\(\int F^{a+b (c+d x)} x (e+f x)^2 \, dx\) [106]

Optimal result

Integrand size = 20, antiderivative size = 129 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=-\frac {6 f^2 F^{a+b c+b d x}}{b^4 d^4 \log ^4(F)}+\frac {2 f F^{a+b c+b d x} (2 e+3 f x)}{b^3 d^3 \log ^3(F)}-\frac {F^{a+b c+b d x} \left (e^2+4 e f x+3 f^2 x^2\right )}{b^2 d^2 \log ^2(F)}+\frac {F^{a+b c+b d x} x (e+f x)^2}{b d \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.71 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\frac {F^{a+b (c+d x)} \left (-6 f^2+2 b d f (2 e+3 f x) \log (F)-b^2 d^2 \left (e^2+4 e f x+3 f^2 x^2\right ) \log ^2(F)+b^3 d^3 x (e+f x)^2 \log ^3(F)\right )}{b^4 d^4 \log ^4(F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.

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

\(\Big \downarrow \) 2626

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-6*f^2*F^(a + b*c + b*d*x))/(b^4*d^4*Log[F]^4) + (4*e*f*F^(a + b*c + b*d* 
x))/(b^3*d^3*Log[F]^3) + (6*f^2*F^(a + b*c + b*d*x)*x)/(b^3*d^3*Log[F]^3) 
- (e^2*F^(a + b*c + b*d*x))/(b^2*d^2*Log[F]^2) - (4*e*f*F^(a + b*c + b*d*x 
)*x)/(b^2*d^2*Log[F]^2) - (3*f^2*F^(a + b*c + b*d*x)*x^2)/(b^2*d^2*Log[F]^ 
2) + (e^2*F^(a + b*c + b*d*x)*x)/(b*d*Log[F]) + (2*e*f*F^(a + b*c + b*d*x) 
*x^2)/(b*d*Log[F]) + (f^2*F^(a + b*c + b*d*x)*x^3)/(b*d*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 = 0.06 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12

method	result	size
gosper	\(\frac {\left (\ln \left (F \right )^{3} b^{3} d^{3} f^{2} x^{3}+2 \ln \left (F \right )^{3} b^{3} d^{3} e f \,x^{2}+\ln \left (F \right )^{3} b^{3} d^{3} e^{2} x -3 \ln \left (F \right )^{2} b^{2} d^{2} f^{2} x^{2}-4 \ln \left (F \right )^{2} b^{2} d^{2} e f x -\ln \left (F \right )^{2} b^{2} d^{2} e^{2}+6 \ln \left (F \right ) b d \,f^{2} x +4 e f \ln \left (F \right ) b d -6 f^{2}\right ) F^{b d x +b c +a}}{\ln \left (F \right )^{4} b^{4} d^{4}}\)	\(144\)
risch	\(\frac {\left (\ln \left (F \right )^{3} b^{3} d^{3} f^{2} x^{3}+2 \ln \left (F \right )^{3} b^{3} d^{3} e f \,x^{2}+\ln \left (F \right )^{3} b^{3} d^{3} e^{2} x -3 \ln \left (F \right )^{2} b^{2} d^{2} f^{2} x^{2}-4 \ln \left (F \right )^{2} b^{2} d^{2} e f x -\ln \left (F \right )^{2} b^{2} d^{2} e^{2}+6 \ln \left (F \right ) b d \,f^{2} x +4 e f \ln \left (F \right ) b d -6 f^{2}\right ) F^{b d x +b c +a}}{\ln \left (F \right )^{4} b^{4} d^{4}}\)	\(144\)
orering	\(\frac {\left (\ln \left (F \right )^{3} b^{3} d^{3} f^{2} x^{3}+2 \ln \left (F \right )^{3} b^{3} d^{3} e f \,x^{2}+\ln \left (F \right )^{3} b^{3} d^{3} e^{2} x -3 \ln \left (F \right )^{2} b^{2} d^{2} f^{2} x^{2}-4 \ln \left (F \right )^{2} b^{2} d^{2} e f x -\ln \left (F \right )^{2} b^{2} d^{2} e^{2}+6 \ln \left (F \right ) b d \,f^{2} x +4 e f \ln \left (F \right ) b d -6 f^{2}\right ) F^{a +b \left (d x +c \right )}}{\ln \left (F \right )^{4} b^{4} d^{4}}\)	\(144\)
meijerg	\(\frac {F^{b c +a} f^{2} \left (6-\frac {\left (-4 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}+12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}-24 b d x \ln \left (F \right )+24\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{4}\right )}{\ln \left (F \right )^{4} b^{4} d^{4}}-\frac {2 F^{b c +a} f e \left (2-\frac {\left (3 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}-6 b d x \ln \left (F \right )+6\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{3}\right )}{b^{3} d^{3} \ln \left (F \right )^{3}}+\frac {F^{b c +a} e^{2} \left (1-\frac {\left (-2 b d x \ln \left (F \right )+2\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{2}\right )}{b^{2} d^{2} \ln \left (F \right )^{2}}\)	\(170\)
norman	\(\frac {f^{2} x^{3} {\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{b d \ln \left (F \right )}+\frac {\left (\ln \left (F \right )^{2} b^{2} d^{2} e^{2}-4 e f \ln \left (F \right ) b d +6 f^{2}\right ) x \,{\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{\ln \left (F \right )^{3} b^{3} d^{3}}+\frac {f \left (2 \ln \left (F \right ) b d e -3 f \right ) x^{2} {\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2} d^{2}}-\frac {\left (\ln \left (F \right )^{2} b^{2} d^{2} e^{2}-4 e f \ln \left (F \right ) b d +6 f^{2}\right ) {\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{\ln \left (F \right )^{4} b^{4} d^{4}}\)	\(177\)
parallelrisch	\(\frac {\ln \left (F \right )^{3} x^{3} F^{b d x +b c +a} b^{3} d^{3} f^{2}+2 \ln \left (F \right )^{3} x^{2} F^{b d x +b c +a} b^{3} d^{3} e f +\ln \left (F \right )^{3} x \,F^{b d x +b c +a} b^{3} d^{3} e^{2}-3 \ln \left (F \right )^{2} x^{2} F^{b d x +b c +a} b^{2} d^{2} f^{2}-4 \ln \left (F \right )^{2} x \,F^{b d x +b c +a} b^{2} d^{2} e f -\ln \left (F \right )^{2} F^{b d x +b c +a} b^{2} d^{2} e^{2}+6 \ln \left (F \right ) x \,F^{b d x +b c +a} b d \,f^{2}+4 \ln \left (F \right ) F^{b d x +b c +a} b d e f -6 F^{b d x +b c +a} f^{2}}{\ln \left (F \right )^{4} b^{4} d^{4}}\)	\(232\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\frac {{\left ({\left (b^{3} d^{3} f^{2} x^{3} + 2 \, b^{3} d^{3} e f x^{2} + b^{3} d^{3} e^{2} x\right )} \log \left (F\right )^{3} - {\left (3 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} d^{2} e f x + b^{2} d^{2} e^{2}\right )} \log \left (F\right )^{2} - 6 \, f^{2} + 2 \, {\left (3 \, b d f^{2} x + 2 \, b d e f\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{b^{4} d^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.54 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )} \left (b^{3} d^{3} e^{2} x \log {\left (F \right )}^{3} + 2 b^{3} d^{3} e f x^{2} \log {\left (F \right )}^{3} + b^{3} d^{3} f^{2} x^{3} \log {\left (F \right )}^{3} - b^{2} d^{2} e^{2} \log {\left (F \right )}^{2} - 4 b^{2} d^{2} e f x \log {\left (F \right )}^{2} - 3 b^{2} d^{2} f^{2} x^{2} \log {\left (F \right )}^{2} + 4 b d e f \log {\left (F \right )} + 6 b d f^{2} x \log {\left (F \right )} - 6 f^{2}\right )}{b^{4} d^{4} \log {\left (F \right )}^{4}} & \text {for}\: b^{4} d^{4} \log {\left (F \right )}^{4} \neq 0 \\\frac {e^{2} x^{2}}{2} + \frac {2 e f x^{3}}{3} + \frac {f^{2} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.52 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\frac {{\left (F^{b c + a} b d x \log \left (F\right ) - F^{b c + a}\right )} F^{b d x} e^{2}}{b^{2} d^{2} \log \left (F\right )^{2}} + \frac {2 \, {\left (F^{b c + a} b^{2} d^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{b c + a} b d x \log \left (F\right ) + 2 \, F^{b c + a}\right )} F^{b d x} e f}{b^{3} d^{3} \log \left (F\right )^{3}} + \frac {{\left (F^{b c + a} b^{3} d^{3} x^{3} \log \left (F\right )^{3} - 3 \, F^{b c + a} b^{2} d^{2} x^{2} \log \left (F\right )^{2} + 6 \, F^{b c + a} b d x \log \left (F\right ) - 6 \, F^{b c + a}\right )} F^{b d x} f^{2}}{b^{4} d^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 4114, normalized size of antiderivative = 31.89 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-(((3*pi^2*b^3*d^3*f^2*x^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*f^2*x^3*log 
(abs(F)) + 2*b^3*d^3*f^2*x^3*log(abs(F))^3 + 6*pi^2*b^3*d^3*e*f*x^2*log(ab 
s(F))*sgn(F) - 6*pi^2*b^3*d^3*e*f*x^2*log(abs(F)) + 4*b^3*d^3*e*f*x^2*log( 
abs(F))^3 + 3*pi^2*b^3*d^3*e^2*x*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*e^2*x 
*log(abs(F)) + 2*b^3*d^3*e^2*x*log(abs(F))^3 - 3*pi^2*b^2*d^2*f^2*x^2*sgn( 
F) + 3*pi^2*b^2*d^2*f^2*x^2 - 6*b^2*d^2*f^2*x^2*log(abs(F))^2 - 4*pi^2*b^2 
*d^2*e*f*x*sgn(F) + 4*pi^2*b^2*d^2*e*f*x - 8*b^2*d^2*e*f*x*log(abs(F))^2 - 
 pi^2*b^2*d^2*e^2*sgn(F) + pi^2*b^2*d^2*e^2 - 2*b^2*d^2*e^2*log(abs(F))^2 
+ 12*b*d*f^2*x*log(abs(F)) + 8*b*d*e*f*log(abs(F)) - 12*f^2)*(pi^4*b^4*d^4 
*sgn(F) - 6*pi^2*b^4*d^4*log(abs(F))^2*sgn(F) - pi^4*b^4*d^4 + 6*pi^2*b^4* 
d^4*log(abs(F))^2 - 2*b^4*d^4*log(abs(F))^4)/((pi^4*b^4*d^4*sgn(F) - 6*pi^ 
2*b^4*d^4*log(abs(F))^2*sgn(F) - pi^4*b^4*d^4 + 6*pi^2*b^4*d^4*log(abs(F)) 
^2 - 2*b^4*d^4*log(abs(F))^4)^2 + 16*(pi^3*b^4*d^4*log(abs(F))*sgn(F) - pi 
*b^4*d^4*log(abs(F))^3*sgn(F) - pi^3*b^4*d^4*log(abs(F)) + pi*b^4*d^4*log( 
abs(F))^3)^2) - 4*(pi^3*b^3*d^3*f^2*x^3*sgn(F) - 3*pi*b^3*d^3*f^2*x^3*log( 
abs(F))^2*sgn(F) - pi^3*b^3*d^3*f^2*x^3 + 3*pi*b^3*d^3*f^2*x^3*log(abs(F)) 
^2 + 2*pi^3*b^3*d^3*e*f*x^2*sgn(F) - 6*pi*b^3*d^3*e*f*x^2*log(abs(F))^2*sg 
n(F) - 2*pi^3*b^3*d^3*e*f*x^2 + 6*pi*b^3*d^3*e*f*x^2*log(abs(F))^2 + pi^3* 
b^3*d^3*e^2*x*sgn(F) - 3*pi*b^3*d^3*e^2*x*log(abs(F))^2*sgn(F) - pi^3*b^3* 
d^3*e^2*x + 3*pi*b^3*d^3*e^2*x*log(abs(F))^2 + 6*pi*b^2*d^2*f^2*x^2*log...
 

Mupad [B] (verification not implemented)

Time = 23.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.11 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\frac {F^{a+b\,c+b\,d\,x}\,\left (b^3\,d^3\,e^2\,x\,{\ln \left (F\right )}^3+2\,b^3\,d^3\,e\,f\,x^2\,{\ln \left (F\right )}^3+b^3\,d^3\,f^2\,x^3\,{\ln \left (F\right )}^3-b^2\,d^2\,e^2\,{\ln \left (F\right )}^2-4\,b^2\,d^2\,e\,f\,x\,{\ln \left (F\right )}^2-3\,b^2\,d^2\,f^2\,x^2\,{\ln \left (F\right )}^2+4\,b\,d\,e\,f\,\ln \left (F\right )+6\,b\,d\,f^2\,x\,\ln \left (F\right )-6\,f^2\right )}{b^4\,d^4\,{\ln \left (F\right )}^4} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.11 \[ \int F^{a+b (c+d x)} x (e+f x)^2 \, dx=\frac {f^{b d x +b c +a} \left (\mathrm {log}\left (f \right )^{3} b^{3} d^{3} e^{2} x +2 \mathrm {log}\left (f \right )^{3} b^{3} d^{3} e f \,x^{2}+\mathrm {log}\left (f \right )^{3} b^{3} d^{3} f^{2} x^{3}-\mathrm {log}\left (f \right )^{2} b^{2} d^{2} e^{2}-4 \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e f x -3 \mathrm {log}\left (f \right )^{2} b^{2} d^{2} f^{2} x^{2}+4 \,\mathrm {log}\left (f \right ) b d e f +6 \,\mathrm {log}\left (f \right ) b d \,f^{2} x -6 f^{2}\right )}{\mathrm {log}\left (f \right )^{4} b^{4} d^{4}} \] Input:

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

Output:

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