3.4.47 \(\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^4} \, dx\) [347]

Optimal. Leaf size=27 \[ -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2240} \begin {gather*} -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^3)/(c + d*x)^4,x]

[Out]

-1/3*F^(a + b/(c + d*x)^3)/(b*d*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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 27, normalized size = 1.00 \begin {gather*} -\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^4,x]

[Out]

-1/3*F^(a + b/(c + d*x)^3)/(b*d*Log[F])

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Maple [A]
time = 0.06, size = 26, normalized size = 0.96

method	result	size
derivativedivides	\(-\frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{3 b d \ln \left (F \right )}\)	\(26\)
default	\(-\frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{3 b d \ln \left (F \right )}\)	\(26\)
risch	\(-\frac {F^{\frac {a \,d^{3} x^{3}+3 a c \,d^{2} x^{2}+3 a \,c^{2} d x +a \,c^{3}+b}{\left (d x +c \right )^{3}}}}{3 b d \ln \left (F \right )}\)	\(56\)
norman	\(\frac {-\frac {c^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 \ln \left (F \right ) b d}-\frac {c^{2} x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}-\frac {d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{3 \ln \left (F \right ) b}-\frac {d c \,x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}}{\left (d x +c \right )^{3}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 25, normalized size = 0.93 \begin {gather*} -\frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{3 \, b d \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (25) = 50\).
time = 0.34, size = 77, normalized size = 2.85 \begin {gather*} -\frac {F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, b d \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
time = 0.17, size = 65, normalized size = 2.41 \begin {gather*} \begin {cases} - \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}}}{3 b d \log {\left (F \right )}} & \text {for}\: b d \log {\left (F \right )} \neq 0 \\- \frac {1}{3 c^{3} d + 9 c^{2} d^{2} x + 9 c d^{3} x^{2} + 3 d^{4} x^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**4,x)

[Out]

Piecewise((-F**(a + b/(c + d*x)**3)/(3*b*d*log(F)), Ne(b*d*log(F), 0)), (-1/(3*c**3*d + 9*c**2*d**2*x + 9*c*d*
*3*x**2 + 3*d**4*x**3), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (25) = 50\).
time = 2.22, size = 77, normalized size = 2.85 \begin {gather*} -\frac {F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, b d \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.64, size = 48, normalized size = 1.78 \begin {gather*} -\frac {F^a\,F^{\frac {b}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3}}}{3\,b\,d\,\ln \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^3)/(c + d*x)^4,x)

[Out]

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

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