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]

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

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Rubi [A]  time = 0.09, antiderivative size = 62, 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 = {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 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]

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])

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.41, size = 100, normalized size = 1.61 \[ \frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - b \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \relax (F)^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{5}}\,{d x} \]

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)

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maple [B]  time = 0.05, size = 185, normalized size = 2.98 \[ \frac {\frac {d^{3} x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )}}{2 b^{2} \ln \relax (F )^{2}}+\frac {2 c \,d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (b \ln \relax (F )-6 c^{2}\right ) d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )}}{2 b^{2} \ln \relax (F )^{2}}-\frac {\left (b \ln \relax (F )-2 c^{2}\right ) c x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (b \ln \relax (F )-c^{2}\right ) c^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )}}{2 b^{2} d \ln \relax (F )^{2}}}{\left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^2*b)/(d*x+c)^5,x)

[Out]

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

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maxima [A]  time = 0.98, size = 101, normalized size = 1.63 \[ \frac {{\left (F^{a} d^{2} x^{2} + 2 \, F^{a} c d x + F^{a} c^{2} - F^{a} b \log \relax (F)\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, {\left (b^{2} d^{3} x^{2} \log \relax (F)^{2} + 2 \, b^{2} c d^{2} x \log \relax (F)^{2} + b^{2} c^{2} d \log \relax (F)^{2}\right )}} \]

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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mupad [B]  time = 3.71, size = 97, normalized size = 1.56 \[ \frac {F^a\,F^{\frac {b}{c^2+2\,c\,d\,x+d^2\,x^2}}\,\left (\frac {x^2}{2\,b^2\,d\,{\ln \relax (F)}^2}-\frac {b\,\ln \relax (F)-c^2}{2\,b^2\,d^3\,{\ln \relax (F)}^2}+\frac {c\,x}{b^2\,d^2\,{\ln \relax (F)}^2}\right )}{x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.31, size = 82, normalized size = 1.32 \[ \frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (- b \log {\relax (F )} + c^{2} + 2 c d x + d^{2} x^{2}\right )}{2 b^{2} c^{2} d \log {\relax (F )}^{2} + 4 b^{2} c d^{2} x \log {\relax (F )}^{2} + 2 b^{2} d^{3} x^{2} \log {\relax (F )}^{2}} \]

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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