Integrand size = 21, antiderivative size = 96 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=-\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^3 d \log ^3(F)}+\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d (c+d x)^6 \log (F)} \]
-2/3*F^(a+b/(d*x+c)^3)/b^3/d/ln(F)^3+2/3*F^(a+b/(d*x+c)^3)/b^2/d/(d*x+c)^3 /ln(F)^2-1/3*F^(a+b/(d*x+c)^3)/b/d/(d*x+c)^6/ln(F)
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^3}} \left (2 (c+d x)^6-2 b (c+d x)^3 \log (F)+b^2 \log ^2(F)\right )}{3 b^3 d (c+d x)^6 \log ^3(F)} \]
-1/3*(F^(a + b/(c + d*x)^3)*(2*(c + d*x)^6 - 2*b*(c + d*x)^3*Log[F] + b^2* Log[F]^2))/(b^3*d*(c + d*x)^6*Log[F]^3)
Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2641, 2641, 2638}
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.
| \(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {2 \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^7}dx}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^6}\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {2 \left (-\frac {\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^4}dx}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^3}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^6}\) |
| \(\Big \downarrow \) 2638 |
| \(\displaystyle -\frac {2 \left (\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^3}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^6}\) |
-1/3*F^(a + b/(c + d*x)^3)/(b*d*(c + d*x)^6*Log[F]) - (2*(F^(a + b/(c + d* x)^3)/(3*b^2*d*Log[F]^2) - F^(a + b/(c + d*x)^3)/(3*b*d*(c + d*x)^3*Log[F] )))/(b*Log[F])
3.4.49.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^n*(F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n *Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] && EqQ [d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F]) Int[(c + d*x)^(m - n)*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/ n)] && LtQ[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n , 0])
Time = 3.00 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.82
| method | result | size |
| risch | \(-\frac {\left (2 d^{6} x^{6}+12 c \,d^{5} x^{5}+30 c^{2} d^{4} x^{4}+40 c^{3} d^{3} x^{3}-2 \ln \left (F \right ) b \,d^{3} x^{3}+30 c^{4} d^{2} x^{2}-6 \ln \left (F \right ) b c \,d^{2} x^{2}+12 c^{5} d x -6 \ln \left (F \right ) b \,c^{2} d x +2 c^{6}-2 \ln \left (F \right ) b \,c^{3}+\ln \left (F \right )^{2} b^{2}\right ) F^{\frac {a \,d^{3} x^{3}+3 a c \,d^{2} x^{2}+3 a \,c^{2} d x +a \,c^{3}+b}{\left (d x +c \right )^{3}}}}{3 b^{3} \ln \left (F \right )^{3} d \left (d x +c \right )^{6}}\) | \(175\) |
| parallelrisch | \(\frac {-2 x^{6} F^{a +\frac {b}{\left (d x +c \right )^{3}}} d^{19}-12 x^{5} F^{a +\frac {b}{\left (d x +c \right )^{3}}} c \,d^{18}-30 x^{4} F^{a +\frac {b}{\left (d x +c \right )^{3}}} c^{2} d^{17}-40 x^{3} F^{a +\frac {b}{\left (d x +c \right )^{3}}} c^{3} d^{16}+2 \ln \left (F \right ) x^{3} F^{a +\frac {b}{\left (d x +c \right )^{3}}} b \,d^{16}-30 x^{2} F^{a +\frac {b}{\left (d x +c \right )^{3}}} c^{4} d^{15}+6 \ln \left (F \right ) x^{2} F^{a +\frac {b}{\left (d x +c \right )^{3}}} b c \,d^{15}-12 x \,F^{a +\frac {b}{\left (d x +c \right )^{3}}} c^{5} d^{14}+6 \ln \left (F \right ) x \,F^{a +\frac {b}{\left (d x +c \right )^{3}}} b \,c^{2} d^{14}-2 F^{a +\frac {b}{\left (d x +c \right )^{3}}} c^{6} d^{13}+2 \ln \left (F \right ) F^{a +\frac {b}{\left (d x +c \right )^{3}}} b \,c^{3} d^{13}-\ln \left (F \right )^{2} F^{a +\frac {b}{\left (d x +c \right )^{3}}} b^{2} d^{13}}{3 \left (d x +c \right )^{6} \ln \left (F \right )^{3} b^{3} d^{14}}\) | \(302\) |
| norman | \(\frac {-\frac {2 d^{8} x^{9} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 \ln \left (F \right )^{3} b^{3}}-\frac {c^{2} \left (6 c^{6}-4 \ln \left (F \right ) b \,c^{3}+\ln \left (F \right )^{2} b^{2}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{b^{3} \ln \left (F \right )^{3}}-\frac {d^{2} \left (168 c^{6}-40 \ln \left (F \right ) b \,c^{3}+\ln \left (F \right )^{2} b^{2}\right ) x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 \ln \left (F \right )^{3} b^{3}}+\frac {2 d^{5} \left (-84 c^{3}+b \ln \left (F \right )\right ) x^{6} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 \ln \left (F \right )^{3} b^{3}}-\frac {24 d^{6} c^{2} x^{7} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {6 d^{7} c \,x^{8} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (2 c^{6}-2 \ln \left (F \right ) b \,c^{3}+\ln \left (F \right )^{2} b^{2}\right ) c^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 b^{3} \ln \left (F \right )^{3} d}-\frac {c d \left (24 c^{6}-10 \ln \left (F \right ) b \,c^{3}+\ln \left (F \right )^{2} b^{2}\right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}+\frac {4 c \,d^{4} \left (-21 c^{3}+b \ln \left (F \right )\right ) x^{5} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}+\frac {2 c^{2} d^{3} \left (-42 c^{3}+5 b \ln \left (F \right )\right ) x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}}{\left (d x +c \right )^{9}}\) | \(434\) |
-1/3*(2*d^6*x^6+12*c*d^5*x^5+30*c^2*d^4*x^4+40*c^3*d^3*x^3-2*ln(F)*b*d^3*x ^3+30*c^4*d^2*x^2-6*ln(F)*b*c*d^2*x^2+12*c^5*d*x-6*ln(F)*b*c^2*d*x+2*c^6-2 *ln(F)*b*c^3+ln(F)^2*b^2)/b^3/ln(F)^3/d/(d*x+c)^6*F^((a*d^3*x^3+3*a*c*d^2* x^2+3*a*c^2*d*x+a*c^3+b)/(d*x+c)^3)
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (90) = 180\).
Time = 0.30 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.76 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=-\frac {{\left (2 \, d^{6} x^{6} + 12 \, c d^{5} x^{5} + 30 \, c^{2} d^{4} x^{4} + 40 \, c^{3} d^{3} x^{3} + 30 \, c^{4} d^{2} x^{2} + 12 \, c^{5} d x + 2 \, c^{6} + b^{2} \log \left (F\right )^{2} - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \left (F\right )^{3}} \]
-1/3*(2*d^6*x^6 + 12*c*d^5*x^5 + 30*c^2*d^4*x^4 + 40*c^3*d^3*x^3 + 30*c^4* d^2*x^2 + 12*c^5*d*x + 2*c^6 + b^2*log(F)^2 - 2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*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))/((b^3*d^7*x^6 + 6*b^3*c*d^6*x^5 + 15*b^3*c^2*d^5*x^4 + 20*b^3*c^3*d^4*x^3 + 15*b^3*c^4*d^3 *x^2 + 6*b^3*c^5*d^2*x + b^3*c^6*d)*log(F)^3)
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (83) = 166\).
Time = 0.22 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.81 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=\frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (- b^{2} \log {\left (F \right )}^{2} + 2 b c^{3} \log {\left (F \right )} + 6 b c^{2} d x \log {\left (F \right )} + 6 b c d^{2} x^{2} \log {\left (F \right )} + 2 b d^{3} x^{3} \log {\left (F \right )} - 2 c^{6} - 12 c^{5} d x - 30 c^{4} d^{2} x^{2} - 40 c^{3} d^{3} x^{3} - 30 c^{2} d^{4} x^{4} - 12 c d^{5} x^{5} - 2 d^{6} x^{6}\right )}{3 b^{3} c^{6} d \log {\left (F \right )}^{3} + 18 b^{3} c^{5} d^{2} x \log {\left (F \right )}^{3} + 45 b^{3} c^{4} d^{3} x^{2} \log {\left (F \right )}^{3} + 60 b^{3} c^{3} d^{4} x^{3} \log {\left (F \right )}^{3} + 45 b^{3} c^{2} d^{5} x^{4} \log {\left (F \right )}^{3} + 18 b^{3} c d^{6} x^{5} \log {\left (F \right )}^{3} + 3 b^{3} d^{7} x^{6} \log {\left (F \right )}^{3}} \]
F**(a + b/(c + d*x)**3)*(-b**2*log(F)**2 + 2*b*c**3*log(F) + 6*b*c**2*d*x* log(F) + 6*b*c*d**2*x**2*log(F) + 2*b*d**3*x**3*log(F) - 2*c**6 - 12*c**5* d*x - 30*c**4*d**2*x**2 - 40*c**3*d**3*x**3 - 30*c**2*d**4*x**4 - 12*c*d** 5*x**5 - 2*d**6*x**6)/(3*b**3*c**6*d*log(F)**3 + 18*b**3*c**5*d**2*x*log(F )**3 + 45*b**3*c**4*d**3*x**2*log(F)**3 + 60*b**3*c**3*d**4*x**3*log(F)**3 + 45*b**3*c**2*d**5*x**4*log(F)**3 + 18*b**3*c*d**6*x**5*log(F)**3 + 3*b* *3*d**7*x**6*log(F)**3)
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (90) = 180\).
Time = 0.23 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.12 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=-\frac {{\left (2 \, F^{a} d^{6} x^{6} + 12 \, F^{a} c d^{5} x^{5} + 30 \, F^{a} c^{2} d^{4} x^{4} + 2 \, F^{a} c^{6} - 2 \, F^{a} b c^{3} \log \left (F\right ) + F^{a} b^{2} \log \left (F\right )^{2} + 2 \, {\left (20 \, F^{a} c^{3} d^{3} - F^{a} b d^{3} \log \left (F\right )\right )} x^{3} + 6 \, {\left (5 \, F^{a} c^{4} d^{2} - F^{a} b c d^{2} \log \left (F\right )\right )} x^{2} + 6 \, {\left (2 \, F^{a} c^{5} d - F^{a} b c^{2} d \log \left (F\right )\right )} x\right )} F^{\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{3} d^{7} x^{6} \log \left (F\right )^{3} + 6 \, b^{3} c d^{6} x^{5} \log \left (F\right )^{3} + 15 \, b^{3} c^{2} d^{5} x^{4} \log \left (F\right )^{3} + 20 \, b^{3} c^{3} d^{4} x^{3} \log \left (F\right )^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} \log \left (F\right )^{3} + 6 \, b^{3} c^{5} d^{2} x \log \left (F\right )^{3} + b^{3} c^{6} d \log \left (F\right )^{3}\right )}} \]
-1/3*(2*F^a*d^6*x^6 + 12*F^a*c*d^5*x^5 + 30*F^a*c^2*d^4*x^4 + 2*F^a*c^6 - 2*F^a*b*c^3*log(F) + F^a*b^2*log(F)^2 + 2*(20*F^a*c^3*d^3 - F^a*b*d^3*log( F))*x^3 + 6*(5*F^a*c^4*d^2 - F^a*b*c*d^2*log(F))*x^2 + 6*(2*F^a*c^5*d - F^ a*b*c^2*d*log(F))*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(b^3* d^7*x^6*log(F)^3 + 6*b^3*c*d^6*x^5*log(F)^3 + 15*b^3*c^2*d^5*x^4*log(F)^3 + 20*b^3*c^3*d^4*x^3*log(F)^3 + 15*b^3*c^4*d^3*x^2*log(F)^3 + 6*b^3*c^5*d^ 2*x*log(F)^3 + b^3*c^6*d*log(F)^3)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{10}} \,d x } \]
Time = 0.80 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.74 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx=-\frac {F^a\,F^{\frac {b}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3}}\,\left (\frac {2\,x^6}{3\,b^3\,d\,{\ln \left (F\right )}^3}+\frac {b^2\,{\ln \left (F\right )}^2-2\,b\,c^3\,\ln \left (F\right )+2\,c^6}{3\,b^3\,d^7\,{\ln \left (F\right )}^3}+\frac {4\,c\,x^5}{b^3\,d^2\,{\ln \left (F\right )}^3}+\frac {10\,c^2\,x^4}{b^3\,d^3\,{\ln \left (F\right )}^3}-\frac {2\,x^3\,\left (b\,\ln \left (F\right )-20\,c^3\right )}{3\,b^3\,d^4\,{\ln \left (F\right )}^3}-\frac {2\,c^2\,x\,\left (b\,\ln \left (F\right )-2\,c^3\right )}{b^3\,d^6\,{\ln \left (F\right )}^3}-\frac {2\,c\,x^2\,\left (b\,\ln \left (F\right )-5\,c^3\right )}{b^3\,d^5\,{\ln \left (F\right )}^3}\right )}{x^6+\frac {c^6}{d^6}+\frac {6\,c\,x^5}{d}+\frac {6\,c^5\,x}{d^5}+\frac {15\,c^2\,x^4}{d^2}+\frac {20\,c^3\,x^3}{d^3}+\frac {15\,c^4\,x^2}{d^4}} \]
-(F^a*F^(b/(c^3 + d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x))*((2*x^6)/(3*b^3*d*lo g(F)^3) + (b^2*log(F)^2 + 2*c^6 - 2*b*c^3*log(F))/(3*b^3*d^7*log(F)^3) + ( 4*c*x^5)/(b^3*d^2*log(F)^3) + (10*c^2*x^4)/(b^3*d^3*log(F)^3) - (2*x^3*(b* log(F) - 20*c^3))/(3*b^3*d^4*log(F)^3) - (2*c^2*x*(b*log(F) - 2*c^3))/(b^3 *d^6*log(F)^3) - (2*c*x^2*(b*log(F) - 5*c^3))/(b^3*d^5*log(F)^3)))/(x^6 + c^6/d^6 + (6*c*x^5)/d + (6*c^5*x)/d^5 + (15*c^2*x^4)/d^2 + (20*c^3*x^3)/d^ 3 + (15*c^4*x^2)/d^4)