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

Optimal. Leaf size=25 \[ -\frac {1}{b d \log (F) \left (a+b F^{c+d x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2246, 32} \[ -\frac {1}{b d \log (F) \left (a+b F^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*d*(a + b*F^(c + d*x))*Log[F]))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2246

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.00 \[ -\frac {1}{b d \log (F) \left (a+b F^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*d*(a + b*F^(c + d*x))*Log[F]))

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fricas [A]  time = 0.43, size = 25, normalized size = 1.00 \[ -\frac {1}{F^{d x + c} b^{2} d \log \relax (F) + a b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 26, normalized size = 1.04 \[ -\frac {1}{{\left (F^{d x} F^{c} b + a\right )} b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((F^(d*x)*F^c*b + a)*b*d*log(F))

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maple [A]  time = 0.00, size = 26, normalized size = 1.04 \[ -\frac {1}{\left (b \,F^{d x +c}+a \right ) b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 25, normalized size = 1.00 \[ -\frac {1}{{\left (F^{d x + c} b + a\right )} b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((F^(d*x + c)*b + a)*b*d*log(F))

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mupad [B]  time = 3.46, size = 31, normalized size = 1.24 \[ \frac {F^{c+d\,x}}{a^2\,d\,\ln \relax (F)+F^{c+d\,x}\,a\,b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.12, size = 26, normalized size = 1.04 \[ - \frac {1}{F^{c + d x} b^{2} d \log {\relax (F )} + a b d \log {\relax (F )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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