Integrand size = 21, antiderivative size = 50 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (-1-m,-\frac {b \log (F)}{c+d x}\right ) \left (-\frac {b \log (F)}{c+d x}\right )^{1+m}}{d} \] Output:
F^a*(d*x+c)^(1+m)*GAMMA(-1-m,-b*ln(F)/(d*x+c))*(-b*ln(F)/(d*x+c))^(1+m)/d
Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (-1-m,-\frac {b \log (F)}{c+d x}\right ) \left (-\frac {b \log (F)}{c+d x}\right )^{1+m}}{d} \] Input:
Integrate[F^(a + b/(c + d*x))*(c + d*x)^m,x]
Output:
(F^a*(c + d*x)^(1 + m)*Gamma[-1 - m, -((b*Log[F])/(c + d*x))]*(-((b*Log[F] )/(c + d*x)))^(1 + m))/d
Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2648}
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.
\(\displaystyle \int (c+d x)^m F^{a+\frac {b}{c+d x}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {F^a (c+d x)^{m+1} \left (-\frac {b \log (F)}{c+d x}\right )^{m+1} \Gamma \left (-m-1,-\frac {b \log (F)}{c+d x}\right )}{d}\) |
Input:
Int[F^(a + b/(c + d*x))*(c + d*x)^m,x]
Output:
(F^a*(c + d*x)^(1 + m)*Gamma[-1 - m, -((b*Log[F])/(c + d*x))]*(-((b*Log[F] )/(c + d*x)))^(1 + m))/d
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(-F^a)*((e + f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[ F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; FreeQ[{F , a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
\[\int F^{a +\frac {b}{d x +c}} \left (d x +c \right )^{m}d x\]
Input:
int(F^(a+b/(d*x+c))*(d*x+c)^m,x)
Output:
int(F^(a+b/(d*x+c))*(d*x+c)^m,x)
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{d x + c}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))*(d*x+c)^m,x, algorithm="fricas")
Output:
integral((d*x + c)^m*F^((a*d*x + a*c + b)/(d*x + c)), x)
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\int F^{a + \frac {b}{c + d x}} \left (c + d x\right )^{m}\, dx \] Input:
integrate(F**(a+b/(d*x+c))*(d*x+c)**m,x)
Output:
Integral(F**(a + b/(c + d*x))*(c + d*x)**m, x)
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{d x + c}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))*(d*x+c)^m,x, algorithm="maxima")
Output:
integrate((d*x + c)^m*F^(a + b/(d*x + c)), x)
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{d x + c}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))*(d*x+c)^m,x, algorithm="giac")
Output:
integrate((d*x + c)^m*F^(a + b/(d*x + c)), x)
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.46 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )}{2\,\left (c+d\,x\right )}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {m}{2}+1,-\frac {m}{2}-\frac {1}{2}}\left (\frac {b\,\ln \left (F\right )}{c+d\,x}\right )\,{\left (\frac {b\,\ln \left (F\right )}{c+d\,x}\right )}^{m/2}}{d\,\left (m+1\right )} \] Input:
int(F^(a + b/(c + d*x))*(c + d*x)^m,x)
Output:
(F^a*exp((b*log(F))/(2*(c + d*x)))*(c + d*x)^(m + 1)*whittakerM(m/2 + 1, - m/2 - 1/2, (b*log(F))/(c + d*x))*((b*log(F))/(c + d*x))^(m/2))/(d*(m + 1) )
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^m \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c))*(d*x+c)^m,x)
Output:
(f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)**3*b**3 + f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)**2*b**2*c*m - 2*f**((a*c + a*d* x + b)/(c + d*x))*(c + d*x)**m*log(f)**2*b**2*c + f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)**2*b**2*d*m*x - 2*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)**2*b**2*d*x + f**((a*c + a*d*x + b)/(c + d*x))*( c + d*x)**m*log(f)*b*c**2*m**2 - 3*f**((a*c + a*d*x + b)/(c + d*x))*(c + d *x)**m*log(f)*b*c**2*m + 2*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*l og(f)*b*c**2 + 2*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*c* d*m**2*x - 6*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*c*d*m* x + 4*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*c*d*x + f**(( a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*d**2*m**2*x**2 - 3*f**(( a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*d**2*m*x**2 + 2*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*log(f)*b*d**2*x**2 + f**((a*c + a*d* x + b)/(c + d*x))*(c + d*x)**m*c**3*m**3 - 3*f**((a*c + a*d*x + b)/(c + d* x))*(c + d*x)**m*c**3*m**2 + 2*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)* *m*c**3*m + 3*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*c**2*d*m**3*x - 9*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*c**2*d*m**2*x + 6*f**((a *c + a*d*x + b)/(c + d*x))*(c + d*x)**m*c**2*d*m*x + 3*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*c*d**2*m**3*x**2 - 9*f**((a*c + a*d*x + b)/(c + d*x))*(c + d*x)**m*c*d**2*m**2*x**2 + 6*f**((a*c + a*d*x + b)/(c + d*x...