3.1.6 \(\int F^{c (a+b x)} \, dx\) [6]

Optimal. Leaf size=20 \[ \frac {F^{c (a+b x)}}{b c \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2225} \begin {gather*} \frac {F^{c (a+b x)}}{b c \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x)),x]

[Out]

F^(c*(a + b*x))/(b*c*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^{c (a+b x)} \, dx &=\frac {F^{c (a+b x)}}{b c \log (F)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 21, normalized size = 1.05 \begin {gather*} \frac {F^{a c+b c x}}{b c \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x)),x]

[Out]

F^(a*c + b*c*x)/(b*c*Log[F])

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Maple [A]
time = 0.02, size = 21, normalized size = 1.05

method	result	size
gosper	\(\frac {F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\)	\(21\)
derivativedivides	\(\frac {F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\)	\(21\)
default	\(\frac {F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\)	\(21\)
risch	\(\frac {F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\)	\(21\)
norman	\(\frac {{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}\)	\(22\)
meijerg	\(-\frac {F^{c a} \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} \frac {F^{{\left (b x + a\right )} c}}{b c \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F^((b*x + a)*c)/(b*c*log(F))

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Fricas [A]
time = 0.43, size = 21, normalized size = 1.05 \begin {gather*} \frac {F^{b c x + a c}}{b c \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F^(b*c*x + a*c)/(b*c*log(F))

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Sympy [A]
time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} \begin {cases} \frac {F^{c \left (a + b x\right )}}{b c \log {\left (F \right )}} & \text {for}\: b c \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**(c*(b*x+a)),x)

[Out]

Piecewise((F**(c*(a + b*x))/(b*c*log(F)), Ne(b*c*log(F), 0)), (x, True))

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Giac [A]
time = 3.43, size = 21, normalized size = 1.05 \begin {gather*} \frac {F^{b c x + a c}}{b c \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F^(b*c*x + a*c)/(b*c*log(F))

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Mupad [B]
time = 3.43, size = 21, normalized size = 1.05 \begin {gather*} \frac {F^{a\,c+b\,c\,x}}{b\,c\,\ln \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x)),x)

[Out]

F^(a*c + b*c*x)/(b*c*log(F))

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