Integrand size = 21, antiderivative size = 49 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a (c+d x)^4 \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}{3 d} \] Output:
1/3*F^a*(d*x+c)^4*GAMMA(-4/3,-b*ln(F)/(d*x+c)^3)*(-b*ln(F)/(d*x+c)^3)^(4/3 )/d
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a (c+d x)^4 \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}{3 d} \] Input:
Integrate[F^(a + b/(c + d*x)^3)*(c + d*x)^3,x]
Output:
(F^a*(c + d*x)^4*Gamma[-4/3, -((b*Log[F])/(c + d*x)^3)]*(-((b*Log[F])/(c + d*x)^3))^(4/3))/(3*d)
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.
\(\displaystyle \int (c+d x)^3 F^{a+\frac {b}{(c+d x)^3}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {F^a (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3} \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d}\) |
Input:
Int[F^(a + b/(c + d*x)^3)*(c + d*x)^3,x]
Output:
(F^a*(c + d*x)^4*Gamma[-4/3, -((b*Log[F])/(c + d*x)^3)]*(-((b*Log[F])/(c + d*x)^3))^(4/3))/(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]
\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{3}d x\]
Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x)
Output:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x)
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 3.63 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=-\frac {3 \, F^{a} b d \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \left (F\right ) - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + 3 \, {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{4 \, d} \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x, algorithm="fricas")
Output:
-1/4*(3*F^a*b*d*(-b*log(F)/d^3)^(1/3)*gamma(2/3, -b*log(F)/(d^3*x^3 + 3*c* d^2*x^2 + 3*c^2*d*x + c^3))*log(F) - (d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^ 2 + 4*c^3*d*x + c^4 + 3*(b*d*x + b*c)*log(F))*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
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{3}\, dx \] Input:
integrate(F**(a+b/(d*x+c)**3)*(d*x+c)**3,x)
Output:
Integral(F**(a + b/(c + d*x)**3)*(c + d*x)**3, x)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x, algorithm="maxima")
Output:
1/4*(F^a*d^3*x^4 + 4*F^a*c*d^2*x^3 + 6*F^a*c^2*d*x^2 + (4*F^a*c^3 + 3*F^a* b*log(F))*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + integrate(- 3/4*(F^a*b*c^4*log(F) - 3*F^a*b^2*d*x*log(F)^2)*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)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x, algorithm="giac")
Output:
integrate((d*x + c)^3*F^(a + b/(d*x + c)^3), x)
Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.61 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^4}{4\,d}-\frac {3\,F^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,\Gamma \left (\frac {2}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,b\,\ln \left (F\right )\,\left (c+d\,x\right )}{4\,d} \] Input:
int(F^(a + b/(c + d*x)^3)*(c + d*x)^3,x)
Output:
(F^a*F^(b/(c + d*x)^3)*(c + d*x)^4)/(4*d) - (3*F^a*gamma(2/3)*(c + d*x)^4* (-(b*log(F))/(c + d*x)^3)^(4/3))/(4*d) + (3*F^a*igamma(2/3, -(b*log(F))/(c + d*x)^3)*(c + d*x)^4*(-(b*log(F))/(c + d*x)^3)^(4/3))/(4*d) + (3*F^a*F^( b/(c + d*x)^3)*b*log(F)*(c + d*x))/(4*d)
\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c)^3)*(d*x+c)^3,x)
Output:
(27*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)**3*b**3*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*c**4 - 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**3*d*x - 108*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**2*d**2*x**2 - 126*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**3*x**3 - 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))*log(f)**2*b**2*d**4*x**4 - 24*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**7 - 114*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**6*d*x - 180*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*d** 2*x**2 - 30*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**3*x**3 + 240*f**((a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b)/...