (a + b*F^(c + d*x))*(-a + (a + b*F^(c + d*x))*Log[a + b*F^(c + d*x)]))/(2*a^2*b*d^2*(a + b*F^(c + d*x))^2*Log[F]^2)
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2621, 2720, 54, 2009}
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.
\(\displaystyle \int \frac {x F^{c+d x}}{\left (a+b F^{c+d x}\right )^3} \, dx\) |
\(\Big \downarrow \) 2621 |
\(\displaystyle \frac {\int \frac {1}{\left (b F^{c+d x}+a\right )^2}dx}{2 b d \log (F)}-\frac {x}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {F^{-c-d x}}{\left (b F^{c+d x}+a\right )^2}dF^{c+d x}}{2 b d^2 \log ^2(F)}-\frac {x}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {F^{-c-d x}}{a^2}-\frac {b}{a^2 \left (b F^{c+d x}+a\right )}-\frac {b}{a \left (b F^{c+d x}+a\right )^2}\right )dF^{c+d x}}{2 b d^2 \log ^2(F)}-\frac {x}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\log \left (a+b F^{c+d x}\right )}{a^2}+\frac {\log \left (F^{c+d x}\right )}{a^2}+\frac {1}{a \left (a+b F^{c+d x}\right )}}{2 b d^2 \log ^2(F)}-\frac {x}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2}\) |
-1/2*x/(b*d*(a + b*F^(c + d*x))^2*Log[F]) + (1/(a*(a + b*F^(c + d*x))) + L og[F^(c + d*x)]/a^2 - Log[a + b*F^(c + d*x)]/a^2)/(2*b*d^2*Log[F]^2)
3.1.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 = 111, normalized size of antiderivative = 1.05
method | result | size |
risch | \(\frac {x}{2 a^{2} b d \ln \left (F \right )}+\frac {c}{2 a^{2} b \,d^{2} \ln \left (F \right )}-\frac {\ln \left (F \right ) a d x -b \,F^{d x +c}-a}{2 d^{2} \ln \left (F \right )^{2} a b \left (a +b \,F^{d x +c}\right )^{2}}-\frac {\ln \left (F^{d x +c}+\frac {a}{b}\right )}{2 a^{2} b \,d^{2} \ln \left (F \right )^{2}}\) | \(111\) |
norman | \(\frac {\frac {{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{2} a \,d^{2}}+\frac {x \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}}{d \ln \left (F \right ) a}+\frac {b x \,{\mathrm e}^{\left (2 d x +2 c \right ) \ln \left (F \right )}}{2 \ln \left (F \right ) a^{2} d}+\frac {1}{2 \ln \left (F \right )^{2} b \,d^{2}}}{\left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )^{2}}-\frac {\ln \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}{2 a^{2} b \,d^{2} \ln \left (F \right )^{2}}\) | \(127\) |
parallelrisch | \(\frac {b^{3} F^{2 d x +2 c} x \ln \left (F \right ) d +2 x \,F^{d x +c} \ln \left (F \right ) a \,b^{2} d -\ln \left (a +b \,F^{d x +c}\right ) F^{2 d x +2 c} b^{3}-2 \ln \left (a +b \,F^{d x +c}\right ) F^{d x +c} a \,b^{2}-\ln \left (a +b \,F^{d x +c}\right ) a^{2} b +F^{d x +c} a \,b^{2}+a^{2} b}{2 \ln \left (F \right )^{2} a^{2} b^{2} d^{2} \left (a +b \,F^{d x +c}\right )^{2}}\) | \(150\) |
1/2*x/a^2/b/d/ln(F)+1/2/a^2/b/d^2/ln(F)*c-1/2*(ln(F)*a*d*x-b*F^(d*x+c)-a)/ d^2/ln(F)^2/a/b/(a+b*F^(d*x+c))^2-1/2/a^2/b/d^2/ln(F)^2*ln(F^(d*x+c)+a/b)
Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.40 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\frac {F^{2 \, d x + 2 \, c} b^{2} d x \log \left (F\right ) + {\left (2 \, a b d x \log \left (F\right ) + a b\right )} F^{d x + c} + a^{2} - {\left (2 \, F^{d x + c} a b + F^{2 \, d x + 2 \, c} b^{2} + a^{2}\right )} \log \left (F^{d x + c} b + a\right )}{2 \, {\left (2 \, F^{d x + c} a^{3} b^{2} d^{2} \log \left (F\right )^{2} + F^{2 \, d x + 2 \, c} a^{2} b^{3} d^{2} \log \left (F\right )^{2} + a^{4} b d^{2} \log \left (F\right )^{2}\right )}} \]
1/2*(F^(2*d*x + 2*c)*b^2*d*x*log(F) + (2*a*b*d*x*log(F) + a*b)*F^(d*x + c) + a^2 - (2*F^(d*x + c)*a*b + F^(2*d*x + 2*c)*b^2 + a^2)*log(F^(d*x + c)*b + a))/(2*F^(d*x + c)*a^3*b^2*d^2*log(F)^2 + F^(2*d*x + 2*c)*a^2*b^3*d^2*l og(F)^2 + a^4*b*d^2*log(F)^2)
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\frac {F^{c + d x} b - a d x \log {\left (F \right )} + a}{4 F^{c + d x} a^{2} b^{2} d^{2} \log {\left (F \right )}^{2} + 2 F^{2 c + 2 d x} a b^{3} d^{2} \log {\left (F \right )}^{2} + 2 a^{3} b d^{2} \log {\left (F \right )}^{2}} + \frac {x}{2 a^{2} b d \log {\left (F \right )}} - \frac {\log {\left (F^{c + d x} + \frac {a}{b} \right )}}{2 a^{2} b d^{2} \log {\left (F \right )}^{2}} \]
(F**(c + d*x)*b - a*d*x*log(F) + a)/(4*F**(c + d*x)*a**2*b**2*d**2*log(F)* *2 + 2*F**(2*c + 2*d*x)*a*b**3*d**2*log(F)**2 + 2*a**3*b*d**2*log(F)**2) + x/(2*a**2*b*d*log(F)) - log(F**(c + d*x) + a/b)/(2*a**2*b*d**2*log(F)**2)
Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.42 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\frac {F^{2 \, d x} F^{2 \, c} b^{2} d x \log \left (F\right ) + {\left (2 \, F^{c} a b d x \log \left (F\right ) + F^{c} a b\right )} F^{d x} + a^{2}}{2 \, {\left (2 \, F^{d x} F^{c} a^{3} b^{2} d^{2} \log \left (F\right )^{2} + F^{2 \, d x} F^{2 \, c} a^{2} b^{3} d^{2} \log \left (F\right )^{2} + a^{4} b d^{2} \log \left (F\right )^{2}\right )}} - \frac {\log \left (\frac {F^{d x} F^{c} b + a}{F^{c} b}\right )}{2 \, a^{2} b d^{2} \log \left (F\right )^{2}} \]
1/2*(F^(2*d*x)*F^(2*c)*b^2*d*x*log(F) + (2*F^c*a*b*d*x*log(F) + F^c*a*b)*F ^(d*x) + a^2)/(2*F^(d*x)*F^c*a^3*b^2*d^2*log(F)^2 + F^(2*d*x)*F^(2*c)*a^2* b^3*d^2*log(F)^2 + a^4*b*d^2*log(F)^2) - 1/2*log((F^(d*x)*F^c*b + a)/(F^c* b))/(a^2*b*d^2*log(F)^2)
\[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=\int { \frac {F^{d x + c} x}{{\left (F^{d x + c} b + a\right )}^{3}} \,d x } \]
Time = 0.36 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46 \[ \int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^3} \, dx=-\frac {\frac {F^c\,F^{d\,x}}{2\,a\,d^2\,{\ln \left (F\right )}^2}-\frac {F^c\,F^{d\,x}\,x}{a\,d\,\ln \left (F\right )}+\frac {F^{2\,c}\,F^{2\,d\,x}\,b}{2\,a^2\,d^2\,{\ln \left (F\right )}^2}-\frac {F^{2\,c}\,F^{2\,d\,x}\,b\,x}{2\,a^2\,d\,\ln \left (F\right )}}{a^2+F^{2\,c}\,F^{2\,d\,x}\,b^2+2\,F^c\,F^{d\,x}\,a\,b}-\frac {\ln \left (a+F^c\,F^{d\,x}\,b\right )}{2\,a^2\,b\,d^2\,{\ln \left (F\right )}^2} \]