Integrand size = 17, antiderivative size = 74 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {x}{d^2}+\frac {1}{b c d \left (d+e \left (F^{c (a+b x)}\right )^n\right ) n \log (F)}-\frac {\log \left (d+e \left (F^{c (a+b x)}\right )^n\right )}{b c d^2 n \log (F)} \] Output:
x/d^2+1/b/c/d/(d+e*(F^(c*(b*x+a)))^n)/n/ln(F)-ln(d+e*(F^(c*(b*x+a)))^n)/b/ c/d^2/n/ln(F)
Time = 0.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.45 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {\frac {1}{b c d \left (d+e \left (F^{c (a+b x)}\right )^n\right ) n}+\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{b c d^2 n}-\frac {\log \left (b c d^3 n \log (F)+b c d^2 e \left (F^{c (a+b x)}\right )^n n \log (F)\right )}{b c d^2 n}}{\log (F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)^(-2),x]
Output:
(1/(b*c*d*(d + e*(F^(c*(a + b*x)))^n)*n) + Log[(F^(c*(a + b*x)))^n]/(b*c*d ^2*n) - Log[b*c*d^3*n*Log[F] + b*c*d^2*e*(F^(c*(a + b*x)))^n*n*Log[F]]/(b* c*d^2*n))/Log[F]
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2720, 798, 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.
\(\displaystyle \int \frac {1}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^2} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^2}dF^{c (a+b x)}}{b c \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^2}d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {F^{-c (a+b x)}}{d^2}-\frac {e}{d^2 \left (e \left (F^{c (a+b x)}\right )^n+d\right )}-\frac {e}{d \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2}\right )d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\log \left (e \left (F^{c (a+b x)}\right )^n+d\right )}{d^2}+\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{d^2}+\frac {1}{d \left (e \left (F^{c (a+b x)}\right )^n+d\right )}}{b c n \log (F)}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)^(-2),x]
Output:
(1/(d*(d + e*(F^(c*(a + b*x)))^n)) + Log[(F^(c*(a + b*x)))^n]/d^2 - Log[d + e*(F^(c*(a + b*x)))^n]/d^2)/(b*c*n*Log[F])
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.00 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}{d^{2}}+\frac {1}{d \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}+\frac {\ln \left (\left (F^{c \left (b x +a \right )}\right )^{n}\right )}{d^{2}}}{\ln \left (F \right ) b c n}\) | \(74\) |
default | \(\frac {-\frac {\ln \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}{d^{2}}+\frac {1}{d \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}+\frac {\ln \left (\left (F^{c \left (b x +a \right )}\right )^{n}\right )}{d^{2}}}{\ln \left (F \right ) b c n}\) | \(74\) |
risch | \(\frac {\ln \left (F^{c \left (b x +a \right )}\right )}{\ln \left (F \right ) b c \,d^{2}}+\frac {1}{b c d \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) n \ln \left (F \right )}-\frac {\ln \left (\left (F^{c \left (b x +a \right )}\right )^{n}+\frac {d}{e}\right )}{\ln \left (F \right ) b c n \,d^{2}}\) | \(96\) |
parallelrisch | \(\frac {e^{2} \left (F^{c \left (b x +a \right )}\right )^{n} x \ln \left (F \right ) b c n +x \ln \left (F \right ) b c d e n -\ln \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (F^{c \left (b x +a \right )}\right )^{n} e^{2}-\ln \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) d e +d e}{\ln \left (F \right ) b c \,d^{2} e n \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}\) | \(124\) |
norman | \(\frac {-\frac {e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) b c n \,d^{2}}+\frac {e \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{d^{2} \ln \left (F \right ) b c}+\frac {\ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}{\ln \left (F \right ) b c d}}{d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}-\frac {\ln \left (d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}\right )}{\ln \left (F \right ) b c n \,d^{2}}\) | \(159\) |
Input:
int(1/(d+e*(F^(c*(b*x+a)))^n)^2,x,method=_RETURNVERBOSE)
Output:
1/ln(F)/b/c/n*(-1/d^2*ln(d+e*(F^(c*(b*x+a)))^n)+1/d/(d+e*(F^(c*(b*x+a)))^n )+1/d^2*ln((F^(c*(b*x+a)))^n))
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {F^{b c n x + a c n} b c e n x \log \left (F\right ) + b c d n x \log \left (F\right ) - {\left (F^{b c n x + a c n} e + d\right )} \log \left (F^{b c n x + a c n} e + d\right ) + d}{F^{b c n x + a c n} b c d^{2} e n \log \left (F\right ) + b c d^{3} n \log \left (F\right )} \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="fricas")
Output:
(F^(b*c*n*x + a*c*n)*b*c*e*n*x*log(F) + b*c*d*n*x*log(F) - (F^(b*c*n*x + a *c*n)*e + d)*log(F^(b*c*n*x + a*c*n)*e + d) + d)/(F^(b*c*n*x + a*c*n)*b*c* d^2*e*n*log(F) + b*c*d^3*n*log(F))
\[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\int \frac {1}{\left (d + e \left (F^{c \left (a + b x\right )}\right )^{n}\right )^{2}}\, dx \] Input:
integrate(1/(d+e*(F**((b*x+a)*c))**n)**2,x)
Output:
Integral((d + e*(F**(c*(a + b*x)))**n)**(-2), x)
Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {b c n x + a c n}{b c d^{2} n} + \frac {1}{{\left (F^{b c n x + a c n} d e + d^{2}\right )} b c n \log \left (F\right )} - \frac {\log \left (F^{b c n x + a c n} e + d\right )}{b c d^{2} n \log \left (F\right )} \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="maxima")
Output:
(b*c*n*x + a*c*n)/(b*c*d^2*n) + 1/((F^(b*c*n*x + a*c*n)*d*e + d^2)*b*c*n*l og(F)) - log(F^(b*c*n*x + a*c*n)*e + d)/(b*c*d^2*n*log(F))
Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {\log \left ({\left | F \right |}^{b c n x} {\left | F \right |}^{a c n}\right )}{b c d^{2} n \log \left (F\right )} - \frac {\log \left ({\left | F^{b c n x} F^{a c n} e + d \right |}\right )}{b c d^{2} n \log \left (F\right )} + \frac {1}{{\left (F^{b c n x} F^{a c n} e + d\right )} b c d n \log \left (F\right )} \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="giac")
Output:
log(abs(F)^(b*c*n*x)*abs(F)^(a*c*n))/(b*c*d^2*n*log(F)) - log(abs(F^(b*c*n *x)*F^(a*c*n)*e + d))/(b*c*d^2*n*log(F)) + 1/((F^(b*c*n*x)*F^(a*c*n)*e + d )*b*c*d*n*log(F))
Time = 0.00 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {x}{d^2}+\frac {1}{b\,c\,d\,n\,\ln \left (F\right )\,\left (d+e\,{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\right )}-\frac {\ln \left (d+e\,{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\right )}{b\,c\,d^2\,n\,\ln \left (F\right )} \] Input:
int(1/(d + e*(F^(c*(a + b*x)))^n)^2,x)
Output:
x/d^2 + 1/(b*c*d*n*log(F)*(d + e*(F^(b*c*x)*F^(a*c))^n)) - log(d + e*(F^(b *c*x)*F^(a*c))^n)/(b*c*d^2*n*log(F))
Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.77 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2} \, dx=\frac {-f^{b c n x +a c n} \mathrm {log}\left (f^{b c n x +a c n} e +d \right ) e +f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c e n x -f^{b c n x +a c n} e -\mathrm {log}\left (f^{b c n x +a c n} e +d \right ) d +\mathrm {log}\left (f \right ) b c d n x}{\mathrm {log}\left (f \right ) b c \,d^{2} n \left (f^{b c n x +a c n} e +d \right )} \] Input:
int(1/(d+e*(F^((b*x+a)*c))^n)^2,x)
Output:
( - f**(a*c*n + b*c*n*x)*log(f**(a*c*n + b*c*n*x)*e + d)*e + f**(a*c*n + b *c*n*x)*log(f)*b*c*e*n*x - f**(a*c*n + b*c*n*x)*e - log(f**(a*c*n + b*c*n* x)*e + d)*d + log(f)*b*c*d*n*x)/(log(f)*b*c*d**2*n*(f**(a*c*n + b*c*n*x)*e + d))