\(\int f^{a+b x+c x^2} (b+2 c x)^3 \, dx\) [450]

Optimal result

Integrand size = 21, antiderivative size = 45 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=-\frac {4 c f^{a+b x+c x^2}}{\log ^2(f)}+\frac {f^{a+b x+c x^2} (b+2 c x)^2}{\log (f)} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2269, 2268} \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\frac {(b+2 c x)^2 f^{a+b x+c x^2}}{\log (f)}-\frac {4 c f^{a+b x+c x^2}}{\log ^2(f)} \]

[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 2268

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 2269

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*} \text {integral}& = \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] (verified)

Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.69 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\frac {f^{a+x (b+c x)} \left (-4 c+(b+2 c x)^2 \log (f)\right )}{\log ^2(f)} \]

[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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {\left (4 \ln \left (f \right ) c^{2} x^{2}+4 b c x \ln \left (f \right )+\ln \left (f \right ) b^{2}-4 c \right ) f^{c \,x^{2}+b x +a}}{\ln \left (f \right )^{2}}\)	\(45\)
risch	\(\frac {\left (4 \ln \left (f \right ) c^{2} x^{2}+4 b c x \ln \left (f \right )+\ln \left (f \right ) b^{2}-4 c \right ) f^{c \,x^{2}+b x +a}}{\ln \left (f \right )^{2}}\)	\(45\)
norman	\(\frac {\left (\ln \left (f \right ) b^{2}-4 c \right ) {\mathrm e}^{\left (c \,x^{2}+b x +a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{2}}+\frac {4 c^{2} x^{2} {\mathrm e}^{\left (c \,x^{2}+b x +a \right ) \ln \left (f \right )}}{\ln \left (f \right )}+\frac {4 c b x \,{\mathrm e}^{\left (c \,x^{2}+b x +a \right ) \ln \left (f \right )}}{\ln \left (f \right )}\)	\(80\)
parallelrisch	\(\frac {4 x^{2} f^{c \,x^{2}+b x +a} c^{2} \ln \left (f \right )+4 x \,f^{c \,x^{2}+b x +a} c b \ln \left (f \right )+\ln \left (f \right ) f^{c \,x^{2}+b x +a} b^{2}-4 f^{c \,x^{2}+b x +a} c}{\ln \left (f \right )^{2}}\)	\(81\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\frac {{\left ({\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right ) - 4 \, c\right )} f^{c x^{2} + b x + a}}{\log \left (f\right )^{2}} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (42) = 84\).

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.89 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\begin {cases} \frac {f^{a + b x + c x^{2}} \left (b^{2} \log {\left (f \right )} + 4 b c x \log {\left (f \right )} + 4 c^{2} x^{2} \log {\left (f \right )} - 4 c\right )}{\log {\left (f \right )}^{2}} & \text {for}\: \log {\left (f \right )}^{2} \neq 0 \\b^{3} x + 3 b^{2} c x^{2} + 4 b c^{2} x^{3} + 2 c^{3} x^{4} & \text {otherwise} \end {cases} \]

[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))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.45 (sec) , antiderivative size = 539, normalized size of antiderivative = 11.98 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=-\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 \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{2}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\right )^{\frac {3}{2}}} - \frac {2 \, c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )}{\left (c \log \left (f\right )\right )^{\frac {3}{2}}}\right )} b^{2} c f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \left (f\right )}} + \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 \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{3}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\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 \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{3}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}\right )^{\frac {3}{2}} \left (c \log \left (f\right )\right )^{\frac {5}{2}}} - \frac {4 \, b c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {5}{2}}}\right )} b c^{2} f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \left (f\right )}} - \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 \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{4}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\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 \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{4}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}\right )^{\frac {3}{2}} \left (c \log \left (f\right )\right )^{\frac {7}{2}}} - \frac {6 \, b^{2} c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )^{3}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}} + \frac {8 \, c^{2} \Gamma \left (2, -\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}}\right )} c^{3} f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \left (f\right )}} + \frac {\sqrt {\pi } b^{3} f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{2 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \]

[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))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 798, normalized size of antiderivative = 17.73 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\text {Too large to display} \]

[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*I*((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 - 2*I*b^2*log(abs(f)) + 8*(-I*c*x^2 - I*b*x)*c*log(abs(f)) + 8*I*c)*
e^(1/2*I*pi*c*x^2*sgn(f) - 1/2*I*pi*c*x^2 + 1/2*I*pi*b*x*sgn(f) - 1/2*I*pi*b*x + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*
a)/(pi^2*sgn(f) + 2*I*pi*log(abs(f))*sgn(f) - pi^2 - 2*I*pi*log(abs(f)) + 2*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 + 2*I*b^2*log(abs(f)) - 8*(-I*c*x^2 - I*b*x)*c*log
(abs(f)) - 8*I*c)*e^(-1/2*I*pi*c*x^2*sgn(f) + 1/2*I*pi*c*x^2 - 1/2*I*pi*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi*a
*sgn(f) + 1/2*I*pi*a)/(pi^2*sgn(f) - 2*I*pi*log(abs(f))*sgn(f) - pi^2 + 2*I*pi*log(abs(f)) + 2*log(abs(f))^2))
*e^((c*x^2 + b*x)*log(abs(f)) + a*log(abs(f)))

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int f^{a+b x+c x^2} (b+2 c x)^3 \, dx=\frac {f^{c\,x^2+b\,x+a}\,\left (\ln \left (f\right )\,b^2+4\,\ln \left (f\right )\,b\,c\,x+4\,\ln \left (f\right )\,c^2\,x^2-4\,c\right )}{{\ln \left (f\right )}^2} \]

[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