Integrand size = 21, antiderivative size = 62 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^2 \log (F)} \] Output:
1/2*F^(a+b/(d*x+c)^2)/b^2/d/ln(F)^2-1/2*F^(a+b/(d*x+c)^2)/b/d/(d*x+c)^2/ln (F)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} \left ((c+d x)^2-b \log (F)\right )}{2 b^2 d (c+d x)^2 \log ^2(F)} \] Input:
Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^5,x]
Output:
(F^(a + b/(c + d*x)^2)*((c + d*x)^2 - b*Log[F]))/(2*b^2*d*(c + d*x)^2*Log[ F]^2)
Time = 0.43 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.
\(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3}dx}{b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2}\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle \frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2}\) |
Input:
Int[F^(a + b/(c + d*x)^2)/(c + d*x)^5,x]
Output:
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])
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 = 0.49 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19
method | result | size |
risch | \(-\frac {\left (-x^{2} d^{2}-2 c d x +b \ln \left (F \right )-c^{2}\right ) F^{\frac {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}{\left (d x +c \right )^{2}}}}{2 d \ln \left (F \right )^{2} b^{2} \left (d x +c \right )^{2}}\) | \(74\) |
parallelrisch | \(\frac {x^{2} F^{a +\frac {b}{\left (d x +c \right )^{2}}} d^{9}+2 x \,F^{a +\frac {b}{\left (d x +c \right )^{2}}} c \,d^{8}-\ln \left (F \right ) F^{a +\frac {b}{\left (d x +c \right )^{2}}} b \,d^{7}+F^{a +\frac {b}{\left (d x +c \right )^{2}}} c^{2} d^{7}}{2 \left (d x +c \right )^{2} \ln \left (F \right )^{2} b^{2} d^{8}}\) | \(102\) |
norman | \(\frac {\frac {d^{3} x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{2} b^{2}}-\frac {c \left (b \ln \left (F \right )-2 c^{2}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}-\frac {c^{2} \left (b \ln \left (F \right )-c^{2}\right ) {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 d \ln \left (F \right )^{2} b^{2}}-\frac {d \left (b \ln \left (F \right )-6 c^{2}\right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{2} b^{2}}+\frac {2 d^{2} c \,x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}}{\left (d x +c \right )^{4}}\) | \(185\) |
Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
-1/2*(-x^2*d^2-2*c*d*x+b*ln(F)-c^2)/d/ln(F)^2/b^2/(d*x+c)^2*F^((a*d^2*x^2+ 2*a*c*d*x+a*c^2+b)/(d*x+c)^2)
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.61 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - b \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (F\right )^{2}} \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="fricas")
Output:
1/2*(d^2*x^2 + 2*c*d*x + c^2 - b*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^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d )*log(F)^2)
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (- b \log {\left (F \right )} + c^{2} + 2 c d x + d^{2} x^{2}\right )}{2 b^{2} c^{2} d \log {\left (F \right )}^{2} + 4 b^{2} c d^{2} x \log {\left (F \right )}^{2} + 2 b^{2} d^{3} x^{2} \log {\left (F \right )}^{2}} \] Input:
integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**5,x)
Output:
F**(a + b/(c + d*x)**2)*(-b*log(F) + c**2 + 2*c*d*x + d**2*x**2)/(2*b**2*c **2*d*log(F)**2 + 4*b**2*c*d**2*x*log(F)**2 + 2*b**2*d**3*x**2*log(F)**2)
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.63 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {{\left (F^{a} d^{2} x^{2} + 2 \, F^{a} c d x + F^{a} c^{2} - F^{a} b \log \left (F\right )\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{2} d^{3} x^{2} \log \left (F\right )^{2} + 2 \, b^{2} c d^{2} x \log \left (F\right )^{2} + b^{2} c^{2} d \log \left (F\right )^{2}\right )}} \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="maxima")
Output:
1/2*(F^a*d^2*x^2 + 2*F^a*c*d*x + F^a*c^2 - F^a*b*log(F))*F^(b/(d^2*x^2 + 2 *c*d*x + c^2))/(b^2*d^3*x^2*log(F)^2 + 2*b^2*c*d^2*x*log(F)^2 + b^2*c^2*d* log(F)^2)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{5}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^5, x)
Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.56 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {F^a\,F^{\frac {b}{c^2+2\,c\,d\,x+d^2\,x^2}}\,\left (\frac {x^2}{2\,b^2\,d\,{\ln \left (F\right )}^2}-\frac {b\,\ln \left (F\right )-c^2}{2\,b^2\,d^3\,{\ln \left (F\right )}^2}+\frac {c\,x}{b^2\,d^2\,{\ln \left (F\right )}^2}\right )}{x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}} \] Input:
int(F^(a + b/(c + d*x)^2)/(c + d*x)^5,x)
Output:
(F^a*F^(b/(c^2 + d^2*x^2 + 2*c*d*x))*(x^2/(2*b^2*d*log(F)^2) - (b*log(F) - c^2)/(2*b^2*d^3*log(F)^2) + (c*x)/(b^2*d^2*log(F)^2)))/(x^2 + c^2/d^2 + ( 2*c*x)/d)
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.50 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^5} \, dx=\frac {f^{\frac {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}{d^{2} x^{2}+2 c d x +c^{2}}} \left (-\mathrm {log}\left (f \right ) b +c^{2}+2 c d x +d^{2} x^{2}\right )}{2 \mathrm {log}\left (f \right )^{2} b^{2} d \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x)
Output:
(f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))* ( - log(f)*b + c**2 + 2*c*d*x + d**2*x**2))/(2*log(f)**2*b**2*d*(c**2 + 2* c*d*x + d**2*x**2))