\(\int F^{c (a+b x)} (d+e x+f x^2+g x^3) \, dx\) [92]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=-\frac {6 F^{c (a+b x)} g}{b^4 c^4 \log ^4(F)}+\frac {2 F^{c (a+b x)} (f+3 g x)}{b^3 c^3 \log ^3(F)}-\frac {F^{c (a+b x)} \left (e+2 f x+3 g x^2\right )}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right )}{b c \log (F)} \] Output:

-6*F^(c*(b*x+a))*g/b^4/c^4/ln(F)^4+2*F^(c*(b*x+a))*(3*g*x+f)/b^3/c^3/ln(F) 
^3-F^(c*(b*x+a))*(3*g*x^2+2*f*x+e)/b^2/c^2/ln(F)^2+F^(c*(b*x+a))*(g*x^3+f* 
x^2+e*x+d)/b/c/ln(F)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\frac {F^{c (a+b x)} \left (-6 g+2 b c (f+3 g x) \log (F)-b^2 c^2 (e+x (2 f+3 g x)) \log ^2(F)+b^3 c^3 (d+x (e+x (f+g x))) \log ^3(F)\right )}{b^4 c^4 \log ^4(F)} \] Input:

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

Output:

(F^(c*(a + b*x))*(-6*g + 2*b*c*(f + 3*g*x)*Log[F] - b^2*c^2*(e + x*(2*f + 
3*g*x))*Log[F]^2 + b^3*c^3*(d + x*(e + x*(f + g*x)))*Log[F]^3))/(b^4*c^4*L 
og[F]^4)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2626, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx\)

\(\Big \downarrow \) 2626

\(\displaystyle \int \left (d F^{c (a+b x)}+e x F^{c (a+b x)}+f x^2 F^{c (a+b x)}+g x^3 F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 g F^{c (a+b x)}}{b^4 c^4 \log ^4(F)}+\frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}+\frac {6 g x F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {3 g x^2 F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e x F^{c (a+b x)}}{b c \log (F)}+\frac {f x^2 F^{c (a+b x)}}{b c \log (F)}+\frac {g x^3 F^{c (a+b x)}}{b c \log (F)}\)

Input:

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

Output:

(-6*F^(c*(a + b*x))*g)/(b^4*c^4*Log[F]^4) + (2*f*F^(c*(a + b*x)))/(b^3*c^3 
*Log[F]^3) + (6*F^(c*(a + b*x))*g*x)/(b^3*c^3*Log[F]^3) - (e*F^(c*(a + b*x 
)))/(b^2*c^2*Log[F]^2) - (2*f*F^(c*(a + b*x))*x)/(b^2*c^2*Log[F]^2) - (3*F 
^(c*(a + b*x))*g*x^2)/(b^2*c^2*Log[F]^2) + (d*F^(c*(a + b*x)))/(b*c*Log[F] 
) + (e*F^(c*(a + b*x))*x)/(b*c*Log[F]) + (f*F^(c*(a + b*x))*x^2)/(b*c*Log[ 
F]) + (F^(c*(a + b*x))*g*x^3)/(b*c*Log[F])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2626

Int[(F_)^(v_)*(Px_), x_Symbol] :> Int[ExpandIntegrand[F^v, Px, x], x] /; Fr 
eeQ[F, x] && PolynomialQ[Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17

method	result	size
gosper	\(\frac {\left (g \,x^{3} \ln \left (F \right )^{3} b^{3} c^{3}+\ln \left (F \right )^{3} b^{3} c^{3} f \,x^{2}+\ln \left (F \right )^{3} b^{3} c^{3} e x +\ln \left (F \right )^{3} b^{3} c^{3} d -3 \ln \left (F \right )^{2} b^{2} c^{2} g \,x^{2}-2 f x \ln \left (F \right )^{2} b^{2} c^{2}-\ln \left (F \right )^{2} b^{2} c^{2} e +6 \ln \left (F \right ) b c g x +2 \ln \left (F \right ) b c f -6 g \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{4} b^{4} c^{4}}\)	\(138\)
risch	\(\frac {\left (g \,x^{3} \ln \left (F \right )^{3} b^{3} c^{3}+\ln \left (F \right )^{3} b^{3} c^{3} f \,x^{2}+\ln \left (F \right )^{3} b^{3} c^{3} e x +\ln \left (F \right )^{3} b^{3} c^{3} d -3 \ln \left (F \right )^{2} b^{2} c^{2} g \,x^{2}-2 f x \ln \left (F \right )^{2} b^{2} c^{2}-\ln \left (F \right )^{2} b^{2} c^{2} e +6 \ln \left (F \right ) b c g x +2 \ln \left (F \right ) b c f -6 g \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{4} b^{4} c^{4}}\)	\(138\)
orering	\(\frac {\left (g \,x^{3} \ln \left (F \right )^{3} b^{3} c^{3}+\ln \left (F \right )^{3} b^{3} c^{3} f \,x^{2}+\ln \left (F \right )^{3} b^{3} c^{3} e x +\ln \left (F \right )^{3} b^{3} c^{3} d -3 \ln \left (F \right )^{2} b^{2} c^{2} g \,x^{2}-2 f x \ln \left (F \right )^{2} b^{2} c^{2}-\ln \left (F \right )^{2} b^{2} c^{2} e +6 \ln \left (F \right ) b c g x +2 \ln \left (F \right ) b c f -6 g \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{4} b^{4} c^{4}}\)	\(138\)
norman	\(\frac {\left (\ln \left (F \right )^{3} b^{3} c^{3} d -\ln \left (F \right )^{2} b^{2} c^{2} e +2 \ln \left (F \right ) b c f -6 g \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4} c^{4}}+\frac {\left (\ln \left (F \right ) b c f -3 g \right ) x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2} c^{2}}+\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} e -2 \ln \left (F \right ) b c f +6 g \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}+\frac {g \,x^{3} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) b c}\)	\(163\)
meijerg	\(\frac {F^{a c} g \left (6-\frac {\left (-4 b^{3} c^{3} x^{3} \ln \left (F \right )^{3}+12 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-24 b c x \ln \left (F \right )+24\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{4}\right )}{\ln \left (F \right )^{4} b^{4} c^{4}}-\frac {F^{a c} f \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {F^{a c} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{\ln \left (F \right )^{2} b^{2} c^{2}}-\frac {F^{a c} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{b c \ln \left (F \right )}\)	\(188\)
parallelrisch	\(\frac {x^{3} F^{c \left (b x +a \right )} g \ln \left (F \right )^{3} b^{3} c^{3}+\ln \left (F \right )^{3} x^{2} F^{c \left (b x +a \right )} b^{3} c^{3} f +\ln \left (F \right )^{3} x \,F^{c \left (b x +a \right )} b^{3} c^{3} e +\ln \left (F \right )^{3} F^{c \left (b x +a \right )} b^{3} c^{3} d -3 \ln \left (F \right )^{2} x^{2} F^{c \left (b x +a \right )} b^{2} c^{2} g -2 \ln \left (F \right )^{2} x \,F^{c \left (b x +a \right )} b^{2} c^{2} f -\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} e +6 \ln \left (F \right ) x \,F^{c \left (b x +a \right )} b c g +2 \ln \left (F \right ) F^{c \left (b x +a \right )} b c f -6 F^{c \left (b x +a \right )} g}{\ln \left (F \right )^{4} b^{4} c^{4}}\)	\(219\)

Input:

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

Output:

(g*x^3*ln(F)^3*b^3*c^3+ln(F)^3*b^3*c^3*f*x^2+ln(F)^3*b^3*c^3*e*x+ln(F)^3*b 
^3*c^3*d-3*ln(F)^2*b^2*c^2*g*x^2-2*f*x*ln(F)^2*b^2*c^2-ln(F)^2*b^2*c^2*e+6 
*ln(F)*b*c*g*x+2*ln(F)*b*c*f-6*g)*F^(c*(b*x+a))/ln(F)^4/b^4/c^4
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\frac {{\left ({\left (b^{3} c^{3} g x^{3} + b^{3} c^{3} f x^{2} + b^{3} c^{3} e x + b^{3} c^{3} d\right )} \log \left (F\right )^{3} - {\left (3 \, b^{2} c^{2} g x^{2} + 2 \, b^{2} c^{2} f x + b^{2} c^{2} e\right )} \log \left (F\right )^{2} + 2 \, {\left (3 \, b c g x + b c f\right )} \log \left (F\right ) - 6 \, g\right )} F^{b c x + a c}}{b^{4} c^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

((b^3*c^3*g*x^3 + b^3*c^3*f*x^2 + b^3*c^3*e*x + b^3*c^3*d)*log(F)^3 - (3*b 
^2*c^2*g*x^2 + 2*b^2*c^2*f*x + b^2*c^2*e)*log(F)^2 + 2*(3*b*c*g*x + b*c*f) 
*log(F) - 6*g)*F^(b*c*x + a*c)/(b^4*c^4*log(F)^4)
 

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.61 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{3} c^{3} d \log {\left (F \right )}^{3} + b^{3} c^{3} e x \log {\left (F \right )}^{3} + b^{3} c^{3} f x^{2} \log {\left (F \right )}^{3} + b^{3} c^{3} g x^{3} \log {\left (F \right )}^{3} - b^{2} c^{2} e \log {\left (F \right )}^{2} - 2 b^{2} c^{2} f x \log {\left (F \right )}^{2} - 3 b^{2} c^{2} g x^{2} \log {\left (F \right )}^{2} + 2 b c f \log {\left (F \right )} + 6 b c g x \log {\left (F \right )} - 6 g\right )}{b^{4} c^{4} \log {\left (F \right )}^{4}} & \text {for}\: b^{4} c^{4} \log {\left (F \right )}^{4} \neq 0 \\d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3} + \frac {g x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

integrate(F**((b*x+a)*c)*(g*x**3+f*x**2+e*x+d),x)
 

Output:

Piecewise((F**(c*(a + b*x))*(b**3*c**3*d*log(F)**3 + b**3*c**3*e*x*log(F)* 
*3 + b**3*c**3*f*x**2*log(F)**3 + b**3*c**3*g*x**3*log(F)**3 - b**2*c**2*e 
*log(F)**2 - 2*b**2*c**2*f*x*log(F)**2 - 3*b**2*c**2*g*x**2*log(F)**2 + 2* 
b*c*f*log(F) + 6*b*c*g*x*log(F) - 6*g)/(b**4*c**4*log(F)**4), Ne(b**4*c**4 
*log(F)**4, 0)), (d*x + e*x**2/2 + f*x**3/3 + g*x**4/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.64 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\frac {F^{b c x + a c} d}{b c \log \left (F\right )} + \frac {{\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} f}{b^{3} c^{3} \log \left (F\right )^{3}} + \frac {{\left (F^{a c} b^{3} c^{3} x^{3} \log \left (F\right )^{3} - 3 \, F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} + 6 \, F^{a c} b c x \log \left (F\right ) - 6 \, F^{a c}\right )} F^{b c x} g}{b^{4} c^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

F^(b*c*x + a*c)*d/(b*c*log(F)) + (F^(a*c)*b*c*x*log(F) - F^(a*c))*F^(b*c*x 
)*e/(b^2*c^2*log(F)^2) + (F^(a*c)*b^2*c^2*x^2*log(F)^2 - 2*F^(a*c)*b*c*x*l 
og(F) + 2*F^(a*c))*F^(b*c*x)*f/(b^3*c^3*log(F)^3) + (F^(a*c)*b^3*c^3*x^3*l 
og(F)^3 - 3*F^(a*c)*b^2*c^2*x^2*log(F)^2 + 6*F^(a*c)*b*c*x*log(F) - 6*F^(a 
*c))*F^(b*c*x)*g/(b^4*c^4*log(F)^4)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 4188, normalized size of antiderivative = 35.49 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-(((3*pi^2*b^3*c^3*g*x^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*g*x^3*log(abs 
(F)) + 2*b^3*c^3*g*x^3*log(abs(F))^3 + 3*pi^2*b^3*c^3*f*x^2*log(abs(F))*sg 
n(F) - 3*pi^2*b^3*c^3*f*x^2*log(abs(F)) + 2*b^3*c^3*f*x^2*log(abs(F))^3 + 
3*pi^2*b^3*c^3*e*x*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*e*x*log(abs(F)) + 2 
*b^3*c^3*e*x*log(abs(F))^3 + 3*pi^2*b^3*c^3*d*log(abs(F))*sgn(F) - 3*pi^2* 
b^3*c^3*d*log(abs(F)) + 2*b^3*c^3*d*log(abs(F))^3 - 3*pi^2*b^2*c^2*g*x^2*s 
gn(F) + 3*pi^2*b^2*c^2*g*x^2 - 6*b^2*c^2*g*x^2*log(abs(F))^2 - 2*pi^2*b^2* 
c^2*f*x*sgn(F) + 2*pi^2*b^2*c^2*f*x - 4*b^2*c^2*f*x*log(abs(F))^2 - pi^2*b 
^2*c^2*e*sgn(F) + pi^2*b^2*c^2*e - 2*b^2*c^2*e*log(abs(F))^2 + 12*b*c*g*x* 
log(abs(F)) + 4*b*c*f*log(abs(F)) - 12*g)*(pi^4*b^4*c^4*sgn(F) - 6*pi^2*b^ 
4*c^4*log(abs(F))^2*sgn(F) - pi^4*b^4*c^4 + 6*pi^2*b^4*c^4*log(abs(F))^2 - 
 2*b^4*c^4*log(abs(F))^4)/((pi^4*b^4*c^4*sgn(F) - 6*pi^2*b^4*c^4*log(abs(F 
))^2*sgn(F) - pi^4*b^4*c^4 + 6*pi^2*b^4*c^4*log(abs(F))^2 - 2*b^4*c^4*log( 
abs(F))^4)^2 + 16*(pi^3*b^4*c^4*log(abs(F))*sgn(F) - pi*b^4*c^4*log(abs(F) 
)^3*sgn(F) - pi^3*b^4*c^4*log(abs(F)) + pi*b^4*c^4*log(abs(F))^3)^2) - 4*( 
pi^3*b^3*c^3*g*x^3*sgn(F) - 3*pi*b^3*c^3*g*x^3*log(abs(F))^2*sgn(F) - pi^3 
*b^3*c^3*g*x^3 + 3*pi*b^3*c^3*g*x^3*log(abs(F))^2 + pi^3*b^3*c^3*f*x^2*sgn 
(F) - 3*pi*b^3*c^3*f*x^2*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3*f*x^2 + 3*pi* 
b^3*c^3*f*x^2*log(abs(F))^2 + pi^3*b^3*c^3*e*x*sgn(F) - 3*pi*b^3*c^3*e*x*l 
og(abs(F))^2*sgn(F) - pi^3*b^3*c^3*e*x + 3*pi*b^3*c^3*e*x*log(abs(F))^2...
 

Mupad [B] (verification not implemented)

Time = 22.58 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (g\,b^3\,c^3\,x^3\,{\ln \left (F\right )}^3+f\,b^3\,c^3\,x^2\,{\ln \left (F\right )}^3+e\,b^3\,c^3\,x\,{\ln \left (F\right )}^3+d\,b^3\,c^3\,{\ln \left (F\right )}^3-3\,g\,b^2\,c^2\,x^2\,{\ln \left (F\right )}^2-2\,f\,b^2\,c^2\,x\,{\ln \left (F\right )}^2-e\,b^2\,c^2\,{\ln \left (F\right )}^2+6\,g\,b\,c\,x\,\ln \left (F\right )+2\,f\,b\,c\,\ln \left (F\right )-6\,g\right )}{b^4\,c^4\,{\ln \left (F\right )}^4} \] Input:

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

Output:

(F^(a*c + b*c*x)*(2*b*c*f*log(F) - 6*g + b^3*c^3*d*log(F)^3 - b^2*c^2*e*lo 
g(F)^2 + b^3*c^3*f*x^2*log(F)^3 - 3*b^2*c^2*g*x^2*log(F)^2 + b^3*c^3*g*x^3 
*log(F)^3 + 6*b*c*g*x*log(F) + b^3*c^3*e*x*log(F)^3 - 2*b^2*c^2*f*x*log(F) 
^2))/(b^4*c^4*log(F)^4)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int F^{c (a+b x)} \left (d+e x+f x^2+g x^3\right ) \, dx=\frac {f^{b c x +a c} \left (\mathrm {log}\left (f \right )^{3} b^{3} c^{3} d +\mathrm {log}\left (f \right )^{3} b^{3} c^{3} e x +\mathrm {log}\left (f \right )^{3} b^{3} c^{3} f \,x^{2}+\mathrm {log}\left (f \right )^{3} b^{3} c^{3} g \,x^{3}-\mathrm {log}\left (f \right )^{2} b^{2} c^{2} e -2 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} f x -3 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} g \,x^{2}+2 \,\mathrm {log}\left (f \right ) b c f +6 \,\mathrm {log}\left (f \right ) b c g x -6 g \right )}{\mathrm {log}\left (f \right )^{4} b^{4} c^{4}} \] Input:

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

Output:

(f**(a*c + b*c*x)*(log(f)**3*b**3*c**3*d + log(f)**3*b**3*c**3*e*x + log(f 
)**3*b**3*c**3*f*x**2 + log(f)**3*b**3*c**3*g*x**3 - log(f)**2*b**2*c**2*e 
 - 2*log(f)**2*b**2*c**2*f*x - 3*log(f)**2*b**2*c**2*g*x**2 + 2*log(f)*b*c 
*f + 6*log(f)*b*c*g*x - 6*g))/(log(f)**4*b**4*c**4)