Integrand size = 21, antiderivative size = 91 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{b^3 d \log ^3(F)}+\frac {F^{a+\frac {b}{(c+d x)^2}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^4 \log (F)} \]
-F^(a+b/(d*x+c)^2)/b^3/d/ln(F)^3+F^(a+b/(d*x+c)^2)/b^2/d/(d*x+c)^2/ln(F)^2 -1/2*F^(a+b/(d*x+c)^2)/b/d/(d*x+c)^4/ln(F)
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^2}} \left (2 (c+d x)^4-2 b (c+d x)^2 \log (F)+b^2 \log ^2(F)\right )}{2 b^3 d (c+d x)^4 \log ^3(F)} \]
-1/2*(F^(a + b/(c + d*x)^2)*(2*(c + d*x)^4 - 2*b*(c + d*x)^2*Log[F] + b^2* Log[F]^2))/(b^3*d*(c + d*x)^4*Log[F]^3)
Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16, 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.
\(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {2 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5}dx}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^4}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {2 \left (-\frac {\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3}dx}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^4}\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle -\frac {2 \left (\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^4}\) |
-1/2*F^(a + b/(c + d*x)^2)/(b*d*(c + d*x)^4*Log[F]) - (2*(F^(a + b/(c + d* x)^2)/(2*b^2*d*Log[F]^2) - F^(a + b/(c + d*x)^2)/(2*b*d*(c + d*x)^2*Log[F] )))/(b*Log[F])
3.4.23.3.1 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 = 1.05 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.40
method | result | size |
risch | \(-\frac {\left (2 d^{4} x^{4}+8 c \,d^{3} x^{3}-2 \ln \left (F \right ) b \,d^{2} x^{2}+12 c^{2} d^{2} x^{2}-4 \ln \left (F \right ) b c d x +8 c^{3} d x +\ln \left (F \right )^{2} b^{2}-2 \ln \left (F \right ) b \,c^{2}+2 c^{4}\right ) F^{\frac {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}{\left (d x +c \right )^{2}}}}{2 b^{3} \ln \left (F \right )^{3} d \left (d x +c \right )^{4}}\) | \(127\) |
parallelrisch | \(\frac {-2 x^{4} F^{a +\frac {b}{\left (d x +c \right )^{2}}} d^{13}-8 x^{3} F^{a +\frac {b}{\left (d x +c \right )^{2}}} c \,d^{12}+2 \ln \left (F \right ) x^{2} F^{a +\frac {b}{\left (d x +c \right )^{2}}} b \,d^{11}-12 x^{2} F^{a +\frac {b}{\left (d x +c \right )^{2}}} c^{2} d^{11}+4 \ln \left (F \right ) x \,F^{a +\frac {b}{\left (d x +c \right )^{2}}} b c \,d^{10}-8 x \,F^{a +\frac {b}{\left (d x +c \right )^{2}}} c^{3} d^{10}-\ln \left (F \right )^{2} F^{a +\frac {b}{\left (d x +c \right )^{2}}} b^{2} d^{9}+2 \ln \left (F \right ) F^{a +\frac {b}{\left (d x +c \right )^{2}}} b \,c^{2} d^{9}-2 F^{a +\frac {b}{\left (d x +c \right )^{2}}} c^{4} d^{9}}{2 \left (d x +c \right )^{4} \ln \left (F \right )^{3} b^{3} d^{10}}\) | \(227\) |
norman | \(\frac {\frac {d^{3} \left (b \ln \left (F \right )-15 c^{2}\right ) x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {d^{5} x^{6} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {c \left (\ln \left (F \right )^{2} b^{2}-4 \ln \left (F \right ) b \,c^{2}+6 c^{4}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{b^{3} \ln \left (F \right )^{3}}-\frac {d \left (\ln \left (F \right )^{2} b^{2}-12 \ln \left (F \right ) b \,c^{2}+30 c^{4}\right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{3} b^{3}}-\frac {6 d^{4} c \,x^{5} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (\ln \left (F \right )^{2} b^{2}-2 \ln \left (F \right ) b \,c^{2}+2 c^{4}\right ) c^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 b^{3} \ln \left (F \right )^{3} d}+\frac {4 c \,d^{2} \left (b \ln \left (F \right )-5 c^{2}\right ) x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}}{\left (d x +c \right )^{6}}\) | \(301\) |
-1/2*(2*d^4*x^4+8*c*d^3*x^3-2*ln(F)*b*d^2*x^2+12*c^2*d^2*x^2-4*ln(F)*b*c*d *x+8*c^3*d*x+ln(F)^2*b^2-2*ln(F)*b*c^2+2*c^4)/b^3/ln(F)^3/d/(d*x+c)^4*F^(( a*d^2*x^2+2*a*c*d*x+a*c^2+b)/(d*x+c)^2)
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (89) = 178\).
Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.98 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=-\frac {{\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 12 \, c^{2} d^{2} x^{2} + 8 \, c^{3} d x + 2 \, c^{4} + b^{2} \log \left (F\right )^{2} - 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{3} d^{5} x^{4} + 4 \, b^{3} c d^{4} x^{3} + 6 \, b^{3} c^{2} d^{3} x^{2} + 4 \, b^{3} c^{3} d^{2} x + b^{3} c^{4} d\right )} \log \left (F\right )^{3}} \]
-1/2*(2*d^4*x^4 + 8*c*d^3*x^3 + 12*c^2*d^2*x^2 + 8*c^3*d*x + 2*c^4 + b^2*l og(F)^2 - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*F^((a*d^2*x^2 + 2*a*c* d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2))/((b^3*d^5*x^4 + 4*b^3*c*d^4*x^ 3 + 6*b^3*c^2*d^3*x^2 + 4*b^3*c^3*d^2*x + b^3*c^4*d)*log(F)^3)
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (76) = 152\).
Time = 0.18 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.08 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=\frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (- b^{2} \log {\left (F \right )}^{2} + 2 b c^{2} \log {\left (F \right )} + 4 b c d x \log {\left (F \right )} + 2 b d^{2} x^{2} \log {\left (F \right )} - 2 c^{4} - 8 c^{3} d x - 12 c^{2} d^{2} x^{2} - 8 c d^{3} x^{3} - 2 d^{4} x^{4}\right )}{2 b^{3} c^{4} d \log {\left (F \right )}^{3} + 8 b^{3} c^{3} d^{2} x \log {\left (F \right )}^{3} + 12 b^{3} c^{2} d^{3} x^{2} \log {\left (F \right )}^{3} + 8 b^{3} c d^{4} x^{3} \log {\left (F \right )}^{3} + 2 b^{3} d^{5} x^{4} \log {\left (F \right )}^{3}} \]
F**(a + b/(c + d*x)**2)*(-b**2*log(F)**2 + 2*b*c**2*log(F) + 4*b*c*d*x*log (F) + 2*b*d**2*x**2*log(F) - 2*c**4 - 8*c**3*d*x - 12*c**2*d**2*x**2 - 8*c *d**3*x**3 - 2*d**4*x**4)/(2*b**3*c**4*d*log(F)**3 + 8*b**3*c**3*d**2*x*lo g(F)**3 + 12*b**3*c**2*d**3*x**2*log(F)**3 + 8*b**3*c*d**4*x**3*log(F)**3 + 2*b**3*d**5*x**4*log(F)**3)
Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (89) = 178\).
Time = 0.25 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.29 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=-\frac {{\left (2 \, F^{a} d^{4} x^{4} + 8 \, F^{a} c d^{3} x^{3} + 2 \, F^{a} c^{4} - 2 \, F^{a} b c^{2} \log \left (F\right ) + F^{a} b^{2} \log \left (F\right )^{2} + 2 \, {\left (6 \, F^{a} c^{2} d^{2} - F^{a} b d^{2} \log \left (F\right )\right )} x^{2} + 4 \, {\left (2 \, F^{a} c^{3} d - F^{a} b c d \log \left (F\right )\right )} x\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{3} d^{5} x^{4} \log \left (F\right )^{3} + 4 \, b^{3} c d^{4} x^{3} \log \left (F\right )^{3} + 6 \, b^{3} c^{2} d^{3} x^{2} \log \left (F\right )^{3} + 4 \, b^{3} c^{3} d^{2} x \log \left (F\right )^{3} + b^{3} c^{4} d \log \left (F\right )^{3}\right )}} \]
-1/2*(2*F^a*d^4*x^4 + 8*F^a*c*d^3*x^3 + 2*F^a*c^4 - 2*F^a*b*c^2*log(F) + F ^a*b^2*log(F)^2 + 2*(6*F^a*c^2*d^2 - F^a*b*d^2*log(F))*x^2 + 4*(2*F^a*c^3* d - F^a*b*c*d*log(F))*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(b^3*d^5*x^4*log( F)^3 + 4*b^3*c*d^4*x^3*log(F)^3 + 6*b^3*c^2*d^3*x^2*log(F)^3 + 4*b^3*c^3*d ^2*x*log(F)^3 + b^3*c^4*d*log(F)^3)
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 1703, normalized size of antiderivative = 18.71 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=\text {Too large to display} \]
-1/2*((2*(pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^2*sgn(F) - pi^3*b ^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)*(pi*b*d^2*sgn(F)/(d*x + c)^2 - pi*b^2 *d^2*log(abs(F))*sgn(F)/(d*x + c)^4 - pi*b*d^2/(d*x + c)^2 + pi*b^2*d^2*lo g(abs(F))/(d*x + c)^4)/((pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^2* sgn(F) - pi^3*b^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)^2 + (3*pi^2*b^3*d^3*lo g(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3)^2 ) + (3*pi^2*b^3*d^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^ 3*d^3*log(abs(F))^3)*(pi^2*b^2*d^2*sgn(F)/(d*x + c)^4 - pi^2*b^2*d^2/(d*x + c)^4 + 4*d^2 - 4*b*d^2*log(abs(F))/(d*x + c)^2 + 2*b^2*d^2*log(abs(F))^2 /(d*x + c)^4)/((pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^2*sgn(F) - pi^3*b^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)^2 + (3*pi^2*b^3*d^3*log(abs(F)) *sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3)^2))*cos(-1 /2*pi*a*sgn(F) + 1/2*pi*a - 1/2*pi*b*sgn(F)/(d^2*x^2 + 2*c*d*x + c^2) + 1/ 2*pi*b/(d^2*x^2 + 2*c*d*x + c^2)) - (2*(3*pi^2*b^3*d^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3)*(pi*b*d^2*sgn(F)/( d*x + c)^2 - pi*b^2*d^2*log(abs(F))*sgn(F)/(d*x + c)^4 - pi*b*d^2/(d*x + c )^2 + pi*b^2*d^2*log(abs(F))/(d*x + c)^4)/((pi^3*b^3*d^3*sgn(F) - 3*pi*b^3 *d^3*log(abs(F))^2*sgn(F) - pi^3*b^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)^2 + (3*pi^2*b^3*d^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d ^3*log(abs(F))^3)^2) - (pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^...
Time = 0.57 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.01 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^7} \, dx=-\frac {F^a\,F^{\frac {b}{c^2+2\,c\,d\,x+d^2\,x^2}}\,\left (\frac {x^4}{b^3\,d\,{\ln \left (F\right )}^3}+\frac {b^2\,{\ln \left (F\right )}^2-2\,b\,c^2\,\ln \left (F\right )+2\,c^4}{2\,b^3\,d^5\,{\ln \left (F\right )}^3}+\frac {4\,c\,x^3}{b^3\,d^2\,{\ln \left (F\right )}^3}-\frac {x^2\,\left (b\,\ln \left (F\right )-6\,c^2\right )}{b^3\,d^3\,{\ln \left (F\right )}^3}-\frac {2\,c\,x\,\left (b\,\ln \left (F\right )-2\,c^2\right )}{b^3\,d^4\,{\ln \left (F\right )}^3}\right )}{x^4+\frac {c^4}{d^4}+\frac {4\,c\,x^3}{d}+\frac {4\,c^3\,x}{d^3}+\frac {6\,c^2\,x^2}{d^2}} \]
-(F^a*F^(b/(c^2 + d^2*x^2 + 2*c*d*x))*(x^4/(b^3*d*log(F)^3) + (b^2*log(F)^ 2 + 2*c^4 - 2*b*c^2*log(F))/(2*b^3*d^5*log(F)^3) + (4*c*x^3)/(b^3*d^2*log( F)^3) - (x^2*(b*log(F) - 6*c^2))/(b^3*d^3*log(F)^3) - (2*c*x*(b*log(F) - 2 *c^2))/(b^3*d^4*log(F)^3)))/(x^4 + c^4/d^4 + (4*c*x^3)/d + (4*c^3*x)/d^3 + (6*c^2*x^2)/d^2)