\(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx\) [321]

Optimal result

Integrand size = 21, antiderivative size = 27 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2240} \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

[In]

Int[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])

Rule 2240

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*} \text {integral}& = -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F)} \]

[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])

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(-\frac {F^{a +\frac {b}{\left (d x +c \right )^{2}}}}{2 b d \ln \left (F \right )}\)	\(26\)
default	\(-\frac {F^{a +\frac {b}{\left (d x +c \right )^{2}}}}{2 b d \ln \left (F \right )}\)	\(26\)
parallelrisch	\(-\frac {F^{a +\frac {b}{\left (d x +c \right )^{2}}}}{2 b d \ln \left (F \right )}\)	\(26\)
risch	\(-\frac {F^{\frac {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}{\left (d x +c \right )^{2}}}}{2 b d \ln \left (F \right )}\)	\(44\)
norman	\(\frac {-\frac {c^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 \ln \left (F \right ) b d}-\frac {c x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}-\frac {d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \left (F \right )}}{2 \ln \left (F \right ) b}}{\left (d x +c \right )^{2}}\)	\(94\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, b d \log \left (F\right )} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=\begin {cases} - \frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}}}{2 b d \log {\left (F \right )}} & \text {for}\: b d \log {\left (F \right )} \neq 0 \\- \frac {1}{2 c^{2} d + 4 c d^{2} x + 2 d^{3} x^{2}} & \text {otherwise} \end {cases} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{2 \, b d \log \left (F\right )} \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, b d \log \left (F\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^3} \, dx=-\frac {F^a\,F^{\frac {b}{c^2+2\,c\,d\,x+d^2\,x^2}}}{2\,b\,d\,\ln \left (F\right )} \]

[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))