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

Optimal. Leaf size=20 \[ -\frac{F^{\frac{1}{a+b x+c x^2}}}{\log (F)} \]

[Out]

-(F^(a + b*x + c*x^2)^(-1)/Log[F])

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Rubi [A]  time = 0.166147, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.03, Rules used = {6706} \[ -\frac{F^{\frac{1}{a+b x+c x^2}}}{\log (F)} \]

Antiderivative was successfully verified.

[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 6706

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*} \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)}\\ \end{align*}

Mathematica [A]  time = 0.384291, size = 19, normalized size = 0.95 \[ -\frac{F^{\frac{1}{a+x (b+c x)}}}{\log (F)} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.042, size = 21, normalized size = 1.1 \begin{align*} -{\frac{{F}^{ \left ( c{x}^{2}+bx+a \right ) ^{-1}}}{\ln \left ( F \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.977901, size = 27, normalized size = 1.35 \begin{align*} -\frac{F^{\left (\frac{1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 0.779014, size = 45, normalized size = 2.25 \begin{align*} -\frac{F^{\left (\frac{1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.679001, size = 32, normalized size = 1.6 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.30072, size = 27, normalized size = 1.35 \begin{align*} -\frac{F^{\left (\frac{1}{c x^{2} + b x + a}\right )}}{\log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)