3.3.60 \(\int F^{a+b (c+d x)^2} (c+d x) \, dx\) [260]

Optimal. Leaf size=27 \[ \frac {F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[Out]

1/2*F^(a+b*(d*x+c)^2)/b/d/ln(F)

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Rubi [A]
time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2240} \begin {gather*} \frac {F^{a+b (c+d x)^2}}{2 b d \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)*(c + d*x),x]

[Out]

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

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Mathematica [A]
time = 0.02, size = 27, normalized size = 1.00 \begin {gather*} \frac {F^{a+b (c+d x)^2}}{2 b d \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)*(c + d*x),x]

[Out]

F^(a + b*(c + d*x)^2)/(2*b*d*Log[F])

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Maple [A]
time = 0.01, size = 26, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {F^{a +b \left (d x +c \right )^{2}}}{2 b d \ln \left (F \right )}\)	\(26\)
default	\(\frac {F^{a +b \left (d x +c \right )^{2}}}{2 b d \ln \left (F \right )}\)	\(26\)
norman	\(\frac {{\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{2 b \ln \left (F \right ) d}\)	\(28\)
gosper	\(\frac {F^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a}}{2 b d \ln \left (F \right )}\)	\(36\)
risch	\(\frac {F^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a}}{2 b d \ln \left (F \right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)*(d*x+c),x,method=_RETURNVERBOSE)

[Out]

1/2*F^(a+b*(d*x+c)^2)/b/d/ln(F)

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Maxima [A]
time = 0.28, size = 25, normalized size = 0.93 \begin {gather*} \frac {F^{{\left (d x + c\right )}^{2} b + a}}{2 \, 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)^2)*(d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 35, normalized size = 1.30 \begin {gather*} \frac {F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, 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)^2)*(d*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.05, size = 34, normalized size = 1.26 \begin {gather*} \begin {cases} \frac {F^{a + b \left (c + d x\right )^{2}}}{2 b d \log {\left (F \right )}} & \text {for}\: b d \log {\left (F \right )} \neq 0 \\c x + \frac {d x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)*(d*x+c),x)

[Out]

Piecewise((F**(a + b*(c + d*x)**2)/(2*b*d*log(F)), Ne(b*d*log(F), 0)), (c*x + d*x**2/2, True))

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Giac [A]
time = 2.49, size = 35, normalized size = 1.30 \begin {gather*} \frac {F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, 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)^2)*(d*x+c),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.52, size = 25, normalized size = 0.93 \begin {gather*} \frac {F^{a+b\,{\left (c+d\,x\right )}^2}}{2\,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)^2)*(c + d*x),x)

[Out]

F^(a + b*(c + d*x)^2)/(2*b*d*log(F))

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