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

Optimal. Leaf size=61 \[ -\frac {2 \text {Int}\left (\frac {1}{x^3 \left (a+b F^{c+d x}\right )},x\right )}{b d \log (F)}-\frac {1}{b d x^2 \log (F) \left (a+b F^{c+d x}\right )} \]

[Out]

-1/b/d/(a+b*F^(d*x+c))/x^2/ln(F)-2*Unintegrable(1/(a+b*F^(d*x+c))/x^3,x)/b/d/ln(F)

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Rubi [A]  time = 0.12, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {F^{c+d x}}{\left (a+b F^{c+d x}\right )^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

-(1/(b*d*(a + b*F^(c + d*x))*x^2*Log[F])) - (2*Defer[Int][1/((a + b*F^(c + d*x))*x^3), x])/(b*d*Log[F])

Rubi steps

\begin {align*} \int \frac {F^{c+d x}}{\left (a+b F^{c+d x}\right )^2 x^2} \, dx &=-\frac {1}{b d \left (a+b F^{c+d x}\right ) x^2 \log (F)}-\frac {2 \int \frac {1}{\left (a+b F^{c+d x}\right ) x^3} \, dx}{b d \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {F^{c+d x}}{\left (a+b F^{c+d x}\right )^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {F^{d x + c}}{2 \, F^{d x + c} a b x^{2} + F^{2 \, d x + 2 \, c} b^{2} x^{2} + a^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{d x + c}}{{\left (F^{d x + c} b + a\right )}^{2} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {F^{d x +c}}{\left (b \,F^{d x +c}+a \right )^{2} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{F^{d x} F^{c} b^{2} d x^{2} \log \relax (F) + a b d x^{2} \log \relax (F)} - 2 \, \int \frac {1}{F^{d x} F^{c} b^{2} d x^{3} \log \relax (F) + a b d x^{3} \log \relax (F)}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(F^(d*x)*F^c*b^2*d*x^2*log(F) + a*b*d*x^2*log(F)) - 2*integrate(1/(F^(d*x)*F^c*b^2*d*x^3*log(F) + a*b*d*x^3
*log(F)), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {F^{c+d\,x}}{x^2\,{\left (a+F^{c+d\,x}\,b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {1}{F^{c + d x} b^{2} d x^{2} \log {\relax (F )} + a b d x^{2} \log {\relax (F )}} - \frac {2 \int \frac {1}{a x^{3} + b x^{3} e^{c \log {\relax (F )}} e^{d x \log {\relax (F )}}}\, dx}{b d \log {\relax (F )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(F**(c + d*x)*b**2*d*x**2*log(F) + a*b*d*x**2*log(F)) - 2*Integral(1/(a*x**3 + b*x**3*exp(c*log(F))*exp(d*x
*log(F))), x)/(b*d*log(F))

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