3.322 \(\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^5} \, dx\)

Optimal. Leaf size=62 \[ \frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0866581, antiderivative size = 62, normalized size of antiderivative = 1., 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 = {2212, 2209} \[ \frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^5,x]

[Out]

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

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

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)^5} \, dx &=-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^2 \log (F)}-\frac{\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^3} \, dx}{b \log (F)}\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b^2 d \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^2 \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0248331, size = 47, normalized size = 0.76 \[ \frac{F^{a+\frac{b}{(c+d x)^2}} \left ((c+d x)^2-b \log (F)\right )}{2 b^2 d \log ^2(F) (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^5,x]

[Out]

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

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Maple [B]  time = 0.037, size = 185, normalized size = 3. \begin{align*}{\frac{1}{ \left ( dx+c \right ) ^{4}} \left ({\frac{{d}^{3}{x}^{4}}{2\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}{{\rm e}^{ \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) \ln \left ( F \right ) }}}-{\frac{c \left ( b\ln \left ( F \right ) -2\,{c}^{2} \right ) x}{ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}{{\rm e}^{ \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) \ln \left ( F \right ) }}}-{\frac{{c}^{2} \left ( b\ln \left ( F \right ) -{c}^{2} \right ) }{2\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d}{{\rm e}^{ \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) \ln \left ( F \right ) }}}-{\frac{d \left ( b\ln \left ( F \right ) -6\,{c}^{2} \right ){x}^{2}}{2\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}{{\rm e}^{ \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) \ln \left ( F \right ) }}}+2\,{\frac{c{d}^{2}{x}^{3}}{ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}{{\rm e}^{ \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) \ln \left ( F \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x)

[Out]

(1/2/ln(F)^2/b^2*d^3*x^4*exp((a+b/(d*x+c)^2)*ln(F))-c*(b*ln(F)-2*c^2)/ln(F)^2/b^2*x*exp((a+b/(d*x+c)^2)*ln(F))
-1/2*c^2*(b*ln(F)-c^2)/d/ln(F)^2/b^2*exp((a+b/(d*x+c)^2)*ln(F))-1/2*d*(b*ln(F)-6*c^2)/ln(F)^2/b^2*x^2*exp((a+b
/(d*x+c)^2)*ln(F))+2*d^2*c/ln(F)^2/b^2*x^3*exp((a+b/(d*x+c)^2)*ln(F)))/(d*x+c)^4

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Maxima [A]  time = 1.07444, size = 136, normalized size = 2.19 \begin{align*} \frac{{\left (F^{a} d^{2} x^{2} + 2 \, F^{a} c d x + F^{a} c^{2} - F^{a} b \log \left (F\right )\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \,{\left (b^{2} d^{3} x^{2} \log \left (F\right )^{2} + 2 \, b^{2} c d^{2} x \log \left (F\right )^{2} + b^{2} c^{2} d \log \left (F\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="maxima")

[Out]

1/2*(F^a*d^2*x^2 + 2*F^a*c*d*x + F^a*c^2 - F^a*b*log(F))*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(b^2*d^3*x^2*log(F)^2
+ 2*b^2*c*d^2*x*log(F)^2 + b^2*c^2*d*log(F)^2)

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Fricas [A]  time = 1.54418, size = 217, normalized size = 3.5 \begin{align*} \frac{{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - b \log \left (F\right )\right )} 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 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="fricas")

[Out]

1/2*(d^2*x^2 + 2*c*d*x + c^2 - b*log(F))*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
^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*log(F)^2)

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Sympy [A]  time = 0.269897, size = 82, normalized size = 1.32 \begin{align*} \frac{F^{a + \frac{b}{\left (c + d x\right )^{2}}} \left (- b \log{\left (F \right )} + c^{2} + 2 c d x + d^{2} x^{2}\right )}{2 b^{2} c^{2} d \log{\left (F \right )}^{2} + 4 b^{2} c d^{2} x \log{\left (F \right )}^{2} + 2 b^{2} d^{3} x^{2} \log{\left (F \right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**5,x)

[Out]

F**(a + b/(c + d*x)**2)*(-b*log(F) + c**2 + 2*c*d*x + d**2*x**2)/(2*b**2*c**2*d*log(F)**2 + 4*b**2*c*d**2*x*lo
g(F)**2 + 2*b**2*d**3*x**2*log(F)**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^5,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^5, x)