Integrand size = 132, antiderivative size = 69 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\frac {F^{a+b x+c x^2+d x^3} \left (\frac {2 c f-b g}{4 c^2-3 b d}-\frac {(3 d f-2 c g) x}{4 c^2-3 b d}\right )}{\log (F)} \] Output:
F^(d*x^3+c*x^2+b*x+a)*((-b*g+2*c*f)/(-3*b*d+4*c^2)-(-2*c*g+3*d*f)*x/(-3*b* d+4*c^2))/ln(F)
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.72 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\frac {F^{a+x (b+x (c+d x))} (-b g-3 d f x+2 c (f+g x))}{\left (4 c^2-3 b d\right ) \log (F)} \] Input:
Integrate[F^(a + b*x + c*x^2 + d*x^3)*(f*x + g*x^2 - (3*(3*d^2*f - 2*c*d*g )*x^3)/(4*c^2 - 3*b*d) + ((-9*d*(3*d^2*f - 2*c*d*g))/(4*c^2 - 3*b*d) + 3*b *d*g*Log[F] + (6*b*c*(3*d^2*f - 2*c*d*g)*Log[F])/(4*c^2 - 3*b*d))/(9*d^2*L og[F])),x]
Output:
(F^(a + x*(b + x*(c + d*x)))*(-(b*g) - 3*d*f*x + 2*c*(f + g*x)))/((4*c^2 - 3*b*d)*Log[F])
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.
\(\displaystyle \int F^{a+b x+c x^2+d x^3} \left (\frac {\frac {6 b c \log (F) \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)}{9 d^2 \log (F)}-\frac {3 x^3 \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+f x+g x^2\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int F^{a+b x+c x^2+d x^3} \left (-\frac {-b \log (F) (2 c f-b g)-2 c g+3 d f}{\log (F) \left (4 c^2-3 b d\right )}-\frac {3 d x^3 (3 d f-2 c g)}{4 c^2-3 b d}+f x+g x^2\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 d x^3 (3 d f-2 c g) F^{a+b x+c x^2+d x^3}}{3 b d-4 c^2}+\frac {F^{a+b x+c x^2+d x^3} (b \log (F) (2 c f-b g)+2 c g-3 d f)}{\log (F) \left (4 c^2-3 b d\right )}+f x F^{a+b x+c x^2+d x^3}+g x^2 F^{a+b x+c x^2+d x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 d (3 d f-2 c g) \int F^{d x^3+c x^2+b x+a} x^3dx}{4 c^2-3 b d}-\frac {(-b \log (F) (2 c f-b g)-2 c g+3 d f) \int F^{d x^3+c x^2+b x+a}dx}{\log (F) \left (4 c^2-3 b d\right )}+f \int F^{d x^3+c x^2+b x+a} xdx+g \int F^{d x^3+c x^2+b x+a} x^2dx\) |
Input:
Int[F^(a + b*x + c*x^2 + d*x^3)*(f*x + g*x^2 - (3*(3*d^2*f - 2*c*d*g)*x^3) /(4*c^2 - 3*b*d) + ((-9*d*(3*d^2*f - 2*c*d*g))/(4*c^2 - 3*b*d) + 3*b*d*g*L og[F] + (6*b*c*(3*d^2*f - 2*c*d*g)*Log[F])/(4*c^2 - 3*b*d))/(9*d^2*Log[F]) ),x]
Output:
$Aborted
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77
method | result | size |
gosper | \(\frac {F^{d \,x^{3}+c \,x^{2}+b x +a} \left (-2 c g x +3 d f x +b g -2 c f \right )}{\left (3 b d -4 c^{2}\right ) \ln \left (F \right )}\) | \(53\) |
risch | \(\frac {F^{d \,x^{3}+c \,x^{2}+b x +a} \left (-2 c g x +3 d f x +b g -2 c f \right )}{\left (3 b d -4 c^{2}\right ) \ln \left (F \right )}\) | \(53\) |
norman | \(\frac {\frac {d \left (b g -2 c f \right ) {\mathrm e}^{\left (d \,x^{3}+c \,x^{2}+b x +a \right ) \ln \left (F \right )}}{\left (3 b d -4 c^{2}\right ) \ln \left (F \right )}-\frac {d \left (2 c g -3 d f \right ) x \,{\mathrm e}^{\left (d \,x^{3}+c \,x^{2}+b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) \left (3 b d -4 c^{2}\right )}}{d}\) | \(99\) |
parallelrisch | \(\frac {-2 x \,F^{d \,x^{3}+c \,x^{2}+b x +a} c g +3 x \,F^{d \,x^{3}+c \,x^{2}+b x +a} d f +F^{d \,x^{3}+c \,x^{2}+b x +a} b g -2 F^{d \,x^{3}+c \,x^{2}+b x +a} c f}{\left (3 b d -4 c^{2}\right ) \ln \left (F \right )}\) | \(104\) |
Input:
int(F^(d*x^3+c*x^2+b*x+a)*(f*x+g*x^2-3*(-2*c*d*g+3*d^2*f)*x^3/(-3*b*d+4*c^ 2)+1/9*(-9*d*(-2*c*d*g+3*d^2*f)/(-3*b*d+4*c^2)+3*b*d*g*ln(F)+6*b*c*(-2*c*d *g+3*d^2*f)*ln(F)/(-3*b*d+4*c^2))/d^2/ln(F)),x,method=_RETURNVERBOSE)
Output:
F^(d*x^3+c*x^2+b*x+a)*(-2*c*g*x+3*d*f*x+b*g-2*c*f)/(3*b*d-4*c^2)/ln(F)
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\frac {{\left (2 \, c f - b g - {\left (3 \, d f - 2 \, c g\right )} x\right )} F^{d x^{3} + c x^{2} + b x + a}}{{\left (4 \, c^{2} - 3 \, b d\right )} \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(f*x+g*x^2-3*(-2*c*d*g+3*d^2*f)*x^3/(-3*b* d+4*c^2)+1/9*(-9*d*(-2*c*d*g+3*d^2*f)/(-3*b*d+4*c^2)+3*b*d*g*log(F)+6*b*c* (-2*c*d*g+3*d^2*f)*log(F)/(-3*b*d+4*c^2))/d^2/log(F)),x, algorithm="fricas ")
Output:
(2*c*f - b*g - (3*d*f - 2*c*g)*x)*F^(d*x^3 + c*x^2 + b*x + a)/((4*c^2 - 3* b*d)*log(F))
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (60) = 120\).
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.29 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\begin {cases} \frac {F^{a + b x + c x^{2} + d x^{3}} \left (b g - 2 c f - 2 c g x + 3 d f x\right )}{3 b d \log {\left (F \right )} - 4 c^{2} \log {\left (F \right )}} & \text {for}\: 3 b d \log {\left (F \right )} - 4 c^{2} \log {\left (F \right )} \neq 0 \\\frac {f x^{2}}{2} + \frac {g x^{3}}{3} + \frac {x^{4} \left (- 6 c d g + 9 d^{2} f\right )}{12 b d - 16 c^{2}} + \frac {x \left (b^{2} g \log {\left (F \right )} - 2 b c f \log {\left (F \right )} - 2 c g + 3 d f\right )}{3 b d \log {\left (F \right )} - 4 c^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(d*x**3+c*x**2+b*x+a)*(f*x+g*x**2-3*(-2*c*d*g+3*d**2*f)*x**3/ (-3*b*d+4*c**2)+1/9*(-9*d*(-2*c*d*g+3*d**2*f)/(-3*b*d+4*c**2)+3*b*d*g*ln(F )+6*b*c*(-2*c*d*g+3*d**2*f)*ln(F)/(-3*b*d+4*c**2))/d**2/ln(F)),x)
Output:
Piecewise((F**(a + b*x + c*x**2 + d*x**3)*(b*g - 2*c*f - 2*c*g*x + 3*d*f*x )/(3*b*d*log(F) - 4*c**2*log(F)), Ne(3*b*d*log(F) - 4*c**2*log(F), 0)), (f *x**2/2 + g*x**3/3 + x**4*(-6*c*d*g + 9*d**2*f)/(12*b*d - 16*c**2) + x*(b* *2*g*log(F) - 2*b*c*f*log(F) - 2*c*g + 3*d*f)/(3*b*d*log(F) - 4*c**2*log(F )), True))
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=-\frac {{\left ({\left (3 \, d f - 2 \, c g\right )} F^{a} x - {\left (2 \, c f - b g\right )} F^{a}\right )} e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )}}{{\left (4 \, c^{2} - 3 \, b d\right )} \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(f*x+g*x^2-3*(-2*c*d*g+3*d^2*f)*x^3/(-3*b* d+4*c^2)+1/9*(-9*d*(-2*c*d*g+3*d^2*f)/(-3*b*d+4*c^2)+3*b*d*g*log(F)+6*b*c* (-2*c*d*g+3*d^2*f)*log(F)/(-3*b*d+4*c^2))/d^2/log(F)),x, algorithm="maxima ")
Output:
-((3*d*f - 2*c*g)*F^a*x - (2*c*f - b*g)*F^a)*e^(d*x^3*log(F) + c*x^2*log(F ) + b*x*log(F))/((4*c^2 - 3*b*d)*log(F))
Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.91 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=-\frac {3 \, F^{a} d f x e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )} - 2 \, F^{a} c g x e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )} - 2 \, F^{a} c f e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )} + F^{a} b g e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )}}{4 \, c^{2} \log \left (F\right ) - 3 \, b d \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(f*x+g*x^2-3*(-2*c*d*g+3*d^2*f)*x^3/(-3*b* d+4*c^2)+1/9*(-9*d*(-2*c*d*g+3*d^2*f)/(-3*b*d+4*c^2)+3*b*d*g*log(F)+6*b*c* (-2*c*d*g+3*d^2*f)*log(F)/(-3*b*d+4*c^2))/d^2/log(F)),x, algorithm="giac")
Output:
-(3*F^a*d*f*x*e^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)) - 2*F^a*c*g*x*e ^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)) - 2*F^a*c*f*e^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)) + F^a*b*g*e^(d*x^3*log(F) + c*x^2*log(F) + b*x* log(F)))/(4*c^2*log(F) - 3*b*d*log(F))
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\frac {F^a\,F^{c\,x^2}\,F^{d\,x^3}\,F^{b\,x}\,\left (b\,g-2\,c\,f-2\,c\,g\,x+3\,d\,f\,x\right )}{\ln \left (F\right )\,\left (3\,b\,d-4\,c^2\right )} \] Input:
int(F^(a + b*x + c*x^2 + d*x^3)*(f*x + g*x^2 + ((d*(3*d^2*f - 2*c*d*g))/(3 *b*d - 4*c^2) + (b*d*g*log(F))/3 - (2*b*c*log(F)*(3*d^2*f - 2*c*d*g))/(3*( 3*b*d - 4*c^2)))/(d^2*log(F)) + (3*x^3*(3*d^2*f - 2*c*d*g))/(3*b*d - 4*c^2 )),x)
Output:
(F^a*F^(c*x^2)*F^(d*x^3)*F^(b*x)*(b*g - 2*c*f - 2*c*g*x + 3*d*f*x))/(log(F )*(3*b*d - 4*c^2))
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int F^{a+b x+c x^2+d x^3} \left (f x+g x^2-\frac {3 \left (3 d^2 f-2 c d g\right ) x^3}{4 c^2-3 b d}+\frac {-\frac {9 d \left (3 d^2 f-2 c d g\right )}{4 c^2-3 b d}+3 b d g \log (F)+\frac {6 b c \left (3 d^2 f-2 c d g\right ) \log (F)}{4 c^2-3 b d}}{9 d^2 \log (F)}\right ) \, dx=\frac {f^{d \,x^{3}+c \,x^{2}+b x +a} \left (-2 c g x +3 d f x +b g -2 c f \right )}{\mathrm {log}\left (f \right ) \left (3 b d -4 c^{2}\right )} \] Input:
int(F^(d*x^3+c*x^2+b*x+a)*(f*x+g*x^2-3*(-2*c*d*g+3*d^2*f)*x^3/(-3*b*d+4*c^ 2)+1/9*(-9*d*(-2*c*d*g+3*d^2*f)/(-3*b*d+4*c^2)+3*b*d*g*log(F)+6*b*c*(-2*c* d*g+3*d^2*f)*log(F)/(-3*b*d+4*c^2))/d^2/log(F)),x)
Output:
(f**(a + b*x + c*x**2 + d*x**3)*(b*g - 2*c*f - 2*c*g*x + 3*d*f*x))/(log(f) *(3*b*d - 4*c**2))