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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2209} \[ -\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(F^(a + b/(c + d*x))/(b*d*Log[F]))

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ -\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(F^(a + b/(c + d*x))/(b*d*Log[F]))

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fricas [A]  time = 0.41, size = 31, normalized size = 1.24 \[ -\frac {F^{\frac {a d x + a c + b}{d x + c}}}{b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 31, normalized size = 1.24 \[ -\frac {F^{\frac {a d x + a c + b}{d x + c}}}{b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 25, normalized size = 1.00 \[ -\frac {F^{a + \frac {b}{d x + c}}}{b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-F^(a + b/(d*x + c))/(b*d*log(F))

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mupad [B]  time = 5.04, size = 25, normalized size = 1.00 \[ -\frac {F^{a+\frac {b}{c+d\,x}}}{b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-F^(a + b/(c + d*x))/(b*d*log(F))

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sympy [A]  time = 0.24, size = 34, normalized size = 1.36 \[ \begin {cases} - \frac {F^{a + \frac {b}{c + d x}}}{b d \log {\relax (F )}} & \text {for}\: b d \log {\relax (F )} \neq 0 \\- \frac {1}{c d + d^{2} x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-F**(a + b/(c + d*x))/(b*d*log(F)), Ne(b*d*log(F), 0)), (-1/(c*d + d**2*x), True))

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