\(\int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{(a+b x+c x^2)^2} \, dx\) [618]

Optimal result

Integrand size = 33, antiderivative size = 20 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\frac {1}{a+b x+c x^2}}}{\log (F)} \]

[Out]

-F^(1/(c*x^2+b*x+a))/ln(F)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6838} \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\frac {1}{a+b x+c x^2}}}{\log (F)} \]

[In]

Int[(F^(a + b*x + c*x^2)^(-1)*(b + 2*c*x))/(a + b*x + c*x^2)^2,x]

[Out]

-(F^(a + b*x + c*x^2)^(-1)/Log[F])

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {F^{\frac {1}{a+b x+c x^2}}}{\log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\frac {1}{a+x (b+c x)}}}{\log (F)} \]

[In]

Integrate[(F^(a + b*x + c*x^2)^(-1)*(b + 2*c*x))/(a + b*x + c*x^2)^2,x]

[Out]

-(F^(a + x*(b + c*x))^(-1)/Log[F])

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {F^{\frac {1}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}\)	\(21\)
default	\(-\frac {F^{\frac {1}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}\)	\(21\)
risch	\(-\frac {F^{\frac {1}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}\)	\(21\)
parallelrisch	\(-\frac {F^{\frac {1}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}\)	\(21\)
norman	\(\frac {-\frac {a \,{\mathrm e}^{\frac {\ln \left (F \right )}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}-\frac {b x \,{\mathrm e}^{\frac {\ln \left (F \right )}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}-\frac {c \,x^{2} {\mathrm e}^{\frac {\ln \left (F \right )}{c \,x^{2}+b x +a}}}{\ln \left (F \right )}}{c \,x^{2}+b x +a}\)	\(88\)

[In]

int(F^(1/(c*x^2+b*x+a))*(2*c*x+b)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-F^(1/(c*x^2+b*x+a))/ln(F)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \]

[In]

integrate(F^(1/(c*x^2+b*x+a))*(2*c*x+b)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-F^(1/(c*x^2 + b*x + a))/log(F)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=\begin {cases} - \frac {F^{\frac {1}{a + b x + c x^{2}}}}{\log {\left (F \right )}} & \text {for}\: \log {\left (F \right )} \neq 0 \\- \frac {1}{a + b x + c x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(1/(c*x**2+b*x+a))*(2*c*x+b)/(c*x**2+b*x+a)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \]

[In]

integrate(F^(1/(c*x^2+b*x+a))*(2*c*x+b)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

-F^(1/(c*x^2 + b*x + a))/log(F)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \]

[In]

integrate(F^(1/(c*x^2+b*x+a))*(2*c*x+b)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

-F^(1/(c*x^2 + b*x + a))/log(F)

Mupad [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{\frac {1}{a+b x+c x^2}} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {F^{\frac {1}{c\,x^2+b\,x+a}}}{\ln \left (F\right )} \]

[In]

int((F^(1/(a + b*x + c*x^2))*(b + 2*c*x))/(a + b*x + c*x^2)^2,x)

[Out]

-F^(1/(a + b*x + c*x^2))/log(F)