Integrand size = 19, antiderivative size = 49 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\frac {F^a (c+d x)^2 \Gamma \left (-\frac {2}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{2/3}}{3 d} \] Output:
1/3*F^a*(d*x+c)^2*GAMMA(-2/3,-b*ln(F)/(d*x+c)^3)*(-b*ln(F)/(d*x+c)^3)^(2/3 )/d
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\frac {F^a (c+d x)^2 \Gamma \left (-\frac {2}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{2/3}}{3 d} \] Input:
Integrate[F^(a + b/(c + d*x)^3)*(c + d*x),x]
Output:
(F^a*(c + d*x)^2*Gamma[-2/3, -((b*Log[F])/(c + d*x)^3)]*(-((b*Log[F])/(c + d*x)^3))^(2/3))/(3*d)
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, 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.
\(\displaystyle \int (c+d x) F^{a+\frac {b}{(c+d x)^3}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {F^a (c+d x)^2 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d}\) |
Input:
Int[F^(a + b/(c + d*x)^3)*(c + d*x),x]
Output:
(F^a*(c + d*x)^2*Gamma[-2/3, -((b*Log[F])/(c + d*x)^3)]*(-((b*Log[F])/(c + d*x)^3))^(2/3))/(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]
\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )d x\]
Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c),x)
Output:
int(F^(a+b/(d*x+c)^3)*(d*x+c),x)
Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (43) = 86\).
Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.90 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=-\frac {F^{a} d^{2} \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{2 \, d} \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c),x, algorithm="fricas")
Output:
-1/2*(F^a*d^2*(-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)) - (d^2*x^2 + 2*c*d*x + c^2)*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)))/d
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )\, dx \] Input:
integrate(F**(a+b/(d*x+c)**3)*(d*x+c),x)
Output:
Integral(F**(a + b/(c + d*x)**3)*(c + d*x), x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\int { {\left (d x + c\right )} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c),x, algorithm="maxima")
Output:
1/2*(F^a*d*x^2 + 2*F^a*c*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3) ) + integrate(3/2*(F^a*b*d^2*x^2*log(F) + 2*F^a*b*c*d*x*log(F))*F^(b/(d^3* x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x ^2 + 4*c^3*d*x + c^4), x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\int { {\left (d x + c\right )} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c),x, algorithm="giac")
Output:
integrate((d*x + c)*F^(a + b/(d*x + c)^3), x)
Time = 0.94 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.18 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^2}{2\,d}-\frac {F^a\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )\,{\left (c+d\,x\right )}^2\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{2/3}}{2\,d}+\frac {\pi \,\sqrt {3}\,F^a\,{\left (c+d\,x\right )}^2\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{2/3}}{3\,d\,\Gamma \left (\frac {2}{3}\right )} \] Input:
int(F^(a + b/(c + d*x)^3)*(c + d*x),x)
Output:
(F^a*F^(b/(c + d*x)^3)*(c + d*x)^2)/(2*d) - (F^a*igamma(1/3, -(b*log(F))/( c + d*x)^3)*(c + d*x)^2*(-(b*log(F))/(c + d*x)^3)^(2/3))/(2*d) + (3^(1/2)* F^a*pi*(c + d*x)^2*(-(b*log(F))/(c + d*x)^3)^(2/3))/(3*d*gamma(2/3))
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x) \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c),x)
Output:
(18*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 + 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))*log(f)**2*b**2*d**2*x**2 + 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)*b*c**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))*log(f)*b*c**4*d*x - 66*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))*lo g(f)*b*c**3*d**2*x**2 - 102*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 - 60*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 - 12*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 - 1 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))*c**8 - 76*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 - 158*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...