3.4.10 \(\int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^4} \, dx\) [310]

Optimal. Leaf size=90 \[ -\frac {2 F^{a+\frac {b}{c+d x}}}{b^3 d \log ^3(F)}+\frac {2 F^{a+\frac {b}{c+d x}}}{b^2 d (c+d x) \log ^2(F)}-\frac {F^{a+\frac {b}{c+d x}}}{b d (c+d x)^2 \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2240} \begin {gather*} -\frac {2 F^{a+\frac {b}{c+d x}}}{b^3 d \log ^3(F)}+\frac {2 F^{a+\frac {b}{c+d x}}}{b^2 d \log ^2(F) (c+d x)}-\frac {F^{a+\frac {b}{c+d x}}}{b d \log (F) (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x))/(c + d*x)^4,x]

[Out]

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

Rule 2243

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

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Mathematica [A]
time = 0.02, size = 60, normalized size = 0.67 \begin {gather*} -\frac {F^{a+\frac {b}{c+d x}} \left (2 (c+d x)^2-2 b (c+d x) \log (F)+b^2 \log ^2(F)\right )}{b^3 d (c+d x)^2 \log ^3(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x))/(c + d*x)^4,x]

[Out]

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

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Maple [A]
time = 0.07, size = 79, normalized size = 0.88

method	result	size
risch	\(-\frac {\left (\ln \left (F \right )^{2} b^{2}-2 b d x \ln \left (F \right )+2 d^{2} x^{2}-2 c b \ln \left (F \right )+4 c d x +2 c^{2}\right ) F^{\frac {x a d +c a +b}{d x +c}}}{b^{3} \ln \left (F \right )^{3} d \left (d x +c \right )^{2}}\)	\(79\)
norman	\(\frac {-\frac {2 d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (\ln \left (F \right )^{2} b^{2}-4 c b \ln \left (F \right )+6 c^{2}\right ) x \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}+\frac {2 d \left (b \ln \left (F \right )-3 c \right ) x^{2} {\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3}}-\frac {\left (\ln \left (F \right )^{2} b^{2}-2 c b \ln \left (F \right )+2 c^{2}\right ) c \,{\mathrm e}^{\left (a +\frac {b}{d x +c}\right ) \ln \left (F \right )}}{b^{3} \ln \left (F \right )^{3} d}}{\left (d x +c \right )^{3}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c))/(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^4, x)

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Fricas [A]
time = 0.39, size = 95, normalized size = 1.06 \begin {gather*} -\frac {{\left (2 \, d^{2} x^{2} + b^{2} \log \left (F\right )^{2} + 4 \, c d x + 2 \, c^{2} - 2 \, {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{{\left (b^{3} d^{3} x^{2} + 2 \, b^{3} c d^{2} x + b^{3} c^{2} d\right )} \log \left (F\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.10, size = 102, normalized size = 1.13 \begin {gather*} \frac {F^{a + \frac {b}{c + d x}} \left (- b^{2} \log {\left (F \right )}^{2} + 2 b c \log {\left (F \right )} + 2 b d x \log {\left (F \right )} - 2 c^{2} - 4 c d x - 2 d^{2} x^{2}\right )}{b^{3} c^{2} d \log {\left (F \right )}^{3} + 2 b^{3} c d^{2} x \log {\left (F \right )}^{3} + b^{3} d^{3} x^{2} \log {\left (F \right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c))/(d*x+c)**4,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c))/(d*x+c)^4,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c))/(d*x + c)^4, x)

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Mupad [B]
time = 5.62, size = 104, normalized size = 1.16 \begin {gather*} -\frac {F^{a+\frac {b}{c+d\,x}}\,\left (\frac {b^2\,{\ln \left (F\right )}^2-2\,b\,c\,\ln \left (F\right )+2\,c^2}{b^3\,d^3\,{\ln \left (F\right )}^3}+\frac {2\,x^2}{b^3\,d\,{\ln \left (F\right )}^3}+\frac {2\,x\,\left (2\,c-b\,\ln \left (F\right )\right )}{b^3\,d^2\,{\ln \left (F\right )}^3}\right )}{x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x))/(c + d*x)^4,x)

[Out]

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

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