Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=-\frac {F^a \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4} \] Output:
-1/3*F^a*GAMMA(-4/3,-b*(d*x+c)^3*ln(F))*(-b*(d*x+c)^3*ln(F))^(4/3)/d/(d*x+ c)^4
Time = 0.60 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=-\frac {F^a \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4} \] Input:
Integrate[F^(a + b*(c + d*x)^3)/(c + d*x)^5,x]
Output:
-1/3*(F^a*Gamma[-4/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^( 4/3))/(d*(c + d*x)^4)
Time = 0.34 (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.
\(\displaystyle \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle -\frac {F^a \left (-b \log (F) (c+d x)^3\right )^{4/3} \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right )}{3 d (c+d x)^4}\) |
Input:
Int[F^(a + b*(c + d*x)^3)/(c + d*x)^5,x]
Output:
-1/3*(F^a*Gamma[-4/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^( 4/3))/(d*(c + d*x)^4)
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]
\[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{5}}d x\]
Input:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x)
Output:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x)
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (43) = 86\).
Time = 0.08 (sec) , antiderivative size = 226, normalized size of antiderivative = 4.61 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\frac {3 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \left (-b d^{3} \log \left (F\right )\right )^{\frac {1}{3}} F^{a} \Gamma \left (\frac {2}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right ) - {\left (3 \, {\left (b d^{4} x^{3} + 3 \, b c d^{3} x^{2} + 3 \, b c^{2} d^{2} x + b c^{3} d\right )} \log \left (F\right ) + d\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{4 \, {\left (d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="fricas")
Output:
1/4*(3*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4) *(-b*d^3*log(F))^(1/3)*F^a*gamma(2/3, -(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^ 2*d*x + b*c^3)*log(F))*log(F) - (3*(b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*d^ 2*x + b*c^3*d)*log(F) + d)*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b* c^3 + a))/(d^6*x^4 + 4*c*d^5*x^3 + 6*c^2*d^4*x^2 + 4*c^3*d^3*x + c^4*d^2)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{3}}}{\left (c + d x\right )^{5}}\, dx \] Input:
integrate(F**(a+b*(d*x+c)**3)/(d*x+c)**5,x)
Output:
Integral(F**(a + b*(c + d*x)**3)/(c + d*x)**5, x)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="maxima")
Output:
integrate(F^((d*x + c)^3*b + a)/(d*x + c)^5, x)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="giac")
Output:
integrate(F^((d*x + c)^3*b + a)/(d*x + c)^5, x)
Time = 0.55 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.65 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\frac {3\,F^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}^{4/3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {3\,F^a\,\Gamma \left (\frac {2}{3},-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}^{4/3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {3\,F^a\,F^{b\,{\left (c+d\,x\right )}^3}\,b\,\ln \left (F\right )}{4\,d\,\left (c+d\,x\right )} \] Input:
int(F^(a + b*(c + d*x)^3)/(c + d*x)^5,x)
Output:
(3*F^a*gamma(2/3)*(-b*log(F)*(c + d*x)^3)^(4/3))/(4*d*(c + d*x)^4) - (F^a* F^(b*(c + d*x)^3))/(4*d*(c + d*x)^4) - (3*F^a*igamma(2/3, -b*log(F)*(c + d *x)^3)*(-b*log(F)*(c + d*x)^3)^(4/3))/(4*d*(c + d*x)^4) - (3*F^a*F^(b*(c + d*x)^3)*b*log(F))/(4*d*(c + d*x))
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\text {too large to display} \] Input:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x)
Output:
(f**(a + b*c**3)*( - 2*f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3) + 27*int(f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(3*log(f)*b*c**8 + 15*log(f)*b*c**7*d*x + 30*log(f)*b*c**6*d**2*x**2 + 30*log(f)*b*c**5*d* *3*x**3 + 15*log(f)*b*c**4*d**4*x**4 + 3*log(f)*b*c**3*d**5*x**5 - 4*c**5 - 20*c**4*d*x - 40*c**3*d**2*x**2 - 40*c**2*d**3*x**3 - 20*c*d**4*x**4 - 4 *d**5*x**5),x)*log(f)**2*b**2*c**10*d + 108*int(f**(3*b*c**2*d*x + 3*b*c*d **2*x**2 + b*d**3*x**3)/(3*log(f)*b*c**8 + 15*log(f)*b*c**7*d*x + 30*log(f )*b*c**6*d**2*x**2 + 30*log(f)*b*c**5*d**3*x**3 + 15*log(f)*b*c**4*d**4*x* *4 + 3*log(f)*b*c**3*d**5*x**5 - 4*c**5 - 20*c**4*d*x - 40*c**3*d**2*x**2 - 40*c**2*d**3*x**3 - 20*c*d**4*x**4 - 4*d**5*x**5),x)*log(f)**2*b**2*c**9 *d**2*x + 162*int(f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(3*log (f)*b*c**8 + 15*log(f)*b*c**7*d*x + 30*log(f)*b*c**6*d**2*x**2 + 30*log(f) *b*c**5*d**3*x**3 + 15*log(f)*b*c**4*d**4*x**4 + 3*log(f)*b*c**3*d**5*x**5 - 4*c**5 - 20*c**4*d*x - 40*c**3*d**2*x**2 - 40*c**2*d**3*x**3 - 20*c*d** 4*x**4 - 4*d**5*x**5),x)*log(f)**2*b**2*c**8*d**3*x**2 + 108*int(f**(3*b*c **2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(3*log(f)*b*c**8 + 15*log(f)*b*c* *7*d*x + 30*log(f)*b*c**6*d**2*x**2 + 30*log(f)*b*c**5*d**3*x**3 + 15*log( f)*b*c**4*d**4*x**4 + 3*log(f)*b*c**3*d**5*x**5 - 4*c**5 - 20*c**4*d*x - 4 0*c**3*d**2*x**2 - 40*c**2*d**3*x**3 - 20*c*d**4*x**4 - 4*d**5*x**5),x)*lo g(f)**2*b**2*c**7*d**4*x**3 + 27*int(f**(3*b*c**2*d*x + 3*b*c*d**2*x**2...