\(\int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx\) [309]

Optimal result

Integrand size = 21, antiderivative size = 57 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x) \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 57, 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 = {2243, 2240} \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)} \]

[In]

Int[F^(a + b/(c + d*x))/(c + d*x)^3,x]

[Out]

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])

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) \log (F)}-\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^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x) \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {F^{a+\frac {b}{c+d x}} (c+d x-b \log (F))}{b^2 d (c+d x) \log ^2(F)} \]

[In]

Integrate[F^(a + b/(c + d*x))/(c + d*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {\left (b \ln \left (F \right )-d x -c \right ) F^{\frac {x a d +c a +b}{d x +c}}}{d \ln \left (F \right )^{2} b^{2} \left (d x +c \right )}\)	\(51\)
parallelrisch	\(\frac {-\ln \left (F \right ) F^{a +\frac {b}{d x +c}} b \,d^{4}+x \,F^{a +\frac {b}{d x +c}} d^{5}+F^{a +\frac {b}{d x +c}} c \,d^{4}}{\left (d x +c \right ) \ln \left (F \right )^{2} b^{2} d^{5}}\)	\(77\)
norman	\(\frac {\frac {d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}-\frac {\left (b \ln \left (F \right )-2 c \right ) x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}-\frac {c \left (b \ln \left (F \right )-c \right ) {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{d \ln \left (F \right )^{2} b^{2}}}{\left (d x +c \right )^{2}}\)	\(106\)

[In]

int(F^(a+b/(d*x+c))/(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {{\left (d x - b \log \left (F\right ) + c\right )} F^{\frac {a d x + a c + b}{d x + c}}}{{\left (b^{2} d^{2} x + b^{2} c d\right )} \log \left (F\right )^{2}} \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^3,x, algorithm="fricas")

[Out]

(d*x - b*log(F) + c)*F^((a*d*x + a*c + b)/(d*x + c))/((b^2*d^2*x + b^2*c*d)*log(F)^2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {F^{a + \frac {b}{c + d x}} \left (- b \log {\left (F \right )} + c + d x\right )}{b^{2} c d \log {\left (F \right )}^{2} + b^{2} d^{2} x \log {\left (F \right )}^{2}} \]

[In]

integrate(F**(a+b/(d*x+c))/(d*x+c)**3,x)

[Out]

F**(a + b/(c + d*x))*(-b*log(F) + c + d*x)/(b**2*c*d*log(F)**2 + b**2*d**2*x*log(F)**2)

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^3, x)

Giac [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^3,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^3, x)

Mupad [B] (verification not implemented)

Time = 2.71 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^3} \, dx=\frac {F^{a+\frac {b}{c+d\,x}}\,\left (c+d\,x-b\,\ln \left (F\right )\right )}{b^2\,d\,{\ln \left (F\right )}^2\,\left (c+d\,x\right )} \]

[In]

int(F^(a + b/(c + d*x))/(c + d*x)^3,x)

[Out]

(F^(a + b/(c + d*x))*(c + d*x - b*log(F)))/(b^2*d*log(F)^2*(c + d*x))