3.321 \(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx\)

Optimal. Leaf size=27 \[ -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 27, 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} \[ -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^3,x]

[Out]

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

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fricas [B]  time = 0.40, size = 54, normalized size = 2.00 \[ -\frac {F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 54, normalized size = 2.00 \[ -\frac {F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 26, normalized size = 0.96 \[ -\frac {F^{a +\frac {b}{\left (d x +c \right )^{2}}}}{2 b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.88, size = 25, normalized size = 0.93 \[ -\frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{2 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.54, size = 37, normalized size = 1.37 \[ -\frac {F^a\,F^{\frac {b}{c^2+2\,c\,d\,x+d^2\,x^2}}}{2\,b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^2)/(c + d*x)^3,x)

[Out]

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

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sympy [A]  time = 0.34, size = 54, normalized size = 2.00 \[ \begin {cases} - \frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}}}{2 b d \log {\relax (F )}} & \text {for}\: 2 b d \log {\relax (F )} \neq 0 \\- \frac {1}{2 c^{2} d + 4 c d^{2} x + 2 d^{3} x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**3,x)

[Out]

Piecewise((-F**(a + b/(c + d*x)**2)/(2*b*d*log(F)), Ne(2*b*d*log(F), 0)), (-1/(2*c**2*d + 4*c*d**2*x + 2*d**3*
x**2), True))

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