\(\int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx\) [311]

Optimal result

Integrand size = 21, antiderivative size = 122 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {6 F^{a+\frac {b}{c+d x}}}{b^4 d \log ^4(F)}-\frac {6 F^{a+\frac {b}{c+d x}}}{b^3 d (c+d x) \log ^3(F)}+\frac {3 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^3 \log (F)} \]

[Out]

6*F^(a+b/(d*x+c))/b^4/d/ln(F)^4-6*F^(a+b/(d*x+c))/b^3/d/(d*x+c)/ln(F)^3+3*F^(a+b/(d*x+c))/b^2/d/(d*x+c)^2/ln(F
)^2-F^(a+b/(d*x+c))/b/d/(d*x+c)^3/ln(F)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2240} \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {6 F^{a+\frac {b}{c+d x}}}{b^4 d \log ^4(F)}-\frac {6 F^{a+\frac {b}{c+d x}}}{b^3 d \log ^3(F) (c+d x)}+\frac {3 F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F) (c+d x)^2}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)^3} \]

[In]

Int[F^(a + b/(c + d*x))/(c + d*x)^5,x]

[Out]

(6*F^(a + b/(c + d*x)))/(b^4*d*Log[F]^4) - (6*F^(a + b/(c + d*x)))/(b^3*d*(c + d*x)*Log[F]^3) + (3*F^(a + b/(c
+ d*x)))/(b^2*d*(c + d*x)^2*Log[F]^2) - F^(a + b/(c + d*x))/(b*d*(c + d*x)^3*Log[F])

Rule 2240

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]

Rule 2243

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*Log[F])), x] - Dist[(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])

Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^3 \log (F)}-\frac {3 \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx}{b \log (F)} \\ & = \frac {3 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^3 \log (F)}+\frac {6 \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx}{b^2 \log ^2(F)} \\ & = -\frac {6 F^{a+\frac {b}{c+d x}}}{b^3 d (c+d x) \log ^3(F)}+\frac {3 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^3 \log (F)}-\frac {6 \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx}{b^3 \log ^3(F)} \\ & = \frac {6 F^{a+\frac {b}{c+d x}}}{b^4 d \log ^4(F)}-\frac {6 F^{a+\frac {b}{c+d x}}}{b^3 d (c+d x) \log ^3(F)}+\frac {3 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^3 \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.62 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {F^{a+\frac {b}{c+d x}} \left (6 (c+d x)^3-6 b (c+d x)^2 \log (F)+3 b^2 (c+d x) \log ^2(F)-b^3 \log ^3(F)\right )}{b^4 d (c+d x)^3 \log ^4(F)} \]

[In]

Integrate[F^(a + b/(c + d*x))/(c + d*x)^5,x]

[Out]

(F^(a + b/(c + d*x))*(6*(c + d*x)^3 - 6*b*(c + d*x)^2*Log[F] + 3*b^2*(c + d*x)*Log[F]^2 - b^3*Log[F]^3))/(b^4*
d*(c + d*x)^3*Log[F]^4)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {\left (\ln \left (F \right )^{3} b^{3}-3 \ln \left (F \right )^{2} b^{2} d x +6 \ln \left (F \right ) b \,d^{2} x^{2}-6 d^{3} x^{3}-3 \ln \left (F \right )^{2} b^{2} c +12 \ln \left (F \right ) b c d x -18 c \,d^{2} x^{2}+6 \ln \left (F \right ) b \,c^{2}-18 c^{2} d x -6 c^{3}\right ) F^{\frac {x a d +c a +b}{d x +c}}}{b^{4} \ln \left (F \right )^{4} d \left (d x +c \right )^{3}}\)	\(125\)
norman	\(\frac {-\frac {\left (\ln \left (F \right )^{3} b^{3}-6 \ln \left (F \right )^{2} b^{2} c +18 \ln \left (F \right ) b \,c^{2}-24 c^{3}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4}}+\frac {6 d^{3} x^{4} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4}}+\frac {3 d \left (\ln \left (F \right )^{2} b^{2}-6 c b \ln \left (F \right )+12 c^{2}\right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4}}-\frac {6 d^{2} \left (b \ln \left (F \right )-4 c \right ) x^{3} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4}}-\frac {\left (\ln \left (F \right )^{3} b^{3}-3 \ln \left (F \right )^{2} b^{2} c +6 \ln \left (F \right ) b \,c^{2}-6 c^{3}\right ) c \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{b^{4} \ln \left (F \right )^{4} d}}{\left (d x +c \right )^{4}}\)	\(243\)
parallelrisch	\(\frac {-\ln \left (F \right )^{3} F^{a +\frac {b}{d x +c}} b^{3} d^{6}+3 \ln \left (F \right )^{2} x \,F^{a +\frac {b}{d x +c}} b^{2} d^{7}-6 \ln \left (F \right ) x^{2} F^{a +\frac {b}{d x +c}} b \,d^{8}+6 x^{3} F^{a +\frac {b}{d x +c}} d^{9}+3 \ln \left (F \right )^{2} F^{a +\frac {b}{d x +c}} b^{2} c \,d^{6}-12 \ln \left (F \right ) x \,F^{a +\frac {b}{d x +c}} b c \,d^{7}+18 x^{2} F^{a +\frac {b}{d x +c}} c \,d^{8}-6 \ln \left (F \right ) F^{a +\frac {b}{d x +c}} b \,c^{2} d^{6}+18 x \,F^{a +\frac {b}{d x +c}} c^{2} d^{7}+6 F^{a +\frac {b}{d x +c}} c^{3} d^{6}}{\left (d x +c \right )^{3} d^{7} \ln \left (F \right )^{4} b^{4}}\)	\(254\)

[In]

int(F^(a+b/(d*x+c))/(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

-(ln(F)^3*b^3-3*ln(F)^2*b^2*d*x+6*ln(F)*b*d^2*x^2-6*d^3*x^3-3*ln(F)^2*b^2*c+12*ln(F)*b*c*d*x-18*c*d^2*x^2+6*ln
(F)*b*c^2-18*c^2*d*x-6*c^3)/b^4/ln(F)^4/d/(d*x+c)^3*F^((a*d*x+a*c+b)/(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.23 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {{\left (6 \, d^{3} x^{3} - b^{3} \log \left (F\right )^{3} + 18 \, c d^{2} x^{2} + 18 \, c^{2} d x + 6 \, c^{3} + 3 \, {\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} - 6 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{{\left (b^{4} d^{4} x^{3} + 3 \, b^{4} c d^{3} x^{2} + 3 \, b^{4} c^{2} d^{2} x + b^{4} c^{3} d\right )} \log \left (F\right )^{4}} \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^5,x, algorithm="fricas")

[Out]

(6*d^3*x^3 - b^3*log(F)^3 + 18*c*d^2*x^2 + 18*c^2*d*x + 6*c^3 + 3*(b^2*d*x + b^2*c)*log(F)^2 - 6*(b*d^2*x^2 +
2*b*c*d*x + b*c^2)*log(F))*F^((a*d*x + a*c + b)/(d*x + c))/((b^4*d^4*x^3 + 3*b^4*c*d^3*x^2 + 3*b^4*c^2*d^2*x +
b^4*c^3*d)*log(F)^4)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.45 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {F^{a + \frac {b}{c + d x}} \left (- b^{3} \log {\left (F \right )}^{3} + 3 b^{2} c \log {\left (F \right )}^{2} + 3 b^{2} d x \log {\left (F \right )}^{2} - 6 b c^{2} \log {\left (F \right )} - 12 b c d x \log {\left (F \right )} - 6 b d^{2} x^{2} \log {\left (F \right )} + 6 c^{3} + 18 c^{2} d x + 18 c d^{2} x^{2} + 6 d^{3} x^{3}\right )}{b^{4} c^{3} d \log {\left (F \right )}^{4} + 3 b^{4} c^{2} d^{2} x \log {\left (F \right )}^{4} + 3 b^{4} c d^{3} x^{2} \log {\left (F \right )}^{4} + b^{4} d^{4} x^{3} \log {\left (F \right )}^{4}} \]

[In]

integrate(F**(a+b/(d*x+c))/(d*x+c)**5,x)

[Out]

F**(a + b/(c + d*x))*(-b**3*log(F)**3 + 3*b**2*c*log(F)**2 + 3*b**2*d*x*log(F)**2 - 6*b*c**2*log(F) - 12*b*c*d
*x*log(F) - 6*b*d**2*x**2*log(F) + 6*c**3 + 18*c**2*d*x + 18*c*d**2*x**2 + 6*d**3*x**3)/(b**4*c**3*d*log(F)**4
+ 3*b**4*c**2*d**2*x*log(F)**4 + 3*b**4*c*d**3*x**2*log(F)**4 + b**4*d**4*x**3*log(F)**4)

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{5}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^5, x)

Giac [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{5}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^5,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^5, x)

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.32 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^5} \, dx=\frac {F^{a+\frac {b}{c+d\,x}}\,\left (\frac {6\,x^3}{b^4\,d\,{\ln \left (F\right )}^4}-\frac {b^3\,{\ln \left (F\right )}^3-3\,b^2\,c\,{\ln \left (F\right )}^2+6\,b\,c^2\,\ln \left (F\right )-6\,c^3}{b^4\,d^4\,{\ln \left (F\right )}^4}+\frac {x^2\,\left (18\,c-6\,b\,\ln \left (F\right )\right )}{b^4\,d^2\,{\ln \left (F\right )}^4}+\frac {3\,x\,\left (b^2\,{\ln \left (F\right )}^2-4\,b\,c\,\ln \left (F\right )+6\,c^2\right )}{b^4\,d^3\,{\ln \left (F\right )}^4}\right )}{x^3+\frac {c^3}{d^3}+\frac {3\,c\,x^2}{d}+\frac {3\,c^2\,x}{d^2}} \]

[In]

int(F^(a + b/(c + d*x))/(c + d*x)^5,x)

[Out]

(F^(a + b/(c + d*x))*((6*x^3)/(b^4*d*log(F)^4) - (b^3*log(F)^3 - 6*c^3 + 6*b*c^2*log(F) - 3*b^2*c*log(F)^2)/(b
^4*d^4*log(F)^4) + (x^2*(18*c - 6*b*log(F)))/(b^4*d^2*log(F)^4) + (3*x*(b^2*log(F)^2 + 6*c^2 - 4*b*c*log(F)))/
(b^4*d^3*log(F)^4)))/(x^3 + c^3/d^3 + (3*c*x^2)/d + (3*c^2*x)/d^2)