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

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 0.44, size = 41, normalized size = 0.91 \[ \frac {{\left ({\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \relax (f) - 4 \, c\right )} f^{c x^{2} + b x + a}}{\log \relax (f)^{2}} \]

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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giac [B]  time = 0.57, size = 780, normalized size = 17.33 \[ \text {result too large to display} \]

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]

(2*((b^2*log(abs(f)) + 4*(c*x^2 + b*x)*c*log(abs(f)) - 4*c)*(pi^2*sgn(f) - pi^2 + 2*log(abs(f))^2)/((pi^2*sgn(
f) - pi^2 + 2*log(abs(f))^2)^2 + 4*(pi*log(abs(f))*sgn(f) - pi*log(abs(f)))^2) + (pi*b^2*sgn(f) + 4*pi*(c*x^2
+ b*x)*c*sgn(f) - pi*b^2 - 4*pi*(c*x^2 + b*x)*c)*(pi*log(abs(f))*sgn(f) - pi*log(abs(f)))/((pi^2*sgn(f) - pi^2
 + 2*log(abs(f))^2)^2 + 4*(pi*log(abs(f))*sgn(f) - pi*log(abs(f)))^2))*cos(-1/2*pi*c*x^2*sgn(f) + 1/2*pi*c*x^2
 - 1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a) + ((pi*b^2*sgn(f) + 4*pi*(c*x^2 + b*x)*c*sgn(f
) - pi*b^2 - 4*pi*(c*x^2 + b*x)*c)*(pi^2*sgn(f) - pi^2 + 2*log(abs(f))^2)/((pi^2*sgn(f) - pi^2 + 2*log(abs(f))
^2)^2 + 4*(pi*log(abs(f))*sgn(f) - pi*log(abs(f)))^2) - 4*(b^2*log(abs(f)) + 4*(c*x^2 + b*x)*c*log(abs(f)) - 4
*c)*(pi*log(abs(f))*sgn(f) - pi*log(abs(f)))/((pi^2*sgn(f) - pi^2 + 2*log(abs(f))^2)^2 + 4*(pi*log(abs(f))*sgn
(f) - pi*log(abs(f)))^2))*sin(-1/2*pi*c*x^2*sgn(f) + 1/2*pi*c*x^2 - 1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*
sgn(f) + 1/2*pi*a))*e^((c*x^2 + b*x)*log(abs(f)) + a*log(abs(f))) - 1/2*((2*b^2*i*log(abs(f)) + 8*(c*x^2 + b*x
)*c*i*log(abs(f)) - pi*b^2*sgn(f) - 4*pi*(c*x^2 + b*x)*c*sgn(f) + pi*b^2 + 4*pi*(c*x^2 + b*x)*c - 8*c*i)*e^(1/
2*(pi*(c*x^2 + b*x)*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(2*pi*i*log(abs(f))*sgn(f) - 2*pi*i*log(abs(f)) + pi^
2*sgn(f) - pi^2 + 2*log(abs(f))^2) + (2*b^2*i*log(abs(f)) + 8*(c*x^2 + b*x)*c*i*log(abs(f)) + pi*b^2*sgn(f) +
4*pi*(c*x^2 + b*x)*c*sgn(f) - pi*b^2 - 4*pi*(c*x^2 + b*x)*c - 8*c*i)*e^(-1/2*(pi*(c*x^2 + b*x)*(sgn(f) - 1) +
pi*a*(sgn(f) - 1))*i)/(2*pi*i*log(abs(f))*sgn(f) - 2*pi*i*log(abs(f)) - pi^2*sgn(f) + pi^2 - 2*log(abs(f))^2))
*e^((c*x^2 + b*x)*log(abs(f)) + a*log(abs(f)))/i

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maple [A]  time = 0.01, size = 45, normalized size = 1.00 \[ \frac {\left (4 c^{2} x^{2} \ln \relax (f )+4 b c x \ln \relax (f )+b^{2} \ln \relax (f )-4 c \right ) f^{c \,x^{2}+b x +a}}{\ln \relax (f )^{2}} \]

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 = 2.35, size = 539, normalized size = 11.98 \[ -\frac {3 \, {\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}}\right ) - 1\right )} \log \relax (f)^{2}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}} \left (c \log \relax (f)\right )^{\frac {3}{2}}} - \frac {2 \, c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \relax (f)}{\left (c \log \relax (f)\right )^{\frac {3}{2}}}\right )} b^{2} c f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \relax (f)}} + \frac {3 \, {\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{2} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}}\right ) - 1\right )} \log \relax (f)^{3}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}} \left (c \log \relax (f)\right )^{\frac {5}{2}}} - \frac {4 \, {\left (2 \, c x + b\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{4 \, c}\right ) \log \relax (f)^{3}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}\right )^{\frac {3}{2}} \left (c \log \relax (f)\right )^{\frac {5}{2}}} - \frac {4 \, b c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \relax (f)^{2}}{\left (c \log \relax (f)\right )^{\frac {5}{2}}}\right )} b c^{2} f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \relax (f)}} - \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{3} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}}\right ) - 1\right )} \log \relax (f)^{4}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}} \left (c \log \relax (f)\right )^{\frac {7}{2}}} - \frac {12 \, {\left (2 \, c x + b\right )}^{3} b \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{4 \, c}\right ) \log \relax (f)^{4}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}\right )^{\frac {3}{2}} \left (c \log \relax (f)\right )^{\frac {7}{2}}} - \frac {6 \, b^{2} c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \relax (f)^{3}}{\left (c \log \relax (f)\right )^{\frac {7}{2}}} + \frac {8 \, c^{2} \Gamma \left (2, -\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{4 \, c}\right ) \log \relax (f)^{2}}{\left (c \log \relax (f)\right )^{\frac {7}{2}}}\right )} c^{3} f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \relax (f)}} + \frac {\sqrt {\pi } b^{3} f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x - \frac {b \log \relax (f)}{2 \, \sqrt {-c \log \relax (f)}}\right )}{2 \, \sqrt {-c \log \relax (f)} f^{\frac {b^{2}}{4 \, c}}} \]

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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mupad [B]  time = 3.81, size = 44, normalized size = 0.98 \[ \frac {f^{c\,x^2+b\,x+a}\,\left (\ln \relax (f)\,b^2+4\,\ln \relax (f)\,b\,c\,x+4\,\ln \relax (f)\,c^2\,x^2-4\,c\right )}{{\ln \relax (f)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.17, size = 85, normalized size = 1.89 \[ \begin {cases} \frac {f^{a + b x + c x^{2}} \left (b^{2} \log {\relax (f )} + 4 b c x \log {\relax (f )} + 4 c^{2} x^{2} \log {\relax (f )} - 4 c\right )}{\log {\relax (f )}^{2}} & \text {for}\: \log {\relax (f )}^{2} \neq 0 \\b^{3} x + 3 b^{2} c x^{2} + 4 b c^{2} x^{3} + 2 c^{3} x^{4} & \text {otherwise} \end {cases} \]

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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