Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \] Output:
1/3*F^a*GAMMA(1/3,-b*ln(F)/(d*x+c)^3)/d/(d*x+c)/(-b*ln(F)/(d*x+c)^3)^(1/3)
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \] Input:
Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^2,x]
Output:
(F^a*Gamma[1/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)*(-((b*Log[F])/( c + d*x)^3))^(1/3))
Time = 0.32 (sec) , antiderivative size = 49, 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.
\(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}}\) |
Input:
Int[F^(a + b/(c + d*x)^3)/(c + d*x)^2,x]
Output:
(F^a*Gamma[1/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)*(-((b*Log[F])/( c + d*x)^3))^(1/3))
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]
\[\int \frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{\left (d x +c \right )^{2}}d x\]
Input:
int(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x)
Output:
int(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=-\frac {F^{a} d \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{3 \, b \log \left (F\right )} \] Input:
integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="fricas")
Output:
-1/3*F^a*d*(-b*log(F)/d^3)^(2/3)*gamma(1/3, -b*log(F)/(d^3*x^3 + 3*c*d^2*x ^2 + 3*c^2*d*x + c^3))/(b*log(F))
\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}}}{\left (c + d x\right )^{2}}\, dx \] Input:
integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**2,x)
Output:
Integral(F**(a + b/(c + d*x)**3)/(c + d*x)**2, x)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="maxima")
Output:
integrate(F^(a + b/(d*x + c)^3)/(d*x + c)^2, x)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c)^3)/(d*x + c)^2, x)
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a\,\left (3\,\Gamma \left (\frac {2}{3}\right )\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )-2\,\pi \,\sqrt {3}\right )}{9\,d\,\Gamma \left (\frac {2}{3}\right )\,\left (c+d\,x\right )\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{1/3}} \] Input:
int(F^(a + b/(c + d*x)^3)/(c + d*x)^2,x)
Output:
(F^a*(3*gamma(2/3)*igamma(1/3, -(b*log(F))/(c + d*x)^3) - 2*3^(1/2)*pi))/( 9*d*gamma(2/3)*(c + d*x)*(-(b*log(F))/(c + d*x)^3)^(1/3))
\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x)
Output:
(6*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))*log(f)**2*b**2*c*d*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))*log(f)**2*b**2*d**2*x**2 + 2*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))*log(f)*b*c**5 - 2*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))*log(f)*b*c**4*d*x - 22*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))*log (f)*b*c**3*d**2*x**2 - 34*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))*log(f)*b* c**2*d**3*x**3 - 20*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))*log(f)*b*c*d**4 *x**4 - 4*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))*log(f)*b*d**5*x**5 - 6*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**8 - 36*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**7*d*x - 90*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**6*d...