Integrand size = 20, antiderivative size = 80 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {F^{c (a+b x)} (e+2 f x)}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} \left (d+e x+f x^2\right )}{b c \log (F)} \] Output:
2*f*F^(c*(b*x+a))/b^3/c^3/ln(F)^3-F^(c*(b*x+a))*(2*f*x+e)/b^2/c^2/ln(F)^2+ F^(c*(b*x+a))*(f*x^2+e*x+d)/b/c/ln(F)
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.70 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{c (a+b x)} \left (2 f-b c (e+2 f x) \log (F)+b^2 c^2 (d+x (e+f x)) \log ^2(F)\right )}{b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*(d + e*x + f*x^2),x]
Output:
(F^(c*(a + b*x))*(2*f - b*c*(e + 2*f*x)*Log[F] + b^2*c^2*(d + x*(e + f*x)) *Log[F]^2))/(b^3*c^3*Log[F]^3)
Time = 0.52 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.69, 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.
\(\displaystyle \int \left (d+e x+f x^2\right ) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 2626 |
\(\displaystyle \int \left (d F^{c (a+b x)}+e x F^{c (a+b x)}+f x^2 F^{c (a+b x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e x F^{c (a+b x)}}{b c \log (F)}+\frac {f x^2 F^{c (a+b x)}}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*(d + e*x + f*x^2),x]
Output:
(2*f*F^(c*(a + b*x)))/(b^3*c^3*Log[F]^3) - (e*F^(c*(a + b*x)))/(b^2*c^2*Lo g[F]^2) - (2*f*F^(c*(a + b*x))*x)/(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])
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]
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00
method | result | size |
gosper | \(\frac {\left (f \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +\ln \left (F \right )^{2} b^{2} c^{2} d -2 \ln \left (F \right ) b c f x -\ln \left (F \right ) b c e +2 f \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(80\) |
risch | \(\frac {\left (f \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +\ln \left (F \right )^{2} b^{2} c^{2} d -2 \ln \left (F \right ) b c f x -\ln \left (F \right ) b c e +2 f \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(80\) |
orering | \(\frac {\left (f \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +\ln \left (F \right )^{2} b^{2} c^{2} d -2 \ln \left (F \right ) b c f x -\ln \left (F \right ) b c e +2 f \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(80\) |
norman | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} d -\ln \left (F \right ) b c e +2 f \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {\left (\ln \left (F \right ) b c e -2 f \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2} c^{2}}+\frac {f \,x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) b c}\) | \(103\) |
meijerg | \(-\frac {F^{a c} f \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {F^{a c} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{\ln \left (F \right )^{2} b^{2} c^{2}}-\frac {F^{a c} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{b c \ln \left (F \right )}\) | \(121\) |
parallelrisch | \(\frac {x^{2} F^{c \left (b x +a \right )} f \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} x \,F^{c \left (b x +a \right )} b^{2} c^{2} e +\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} d -2 \ln \left (F \right ) x \,F^{c \left (b x +a \right )} b c f -\ln \left (F \right ) F^{c \left (b x +a \right )} b c e +2 F^{c \left (b x +a \right )} f}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(125\) |
Input:
int(F^(c*(b*x+a))*(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
(f*x^2*ln(F)^2*b^2*c^2+ln(F)^2*b^2*c^2*e*x+ln(F)^2*b^2*c^2*d-2*ln(F)*b*c*f *x-ln(F)*b*c*e+2*f)*F^(c*(b*x+a))/ln(F)^3/b^3/c^3
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {{\left ({\left (b^{2} c^{2} f x^{2} + b^{2} c^{2} e x + b^{2} c^{2} d\right )} \log \left (F\right )^{2} - {\left (2 \, b c f x + b c e\right )} \log \left (F\right ) + 2 \, f\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(f*x^2+e*x+d),x, algorithm="fricas")
Output:
((b^2*c^2*f*x^2 + b^2*c^2*e*x + b^2*c^2*d)*log(F)^2 - (2*b*c*f*x + b*c*e)* log(F) + 2*f)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3)
Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d \log {\left (F \right )}^{2} + b^{2} c^{2} e x \log {\left (F \right )}^{2} + b^{2} c^{2} f x^{2} \log {\left (F \right )}^{2} - b c e \log {\left (F \right )} - 2 b c f x \log {\left (F \right )} + 2 f\right )}{b^{3} c^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} c^{3} \log {\left (F \right )}^{3} \neq 0 \\d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(F**((b*x+a)*c)*(f*x**2+e*x+d),x)
Output:
Piecewise((F**(c*(a + b*x))*(b**2*c**2*d*log(F)**2 + b**2*c**2*e*x*log(F)* *2 + b**2*c**2*f*x**2*log(F)**2 - b*c*e*log(F) - 2*b*c*f*x*log(F) + 2*f)/( b**3*c**3*log(F)**3), Ne(b**3*c**3*log(F)**3, 0)), (d*x + e*x**2/2 + f*x** 3/3, True))
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.46 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{b c x + a c} d}{b c \log \left (F\right )} + \frac {{\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} f}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(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)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 2068, normalized size of antiderivative = 25.85 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(F^((b*x+a)*c)*(f*x^2+e*x+d),x, algorithm="giac")
Output:
(((pi^2*b^2*c^2*f*x^2*sgn(F) - pi^2*b^2*c^2*f*x^2 + 2*b^2*c^2*f*x^2*log(ab s(F))^2 + pi^2*b^2*c^2*e*x*sgn(F) - pi^2*b^2*c^2*e*x + 2*b^2*c^2*e*x*log(a bs(F))^2 + pi^2*b^2*c^2*d*sgn(F) - pi^2*b^2*c^2*d + 2*b^2*c^2*d*log(abs(F) )^2 - 4*b*c*f*x*log(abs(F)) - 2*b*c*e*log(abs(F)) + 4*f)*(3*pi^2*b^3*c^3*l og(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)/ ((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi ^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)^2) - (pi^3*b^3*c^3*sgn(F ) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(ab s(F))^2)*(2*pi*b^2*c^2*f*x^2*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*f*x^2*log(a bs(F)) + 2*pi*b^2*c^2*e*x*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*e*x*log(abs(F) ) + 2*pi*b^2*c^2*d*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*d*log(abs(F)) - 2*pi* b*c*f*x*sgn(F) + 2*pi*b*c*f*x - pi*b*c*e*sgn(F) + pi*b*c*e)/((pi^3*b^3*c^3 *sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3* log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log (abs(F)) + 2*b^3*c^3*log(abs(F))^3)^2))*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi* b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c) + ((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3 *c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2)*(pi ^2*b^2*c^2*f*x^2*sgn(F) - pi^2*b^2*c^2*f*x^2 + 2*b^2*c^2*f*x^2*log(abs(F)) ^2 + pi^2*b^2*c^2*e*x*sgn(F) - pi^2*b^2*c^2*e*x + 2*b^2*c^2*e*x*log(abs...
Time = 23.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (f\,b^2\,c^2\,x^2\,{\ln \left (F\right )}^2+e\,b^2\,c^2\,x\,{\ln \left (F\right )}^2+d\,b^2\,c^2\,{\ln \left (F\right )}^2-2\,f\,b\,c\,x\,\ln \left (F\right )-e\,b\,c\,\ln \left (F\right )+2\,f\right )}{b^3\,c^3\,{\ln \left (F\right )}^3} \] Input:
int(F^(c*(a + b*x))*(d + e*x + f*x^2),x)
Output:
(F^(a*c + b*c*x)*(2*f - b*c*e*log(F) + b^2*c^2*d*log(F)^2 + b^2*c^2*f*x^2* log(F)^2 - 2*b*c*f*x*log(F) + b^2*c^2*e*x*log(F)^2))/(b^3*c^3*log(F)^3)
Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {f^{b c x +a c} \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2} d +\mathrm {log}\left (f \right )^{2} b^{2} c^{2} e x +\mathrm {log}\left (f \right )^{2} b^{2} c^{2} f \,x^{2}-\mathrm {log}\left (f \right ) b c e -2 \,\mathrm {log}\left (f \right ) b c f x +2 f \right )}{\mathrm {log}\left (f \right )^{3} b^{3} c^{3}} \] Input:
int(F^((b*x+a)*c)*(f*x^2+e*x+d),x)
Output:
(f**(a*c + b*c*x)*(log(f)**2*b**2*c**2*d + log(f)**2*b**2*c**2*e*x + log(f )**2*b**2*c**2*f*x**2 - log(f)*b*c*e - 2*log(f)*b*c*f*x + 2*f))/(log(f)**3 *b**3*c**3)