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

Optimal. Leaf size=57 \[ \frac {F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 57, 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} \[ \frac {F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.72 \[ \frac {F^{a+\frac {b}{c+d x}} (-b \log (F)+c+d x)}{b^2 d \log ^2(F) (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 51, normalized size = 0.89 \[ \frac {{\left (d x - b \log \relax (F) + c\right )} F^{\frac {a d x + a c + b}{d x + c}}}{{\left (b^{2} d^{2} x + b^{2} c d\right )} \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 106, normalized size = 1.86 \[ \frac {\frac {d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (b \ln \relax (F )-2 c \right ) x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (b \ln \relax (F )-c \right ) c \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \relax (F )}}{b^{2} d \ln \relax (F )^{2}}}{\left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{d x + c}}}{{\left (d x + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.23, size = 41, normalized size = 0.72 \[ \frac {F^{a+\frac {b}{c+d\,x}}\,\left (c+d\,x-b\,\ln \relax (F)\right )}{b^2\,d\,{\ln \relax (F)}^2\,\left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.22, size = 44, normalized size = 0.77 \[ \frac {F^{a + \frac {b}{c + d x}} \left (- b \log {\relax (F )} + c + d x\right )}{b^{2} c d \log {\relax (F )}^{2} + b^{2} d^{2} x \log {\relax (F )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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