3.450 \(\int f^{a+b x+c x^2} (b+2 c x)^3 \, dx\)

Optimal. Leaf size=45 \[ \frac{(b+2 c x)^2 f^{a+b x+c x^2}}{\log (f)}-\frac{4 c f^{a+b x+c x^2}}{\log ^2(f)} \]

[Out]

(-4*c*f^(a + b*x + c*x^2))/Log[f]^2 + (f^(a + b*x + c*x^2)*(b + 2*c*x)^2)/Log[f]

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Rubi [A]  time = 0.0556069, antiderivative size = 45, 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 = {2237, 2236} \[ \frac{(b+2 c x)^2 f^{a+b x+c x^2}}{\log (f)}-\frac{4 c f^{a+b x+c x^2}}{\log ^2(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*c*f^(a + b*x + c*x^2))/Log[f]^2 + (f^(a + b*x + c*x^2)*(b + 2*c*x)^2)/Log[f]

Rule 2237

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

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin{align*} \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx &=\frac{f^{a+b x+c x^2} (b+2 c x)^2}{\log (f)}-\frac{(4 c) \int f^{a+b x+c x^2} (b+2 c x) \, dx}{\log (f)}\\ &=-\frac{4 c f^{a+b x+c x^2}}{\log ^2(f)}+\frac{f^{a+b x+c x^2} (b+2 c x)^2}{\log (f)}\\ \end{align*}

Mathematica [A]  time = 0.124839, size = 31, normalized size = 0.69 \[ \frac{f^{a+x (b+c x)} \left (\log (f) (b+2 c x)^2-4 c\right )}{\log ^2(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a + x*(b + c*x))*(-4*c + (b + 2*c*x)^2*Log[f]))/Log[f]^2

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Maple [A]  time = 0.004, size = 45, normalized size = 1. \begin{align*}{\frac{ \left ( 4\,\ln \left ( f \right ){c}^{2}{x}^{2}+4\,bcx\ln \left ( f \right ) +\ln \left ( f \right ){b}^{2}-4\,c \right ){f}^{c{x}^{2}+bx+a}}{ \left ( \ln \left ( f \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*ln(f)*c^2*x^2+4*b*c*x*ln(f)+ln(f)*b^2-4*c)*f^(c*x^2+b*x+a)/ln(f)^2

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Maxima [C]  time = 1.32112, size = 728, normalized size = 16.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*(sqrt(pi)*(2*c*x + b)*b*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^2/(sqrt(-(2*c*x + b)^2*log(f)
/c)*(c*log(f))^(3/2)) - 2*c*f^(1/4*(2*c*x + b)^2/c)*log(f)/(c*log(f))^(3/2))*b^2*c*f^(a - 1/4*b^2/c)/sqrt(c*lo
g(f)) + 3/2*(sqrt(pi)*(2*c*x + b)*b^2*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^3/(sqrt(-(2*c*x + b)
^2*log(f)/c)*(c*log(f))^(5/2)) - 4*(2*c*x + b)^3*gamma(3/2, -1/4*(2*c*x + b)^2*log(f)/c)*log(f)^3/((-(2*c*x +
b)^2*log(f)/c)^(3/2)*(c*log(f))^(5/2)) - 4*b*c*f^(1/4*(2*c*x + b)^2/c)*log(f)^2/(c*log(f))^(5/2))*b*c^2*f^(a -
1/4*b^2/c)/sqrt(c*log(f)) - 1/2*(sqrt(pi)*(2*c*x + b)*b^3*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)
^4/(sqrt(-(2*c*x + b)^2*log(f)/c)*(c*log(f))^(7/2)) - 12*(2*c*x + b)^3*b*gamma(3/2, -1/4*(2*c*x + b)^2*log(f)/
c)*log(f)^4/((-(2*c*x + b)^2*log(f)/c)^(3/2)*(c*log(f))^(7/2)) - 6*b^2*c*f^(1/4*(2*c*x + b)^2/c)*log(f)^3/(c*l
og(f))^(7/2) + 8*c^2*gamma(2, -1/4*(2*c*x + b)^2*log(f)/c)*log(f)^2/(c*log(f))^(7/2))*c^3*f^(a - 1/4*b^2/c)/sq
rt(c*log(f)) + 1/2*sqrt(pi)*b^3*f^a*erf(sqrt(-c*log(f))*x - 1/2*b*log(f)/sqrt(-c*log(f)))/(sqrt(-c*log(f))*f^(
1/4*b^2/c))

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Fricas [A]  time = 1.51865, size = 99, normalized size = 2.2 \begin{align*} \frac{{\left ({\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right ) - 4 \, c\right )} f^{c x^{2} + b x + a}}{\log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*c^2*x^2 + 4*b*c*x + b^2)*log(f) - 4*c)*f^(c*x^2 + b*x + a)/log(f)^2

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Sympy [A]  time = 0.152633, size = 85, normalized size = 1.89 \begin{align*} \begin{cases} \frac{f^{a + b x + c x^{2}} \left (b^{2} \log{\left (f \right )} + 4 b c x \log{\left (f \right )} + 4 c^{2} x^{2} \log{\left (f \right )} - 4 c\right )}{\log{\left (f \right )}^{2}} & \text{for}\: \log{\left (f \right )}^{2} \neq 0 \\b^{3} x + 3 b^{2} c x^{2} + 4 b c^{2} x^{3} + 2 c^{3} x^{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(c*x**2+b*x+a)*(2*c*x+b)**3,x)

[Out]

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

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Giac [A]  time = 1.29599, size = 59, normalized size = 1.31 \begin{align*} \frac{{\left (c^{2}{\left (2 \, x + \frac{b}{c}\right )}^{2} \log \left (f\right ) - 4 \, c\right )} e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right )\right )}}{\log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*(2*x + b/c)^2*log(f) - 4*c)*e^(c*x^2*log(f) + b*x*log(f) + a*log(f))/log(f)^2