3.1.86 \(\int \frac {F^{c+d x}}{(a+b F^{c+d x})^2 x} \, dx\) [86]

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

[Out]

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

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Rubi [A]
time = 0.09, 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 = {} \begin {gather*} \int \frac {F^{c+d x}}{\left (a+b F^{c+d x}\right )^2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-(1/(b*d*(a + b*F^(c + d*x))*x*Log[F])) - Defer[Int][1/((a + b*F^(c + d*x))*x^2), 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} \, dx &=-\frac {1}{b d \left (a+b F^{c+d x}\right ) x \log (F)}-\frac {\int \frac {1}{\left (a+b F^{c+d x}\right ) x^2} \, dx}{b d \log (F)}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{c+d x}}{\left (a+b F^{c+d x}\right )^2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {F^{d x +c}}{\left (a +b \,F^{d x +c}\right )^{2} x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {1}{F^{c + d x} b^{2} d x \log {\left (F \right )} + a b d x \log {\left (F \right )}} - \frac {\int \frac {1}{a x^{2} + b x^{2} e^{c \log {\left (F \right )}} e^{d x \log {\left (F \right )}}}\, dx}{b d \log {\left (F \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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