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

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Rubi [A]  time = 0.17, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, 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*}

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Mathematica [A]  time = 0.42, 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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fricas [A]  time = 0.41, size = 20, normalized size = 1.00 \[ -\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \relax (F)} \]

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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giac [A]  time = 0.22, size = 20, normalized size = 1.00 \[ -\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \relax (F)} \]

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)

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maple [A]  time = 0.03, size = 21, normalized size = 1.05 \[ -\frac {F^{\frac {1}{c \,x^{2}+b x +a}}}{\ln \relax (F )} \]

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 = 0.95, size = 20, normalized size = 1.00 \[ -\frac {F^{\left (\frac {1}{c x^{2} + b x + a}\right )}}{\log \relax (F)} \]

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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mupad [B]  time = 4.00, size = 20, normalized size = 1.00 \[ -\frac {F^{\frac {1}{c\,x^2+b\,x+a}}}{\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.57, size = 32, normalized size = 1.60 \[ \begin {cases} - \frac {F^{\frac {1}{a + b x + c x^{2}}}}{\log {\relax (F )}} & \text {for}\: \log {\relax (F )} \neq 0 \\- \frac {1}{a + b x + c x^{2}} & \text {otherwise} \end {cases} \]

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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