Integrand size = 21, antiderivative size = 61 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{3} (-1-m),-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{\frac {1+m}{3}}}{3 d} \] Output:
1/3*F^a*(d*x+c)^(1+m)*GAMMA(-1/3-1/3*m,-b*ln(F)/(d*x+c)^3)*(-b*ln(F)/(d*x+ c)^3)^(1/3+1/3*m)/d
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{3} (-1-m),-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{\frac {1+m}{3}}}{3 d} \] Input:
Integrate[F^(a + b/(c + d*x)^3)*(c + d*x)^m,x]
Output:
(F^a*(c + d*x)^(1 + m)*Gamma[(-1 - m)/3, -((b*Log[F])/(c + d*x)^3)]*(-((b* Log[F])/(c + d*x)^3))^((1 + m)/3))/(3*d)
Time = 0.34 (sec) , antiderivative size = 61, 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.
\(\displaystyle \int (c+d x)^m F^{a+\frac {b}{(c+d x)^3}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {F^a (c+d x)^{m+1} \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{\frac {m+1}{3}} \Gamma \left (\frac {1}{3} (-m-1),-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d}\) |
Input:
Int[F^(a + b/(c + d*x)^3)*(c + d*x)^m,x]
Output:
(F^a*(c + d*x)^(1 + m)*Gamma[(-1 - m)/3, -((b*Log[F])/(c + d*x)^3)]*(-((b* Log[F])/(c + d*x)^3))^((1 + m)/3))/(3*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]
\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{m}d x\]
Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x)
Output:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x, algorithm="fricas")
Output:
integral((d*x + c)^m*F^((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)), x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{m}\, dx \] Input:
integrate(F**(a+b/(d*x+c)**3)*(d*x+c)**m,x)
Output:
Integral(F**(a + b/(c + d*x)**3)*(c + d*x)**m, x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x, algorithm="maxima")
Output:
integrate((d*x + c)^m*F^(a + b/(d*x + c)^3), x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x, algorithm="giac")
Output:
integrate((d*x + c)^m*F^(a + b/(d*x + c)^3), x)
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )}{2\,{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {m}{6}+\frac {2}{3},-\frac {m}{6}-\frac {1}{6}}\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )\,{\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{\frac {m}{6}-\frac {1}{3}}}{d\,\left (m+1\right )} \] Input:
int(F^(a + b/(c + d*x)^3)*(c + d*x)^m,x)
Output:
(F^a*exp((b*log(F))/(2*(c + d*x)^3))*(c + d*x)^(m + 1)*whittakerM(m/6 + 2/ 3, - m/6 - 1/6, (b*log(F))/(c + d*x)^3)*((b*log(F))/(c + d*x)^3)^(m/6 - 1/ 3))/(d*(m + 1))
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^m \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^m,x)
Output:
(9*f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)**2*b**2 + 3* f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3* c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c**3*m - 15*f **((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c **2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c**3 + 9*f**(( a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c**2* d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c**2*d*m*x - 45*f* *((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c* *2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c**2*d*x + 9*f* *((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c* *2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c*d**2*m*x**2 - 45*f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*c*d**2*x* *2 + 3*f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/(c* *3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*d**3*m *x**3 - 15*f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b) /(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*log(f)*b*d* *3*x**3 + f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/ (c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))*(c + d*x)**m*c**6*m**2...