Integrand size = 25, antiderivative size = 100 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=-\frac {F^{a+b (c+d x)^n} (c+d x)^{-2 n}}{2 d n}-\frac {b F^{a+b (c+d x)^n} (c+d x)^{-n} \log (F)}{2 d n}+\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right ) \log ^2(F)}{2 d n} \] Output:
-1/2*F^(a+b*(d*x+c)^n)/d/n/((d*x+c)^(2*n))-1/2*b*F^(a+b*(d*x+c)^n)*ln(F)/d /n/((d*x+c)^n)+1/2*b^2*F^a*Ei(b*(d*x+c)^n*ln(F))*ln(F)^2/d/n
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.32 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=-\frac {b^2 F^a \Gamma \left (-2,-b (c+d x)^n \log (F)\right ) \log ^2(F)}{d n} \] Input:
Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 2*n),x]
Output:
-((b^2*F^a*Gamma[-2, -(b*(c + d*x)^n*Log[F])]*Log[F]^2)/(d*n))
Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2644, 2644, 2639}
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.
\(\displaystyle \int (c+d x)^{-2 n-1} F^{a+b (c+d x)^n} \, dx\) |
\(\Big \downarrow \) 2644 |
\(\displaystyle \frac {1}{2} b \log (F) \int F^{b (c+d x)^n+a} (c+d x)^{-n-1}dx-\frac {(c+d x)^{-2 n} F^{a+b (c+d x)^n}}{2 d n}\) |
\(\Big \downarrow \) 2644 |
\(\displaystyle \frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{b (c+d x)^n+a}}{c+d x}dx-\frac {(c+d x)^{-n} F^{a+b (c+d x)^n}}{d n}\right )-\frac {(c+d x)^{-2 n} F^{a+b (c+d x)^n}}{2 d n}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{2} b \log (F) \left (\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right )}{d n}-\frac {(c+d x)^{-n} F^{a+b (c+d x)^n}}{d n}\right )-\frac {(c+d x)^{-2 n} F^{a+b (c+d x)^n}}{2 d n}\) |
Input:
Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 2*n),x]
Output:
-1/2*F^(a + b*(c + d*x)^n)/(d*n*(c + d*x)^(2*n)) + (b*Log[F]*(-(F^(a + b*( c + d*x)^n)/(d*n*(c + d*x)^n)) + (b*F^a*ExpIntegralEi[b*(c + d*x)^n*Log[F] ]*Log[F])/(d*n)))/2
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_ Symbol] :> Simp[F^a*(ExpIntegralEi[b*(c + d*x)^n*Log[F]]/(f*n)), x] /; Free Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) Int[(c + d*x)^Simplify[m + n]*F^(a + b*( c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simpli fy[(m + 1)/n]] && LtQ[-4, Simplify[(m + 1)/n], 5] && !RationalQ[m] && SumS implerQ[m, n]
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {F^{a} F^{b \left (d x +c \right )^{n}} \left (d x +c \right )^{-2 n}}{2 n d}-\frac {\ln \left (F \right ) b \,F^{a} F^{b \left (d x +c \right )^{n}} \left (d x +c \right )^{-n}}{2 n d}-\frac {\ln \left (F \right )^{2} b^{2} F^{a} \operatorname {expIntegral}_{1}\left (-b \left (d x +c \right )^{n} \ln \left (F \right )\right )}{2 n d}\) | \(99\) |
Input:
int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x,method=_RETURNVERBOSE)
Output:
-1/2/n/d*F^a*F^(b*(d*x+c)^n)/((d*x+c)^n)^2-1/2/n/d*ln(F)*b*F^a*F^(b*(d*x+c )^n)/((d*x+c)^n)-1/2/n/d*ln(F)^2*b^2*F^a*Ei(1,-b*(d*x+c)^n*ln(F))
Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=\frac {{\left (d x + c\right )}^{2 \, n} F^{a} b^{2} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \left (F\right )\right ) \log \left (F\right )^{2} - {\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + 1\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{2 \, {\left (d x + c\right )}^{2 \, n} d n} \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="fricas")
Output:
1/2*((d*x + c)^(2*n)*F^a*b^2*Ei((d*x + c)^n*b*log(F))*log(F)^2 - ((d*x + c )^n*b*log(F) + 1)*e^((d*x + c)^n*b*log(F) + a*log(F)))/((d*x + c)^(2*n)*d* n)
\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=\int F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{- 2 n - 1}\, dx \] Input:
integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1-2*n),x)
Output:
Integral(F**(a + b*(c + d*x)**n)*(c + d*x)**(-2*n - 1), x)
\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=\int { {\left (d x + c\right )}^{-2 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="maxima")
Output:
integrate((d*x + c)^(-2*n - 1)*F^((d*x + c)^n*b + a), x)
\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=\int { {\left (d x + c\right )}^{-2 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="giac")
Output:
integrate((d*x + c)^(-2*n - 1)*F^((d*x + c)^n*b + a), x)
Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=\int \frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{{\left (c+d\,x\right )}^{2\,n+1}} \,d x \] Input:
int(F^(a + b*(c + d*x)^n)/(c + d*x)^(2*n + 1),x)
Output:
int(F^(a + b*(c + d*x)^n)/(c + d*x)^(2*n + 1), x)
\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx=f^{a} \left (\int \frac {f^{\left (d x +c \right )^{n} b}}{\left (d x +c \right )^{2 n} c +\left (d x +c \right )^{2 n} d x}d x \right ) \] Input:
int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x)
Output:
f**a*int(f**((c + d*x)**n*b)/((c + d*x)**(2*n)*c + (c + d*x)**(2*n)*d*x),x )