3.259 \(\int F^{a+b (c+d x)^2} (c+d x)^3 \, dx\)

Optimal. Leaf size=62 \[ \frac{(c+d x)^2 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac{F^{a+b (c+d x)^2}}{2 b^2 d \log ^2(F)} \]

[Out]

-F^(a + b*(c + d*x)^2)/(2*b^2*d*Log[F]^2) + (F^(a + b*(c + d*x)^2)*(c + d*x)^2)/(2*b*d*Log[F])

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Rubi [A]  time = 0.104733, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2212, 2209} \[ \frac{(c+d x)^2 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac{F^{a+b (c+d x)^2}}{2 b^2 d \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)*(c + d*x)^3,x]

[Out]

-F^(a + b*(c + d*x)^2)/(2*b^2*d*Log[F]^2) + (F^(a + b*(c + d*x)^2)*(c + d*x)^2)/(2*b*d*Log[F])

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

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)^3 \, dx &=\frac{F^{a+b (c+d x)^2} (c+d x)^2}{2 b d \log (F)}-\frac{\int F^{a+b (c+d x)^2} (c+d x) \, dx}{b \log (F)}\\ &=-\frac{F^{a+b (c+d x)^2}}{2 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^2} (c+d x)^2}{2 b d \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0218805, size = 40, normalized size = 0.65 \[ \frac{F^{a+b (c+d x)^2} \left (b \log (F) (c+d x)^2-1\right )}{2 b^2 d \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^3,x]

[Out]

(F^(a + b*(c + d*x)^2)*(-1 + b*(c + d*x)^2*Log[F]))/(2*b^2*d*Log[F]^2)

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Maple [A]  time = 0.007, size = 63, normalized size = 1. \begin{align*}{\frac{ \left ( \ln \left ( F \right ) b{d}^{2}{x}^{2}+2\,\ln \left ( F \right ) bcdx+\ln \left ( F \right ) b{c}^{2}-1 \right ){F}^{b{d}^{2}{x}^{2}+2\,bcdx+{c}^{2}b+a}}{2\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)*(d*x+c)^3,x)

[Out]

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

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Maxima [C]  time = 1.68766, size = 965, normalized size = 15.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^3,x, algorithm="maxima")

[Out]

-3/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*d*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^2/((b*d^2*l
og(F))^(3/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*d^2*log(F)/(b*d^2*
log(F))^(3/2))*F^a*c^2*d/sqrt(b*d^2*log(F)) + 3/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*d^2*(erf(sqrt(-(b*d^2*x
+ b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^3/((b*d^2*log(F))^(5/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2
*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^2*c*d^3*log(F)^2/(b*d^2*log(F))^(5/2) - (b*d^2*x + b*c*d)^3*gamma(3/2, -(b*
d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*d^2*log(F))^(5/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*
F^a*c*d^2/sqrt(b*d^2*log(F)) - 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*d^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(
F)/(b*d^2))) - 1)*log(F)^4/((b*d^2*log(F))^(7/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 3*F^((b*d^2*x +
b*c*d)^2/(b*d^2))*b^3*c^2*d^4*log(F)^3/(b*d^2*log(F))^(7/2) - 3*(b*d^2*x + b*c*d)^3*b*c*d*gamma(3/2, -(b*d^2*x
+ b*c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*d^2*log(F))^(7/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + b^2*
d^4*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^2/(b*d^2*log(F))^(7/2))*F^a*d^3/sqrt(b*d^2*log(F)) +
1/2*sqrt(pi)*F^(b*c^2 + a)*c^3*erf(sqrt(-b*log(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log(F))*F^(b*c^2
)*d)

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Fricas [A]  time = 1.49666, size = 142, normalized size = 2.29 \begin{align*} \frac{{\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) - 1\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b^{2} d \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.171906, size = 100, normalized size = 1.61 \begin{align*} \begin{cases} \frac{F^{a + b \left (c + d x\right )^{2}} \left (b c^{2} \log{\left (F \right )} + 2 b c d x \log{\left (F \right )} + b d^{2} x^{2} \log{\left (F \right )} - 1\right )}{2 b^{2} d \log{\left (F \right )}^{2}} & \text{for}\: 2 b^{2} d \log{\left (F \right )}^{2} \neq 0 \\c^{3} x + \frac{3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac{d^{3} x^{4}}{4} & \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)**3,x)

[Out]

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

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Giac [A]  time = 1.27073, size = 82, normalized size = 1.32 \begin{align*} \frac{{\left (b d^{2}{\left (x + \frac{c}{d}\right )}^{2} \log \left (F\right ) - 1\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{2 \, b^{2} d \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^3,x, algorithm="giac")

[Out]

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