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

Optimal. Leaf size=122 \[ \frac{3 F^{a+\frac{b}{c+d x}}}{b^2 d \log ^2(F) (c+d x)^2}-\frac{6 F^{a+\frac{b}{c+d x}}}{b^3 d \log ^3(F) (c+d x)}+\frac{6 F^{a+\frac{b}{c+d x}}}{b^4 d \log ^4(F)}-\frac{F^{a+\frac{b}{c+d x}}}{b d \log (F) (c+d x)^3} \]

[Out]

(6*F^(a + b/(c + d*x)))/(b^4*d*Log[F]^4) - (6*F^(a + b/(c + d*x)))/(b^3*d*(c + d*x)*Log[F]^3) + (3*F^(a + b/(c
+ d*x)))/(b^2*d*(c + d*x)^2*Log[F]^2) - F^(a + b/(c + d*x))/(b*d*(c + d*x)^3*Log[F])

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Rubi [A]  time = 0.18513, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2212, 2209} \[ \frac{3 F^{a+\frac{b}{c+d x}}}{b^2 d \log ^2(F) (c+d x)^2}-\frac{6 F^{a+\frac{b}{c+d x}}}{b^3 d \log ^3(F) (c+d x)}+\frac{6 F^{a+\frac{b}{c+d x}}}{b^4 d \log ^4(F)}-\frac{F^{a+\frac{b}{c+d x}}}{b d \log (F) (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*F^(a + b/(c + d*x)))/(b^4*d*Log[F]^4) - (6*F^(a + b/(c + d*x)))/(b^3*d*(c + d*x)*Log[F]^3) + (3*F^(a + b/(c
+ d*x)))/(b^2*d*(c + d*x)^2*Log[F]^2) - F^(a + b/(c + d*x))/(b*d*(c + d*x)^3*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}}}{(c+d x)^5} \, dx &=-\frac{F^{a+\frac{b}{c+d x}}}{b d (c+d x)^3 \log (F)}-\frac{3 \int \frac{F^{a+\frac{b}{c+d x}}}{(c+d x)^4} \, dx}{b \log (F)}\\ &=\frac{3 F^{a+\frac{b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac{F^{a+\frac{b}{c+d x}}}{b d (c+d x)^3 \log (F)}+\frac{6 \int \frac{F^{a+\frac{b}{c+d x}}}{(c+d x)^3} \, dx}{b^2 \log ^2(F)}\\ &=-\frac{6 F^{a+\frac{b}{c+d x}}}{b^3 d (c+d x) \log ^3(F)}+\frac{3 F^{a+\frac{b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac{F^{a+\frac{b}{c+d x}}}{b d (c+d x)^3 \log (F)}-\frac{6 \int \frac{F^{a+\frac{b}{c+d x}}}{(c+d x)^2} \, dx}{b^3 \log ^3(F)}\\ &=\frac{6 F^{a+\frac{b}{c+d x}}}{b^4 d \log ^4(F)}-\frac{6 F^{a+\frac{b}{c+d x}}}{b^3 d (c+d x) \log ^3(F)}+\frac{3 F^{a+\frac{b}{c+d x}}}{b^2 d (c+d x)^2 \log ^2(F)}-\frac{F^{a+\frac{b}{c+d x}}}{b d (c+d x)^3 \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0341799, size = 76, normalized size = 0.62 \[ \frac{F^{a+\frac{b}{c+d x}} \left (3 b^2 \log ^2(F) (c+d x)-b^3 \log ^3(F)-6 b \log (F) (c+d x)^2+6 (c+d x)^3\right )}{b^4 d \log ^4(F) (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(a + b/(c + d*x))*(6*(c + d*x)^3 - 6*b*(c + d*x)^2*Log[F] + 3*b^2*(c + d*x)*Log[F]^2 - b^3*Log[F]^3))/(b^4*
d*(c + d*x)^3*Log[F]^4)

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Maple [A]  time = 0.038, size = 243, normalized size = 2. \begin{align*}{\frac{1}{ \left ( dx+c \right ) ^{4}} \left ( -{\frac{ \left ( \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}-6\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}c+18\,\ln \left ( F \right ) b{c}^{2}-24\,{c}^{3} \right ) x}{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{dx+c}} \right ) \ln \left ( F \right ) }}}+6\,{\frac{{d}^{3}{x}^{4}}{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{dx+c}} \right ) \ln \left ( F \right ) }}}+3\,{\frac{d \left ( \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}-6\,bc\ln \left ( F \right ) +12\,{c}^{2} \right ){x}^{2}}{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{dx+c}} \right ) \ln \left ( F \right ) }}}-6\,{\frac{{d}^{2} \left ( b\ln \left ( F \right ) -4\,c \right ){x}^{3}}{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{dx+c}} \right ) \ln \left ( F \right ) }}}-{\frac{ \left ( \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}-3\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}c+6\,\ln \left ( F \right ) b{c}^{2}-6\,{c}^{3} \right ) c}{d{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{dx+c}} \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))/(d*x+c)^5,x)

[Out]

(-(ln(F)^3*b^3-6*ln(F)^2*b^2*c+18*ln(F)*b*c^2-24*c^3)/ln(F)^4/b^4*x*exp((a+b/(d*x+c))*ln(F))+6/ln(F)^4/b^4*d^3
*x^4*exp((a+b/(d*x+c))*ln(F))+3*d*(ln(F)^2*b^2-6*b*c*ln(F)+12*c^2)/ln(F)^4/b^4*x^2*exp((a+b/(d*x+c))*ln(F))-6*
d^2*(b*ln(F)-4*c)/ln(F)^4/b^4*x^3*exp((a+b/(d*x+c))*ln(F))-(ln(F)^3*b^3-3*ln(F)^2*b^2*c+6*ln(F)*b*c^2-6*c^3)*c
/b^4/ln(F)^4/d*exp((a+b/(d*x+c))*ln(F)))/(d*x+c)^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{d x + c}}}{{\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))/(d*x+c)^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57381, size = 328, normalized size = 2.69 \begin{align*} \frac{{\left (6 \, d^{3} x^{3} - b^{3} \log \left (F\right )^{3} + 18 \, c d^{2} x^{2} + 18 \, c^{2} d x + 6 \, c^{3} + 3 \,{\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} - 6 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac{a d x + a c + b}{d x + c}}}{{\left (b^{4} d^{4} x^{3} + 3 \, b^{4} c d^{3} x^{2} + 3 \, b^{4} c^{2} d^{2} x + b^{4} c^{3} d\right )} \log \left (F\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*d^3*x^3 - b^3*log(F)^3 + 18*c*d^2*x^2 + 18*c^2*d*x + 6*c^3 + 3*(b^2*d*x + b^2*c)*log(F)^2 - 6*(b*d^2*x^2 +
2*b*c*d*x + b*c^2)*log(F))*F^((a*d*x + a*c + b)/(d*x + c))/((b^4*d^4*x^3 + 3*b^4*c*d^3*x^2 + 3*b^4*c^2*d^2*x +
b^4*c^3*d)*log(F)^4)

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Sympy [A]  time = 0.275247, size = 177, normalized size = 1.45 \begin{align*} \frac{F^{a + \frac{b}{c + d x}} \left (- b^{3} \log{\left (F \right )}^{3} + 3 b^{2} c \log{\left (F \right )}^{2} + 3 b^{2} d x \log{\left (F \right )}^{2} - 6 b c^{2} \log{\left (F \right )} - 12 b c d x \log{\left (F \right )} - 6 b d^{2} x^{2} \log{\left (F \right )} + 6 c^{3} + 18 c^{2} d x + 18 c d^{2} x^{2} + 6 d^{3} x^{3}\right )}{b^{4} c^{3} d \log{\left (F \right )}^{4} + 3 b^{4} c^{2} d^{2} x \log{\left (F \right )}^{4} + 3 b^{4} c d^{3} x^{2} \log{\left (F \right )}^{4} + b^{4} d^{4} x^{3} \log{\left (F \right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F**(a + b/(c + d*x))*(-b**3*log(F)**3 + 3*b**2*c*log(F)**2 + 3*b**2*d*x*log(F)**2 - 6*b*c**2*log(F) - 12*b*c*d
*x*log(F) - 6*b*d**2*x**2*log(F) + 6*c**3 + 18*c**2*d*x + 18*c*d**2*x**2 + 6*d**3*x**3)/(b**4*c**3*d*log(F)**4
+ 3*b**4*c**2*d**2*x*log(F)**4 + 3*b**4*c*d**3*x**2*log(F)**4 + b**4*d**4*x**3*log(F)**4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{d x + c}}}{{\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))/(d*x+c)^5,x, algorithm="giac")

[Out]

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