3.1.27 \(\int F^{2+5 x} \, dx\) [27]

Optimal. Leaf size=15 \[ \frac {F^{2+5 x}}{5 \log (F)} \]

[Out]

1/5*F^(2+5*x)/ln(F)

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2225} \begin {gather*} \frac {F^{5 x+2}}{5 \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(2 + 5*x),x]

[Out]

F^(2 + 5*x)/(5*Log[F])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {align*} \int F^{2+5 x} \, dx &=\frac {F^{2+5 x}}{5 \log (F)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \frac {F^{2+5 x}}{5 \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(2 + 5*x),x]

[Out]

F^(2 + 5*x)/(5*Log[F])

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Maple [A]
time = 0.01, size = 14, normalized size = 0.93

method	result	size
gosper	\(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\)	\(14\)
derivativedivides	\(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\)	\(14\)
default	\(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\)	\(14\)
risch	\(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\)	\(14\)
norman	\(\frac {{\mathrm e}^{\left (2+5 x \right ) \ln \left (F \right )}}{5 \ln \left (F \right )}\)	\(16\)
meijerg	\(-\frac {F^{2} \left (1-{\mathrm e}^{5 x \ln \left (F \right )}\right )}{5 \ln \left (F \right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(2+5*x),x,method=_RETURNVERBOSE)

[Out]

1/5*F^(2+5*x)/ln(F)

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Maxima [A]
time = 0.28, size = 13, normalized size = 0.87 \begin {gather*} \frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(2+5*x),x, algorithm="maxima")

[Out]

1/5*F^(5*x + 2)/log(F)

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Fricas [A]
time = 0.45, size = 13, normalized size = 0.87 \begin {gather*} \frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(2+5*x),x, algorithm="fricas")

[Out]

1/5*F^(5*x + 2)/log(F)

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Sympy [A]
time = 0.02, size = 14, normalized size = 0.93 \begin {gather*} \begin {cases} \frac {F^{5 x + 2}}{5 \log {\left (F \right )}} & \text {for}\: \log {\left (F \right )} \neq 0 \\x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(2+5*x),x)

[Out]

Piecewise((F**(5*x + 2)/(5*log(F)), Ne(log(F), 0)), (x, True))

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Giac [A]
time = 2.53, size = 13, normalized size = 0.87 \begin {gather*} \frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(2+5*x),x, algorithm="giac")

[Out]

1/5*F^(5*x + 2)/log(F)

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Mupad [B]
time = 3.54, size = 13, normalized size = 0.87 \begin {gather*} \frac {F^{5\,x+2}}{5\,\ln \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(5*x + 2),x)

[Out]

F^(5*x + 2)/(5*log(F))

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