Integrand size = 21, antiderivative size = 90 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=-\frac {2 F^{a+\frac {b}{c+d x}}}{b^3 d \log ^3(F)}+\frac {2 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x) \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^2 \log (F)} \] Output:
-2*F^(a+b/(d*x+c))/b^3/d/ln(F)^3+2*F^(a+b/(d*x+c))/b^2/d/(d*x+c)/ln(F)^2-F ^(a+b/(d*x+c))/b/d/(d*x+c)^2/ln(F)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=-\frac {F^{a+\frac {b}{c+d x}} \left (2 (c+d x)^2-2 b (c+d x) \log (F)+b^2 \log ^2(F)\right )}{b^3 d (c+d x)^2 \log ^3(F)} \] Input:
Integrate[F^(a + b/(c + d*x))/(c + d*x)^4,x]
Output:
-((F^(a + b/(c + d*x))*(2*(c + d*x)^2 - 2*b*(c + d*x)*Log[F] + b^2*Log[F]^ 2))/(b^3*d*(c + d*x)^2*Log[F]^3))
Time = 0.56 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {2 \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3}dx}{b \log (F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)^2}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {2 \left (-\frac {\int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2}dx}{b \log (F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)^2}\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle -\frac {2 \left (\frac {F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)}\right )}{b \log (F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)^2}\) |
Input:
Int[F^(a + b/(c + d*x))/(c + d*x)^4,x]
Output:
-(F^(a + b/(c + d*x))/(b*d*(c + d*x)^2*Log[F])) - (2*(F^(a + b/(c + d*x))/ (b^2*d*Log[F]^2) - F^(a + b/(c + d*x))/(b*d*(c + d*x)*Log[F])))/(b*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 = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {\left (\ln \left (F \right )^{2} b^{2}-2 \ln \left (F \right ) b d x +2 x^{2} d^{2}-2 b c \ln \left (F \right )+4 c d x +2 c^{2}\right ) F^{\frac {a d x +a c +b}{d x +c}}}{b^{3} \ln \left (F \right )^{3} d \left (d x +c \right )^{2}}\) | \(79\) |
parallelrisch | \(\frac {-\ln \left (F \right )^{2} F^{a +\frac {b}{d x +c}} b^{2} d^{5}+2 \ln \left (F \right ) x \,F^{a +\frac {b}{d x +c}} b \,d^{6}-2 x^{2} F^{a +\frac {b}{d x +c}} d^{7}+2 \ln \left (F \right ) F^{a +\frac {b}{d x +c}} b c \,d^{5}-4 x \,F^{a +\frac {b}{d x +c}} c \,d^{6}-2 F^{a +\frac {b}{d x +c}} c^{2} d^{5}}{\left (d x +c \right )^{2} b^{3} \ln \left (F \right )^{3} d^{6}}\) | \(151\) |
norman | \(\frac {-\frac {2 d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (\ln \left (F \right )^{2} b^{2}-4 b c \ln \left (F \right )+6 c^{2}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}+\frac {2 d \left (b \ln \left (F \right )-3 c \right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (\ln \left (F \right )^{2} b^{2}-2 b c \ln \left (F \right )+2 c^{2}\right ) c \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{b^{3} \ln \left (F \right )^{3} d}}{\left (d x +c \right )^{3}}\) | \(169\) |
Input:
int(F^(a+b/(d*x+c))/(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-(ln(F)^2*b^2-2*ln(F)*b*d*x+2*x^2*d^2-2*b*c*ln(F)+4*c*d*x+2*c^2)/b^3/ln(F) ^3/d/(d*x+c)^2*F^((a*d*x+a*c+b)/(d*x+c))
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.06 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=-\frac {{\left (2 \, d^{2} x^{2} + b^{2} \log \left (F\right )^{2} + 4 \, c d x + 2 \, c^{2} - 2 \, {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{{\left (b^{3} d^{3} x^{2} + 2 \, b^{3} c d^{2} x + b^{3} c^{2} d\right )} \log \left (F\right )^{3}} \] Input:
integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="fricas")
Output:
-(2*d^2*x^2 + b^2*log(F)^2 + 4*c*d*x + 2*c^2 - 2*(b*d*x + b*c)*log(F))*F^( (a*d*x + a*c + b)/(d*x + c))/((b^3*d^3*x^2 + 2*b^3*c*d^2*x + b^3*c^2*d)*lo g(F)^3)
Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.13 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=\frac {F^{a + \frac {b}{c + d x}} \left (- b^{2} \log {\left (F \right )}^{2} + 2 b c \log {\left (F \right )} + 2 b d x \log {\left (F \right )} - 2 c^{2} - 4 c d x - 2 d^{2} x^{2}\right )}{b^{3} c^{2} d \log {\left (F \right )}^{3} + 2 b^{3} c d^{2} x \log {\left (F \right )}^{3} + b^{3} d^{3} x^{2} \log {\left (F \right )}^{3}} \] Input:
integrate(F**(a+b/(d*x+c))/(d*x+c)**4,x)
Output:
F**(a + b/(c + d*x))*(-b**2*log(F)**2 + 2*b*c*log(F) + 2*b*d*x*log(F) - 2* c**2 - 4*c*d*x - 2*d**2*x**2)/(b**3*c**2*d*log(F)**3 + 2*b**3*c*d**2*x*log (F)**3 + b**3*d**3*x**2*log(F)**3)
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="maxima")
Output:
integrate(F^(a + b/(d*x + c))/(d*x + c)^4, x)
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c))/(d*x + c)^4, x)
Time = 1.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=-\frac {F^{a+\frac {b}{c+d\,x}}\,\left (\frac {b^2\,{\ln \left (F\right )}^2-2\,b\,c\,\ln \left (F\right )+2\,c^2}{b^3\,d^3\,{\ln \left (F\right )}^3}+\frac {2\,x^2}{b^3\,d\,{\ln \left (F\right )}^3}+\frac {2\,x\,\left (2\,c-b\,\ln \left (F\right )\right )}{b^3\,d^2\,{\ln \left (F\right )}^3}\right )}{x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}} \] Input:
int(F^(a + b/(c + d*x))/(c + d*x)^4,x)
Output:
-(F^(a + b/(c + d*x))*((b^2*log(F)^2 + 2*c^2 - 2*b*c*log(F))/(b^3*d^3*log( F)^3) + (2*x^2)/(b^3*d*log(F)^3) + (2*x*(2*c - b*log(F)))/(b^3*d^2*log(F)^ 3)))/(x^2 + c^2/d^2 + (2*c*x)/d)
Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx=\frac {f^{\frac {a d x +a c +b}{d x +c}} \left (-\mathrm {log}\left (f \right )^{2} b^{2}+2 \,\mathrm {log}\left (f \right ) b c +2 \,\mathrm {log}\left (f \right ) b d x -2 c^{2}-4 c d x -2 d^{2} x^{2}\right )}{\mathrm {log}\left (f \right )^{3} b^{3} d \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int(F^(a+b/(d*x+c))/(d*x+c)^4,x)
Output:
(f**((a*c + a*d*x + b)/(c + d*x))*( - log(f)**2*b**2 + 2*log(f)*b*c + 2*lo g(f)*b*d*x - 2*c**2 - 4*c*d*x - 2*d**2*x**2))/(log(f)**3*b**3*d*(c**2 + 2* c*d*x + d**2*x**2))