3.260 \(\int F^{a+b (c+d x)^2} (c+d x) \, dx\)

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

[Out]

F^(a + b*(c + d*x)^2)/(2*b*d*Log[F])

________________________________________________________________________________________

Rubi [A]  time = 0.0362265, antiderivative size = 27, normalized size of antiderivative = 1., 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 = {2209} \[ \frac{F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

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 2209

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*}

Mathematica [A]  time = 0.0073765, size = 27, normalized size = 1. \[ \frac{F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 36, normalized size = 1.3 \begin{align*}{\frac{{F}^{b{d}^{2}{x}^{2}+2\,bcdx+{c}^{2}b+a}}{2\,bd\ln \left ( F \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01561, size = 34, normalized size = 1.26 \begin{align*} \frac{F^{{\left (d x + c\right )}^{2} b + a}}{2 \, b d \log \left (F\right )} \end{align*}

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))

________________________________________________________________________________________

Fricas [A]  time = 1.53574, size = 76, normalized size = 2.81 \begin{align*} \frac{F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b d \log \left (F\right )} \end{align*}

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))

________________________________________________________________________________________

Sympy [A]  time = 0.136847, size = 36, normalized size = 1.33 \begin{align*} \begin{cases} \frac{F^{a + b \left (c + d x\right )^{2}}}{2 b d \log{\left (F \right )}} & \text{for}\: 2 b d \log{\left (F \right )} \neq 0 \\c x + \frac{d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.22253, size = 34, normalized size = 1.26 \begin{align*} \frac{F^{{\left (d x + c\right )}^{2} b + a}}{2 \, b d \log \left (F\right )} \end{align*}

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^((d*x + c)^2*b + a)/(b*d*log(F))