\(\int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx\) [308]

Optimal result

Integrand size = 21, antiderivative size = 25 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2240} \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \]

[In]

Int[F^(a + b/(c + d*x))/(c + d*x)^2,x]

[Out]

-(F^(a + b/(c + d*x))/(b*d*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \]

[In]

Integrate[F^(a + b/(c + d*x))/(c + d*x)^2,x]

[Out]

-(F^(a + b/(c + d*x))/(b*d*Log[F]))

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(-\frac {F^{a +\frac {b}{d x +c}}}{b d \ln \left (F \right )}\)	\(26\)
default	\(-\frac {F^{a +\frac {b}{d x +c}}}{b d \ln \left (F \right )}\)	\(26\)
parallelrisch	\(-\frac {F^{a +\frac {b}{d x +c}}}{b d \ln \left (F \right )}\)	\(26\)
risch	\(-\frac {F^{\frac {x a d +c a +b}{d x +c}}}{b d \ln \left (F \right )}\)	\(32\)
norman	\(\frac {-\frac {x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{b \ln \left (F \right )}-\frac {c \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b d}}{d x +c}\)	\(63\)

[In]

int(F^(a+b/(d*x+c))/(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{\frac {a d x + a c + b}{d x + c}}}{b d \log \left (F\right )} \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=\begin {cases} - \frac {F^{a + \frac {b}{c + d x}}}{b d \log {\left (F \right )}} & \text {for}\: b d \log {\left (F \right )} \neq 0 \\- \frac {1}{c d + d^{2} x} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b/(d*x+c))/(d*x+c)**2,x)

[Out]

Piecewise((-F**(a + b/(c + d*x))/(b*d*log(F)), Ne(b*d*log(F), 0)), (-1/(c*d + d**2*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{a + \frac {b}{d x + c}}}{b d \log \left (F\right )} \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^2,x, algorithm="maxima")

[Out]

-F^(a + b/(d*x + c))/(b*d*log(F))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{\frac {a d x + a c + b}{d x + c}}}{b d \log \left (F\right )} \]

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx=-\frac {F^{a+\frac {b}{c+d\,x}}}{b\,d\,\ln \left (F\right )} \]

[In]

int(F^(a + b/(c + d*x))/(c + d*x)^2,x)

[Out]

-F^(a + b/(c + d*x))/(b*d*log(F))