Integrand size = 84, antiderivative size = 42 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {F^{a+b x+c x^2+d x^3} (3 d g-2 c h+3 d h x)}{9 d^2 \log (F)} \] Output:
1/9*F^(d*x^3+c*x^2+b*x+a)*(3*d*h*x-2*c*h+3*d*g)/d^2/ln(F)
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {F^{a+x (b+x (c+d x))} (-2 c h+3 d (g+h x))}{9 d^2 \log (F)} \] Input:
Integrate[F^(a + b*x + c*x^2 + d*x^3)*(((6*c*d*g - 4*c^2*h + 3*b*d*h)*x)/( 9*d^2) + g*x^2 + h*x^3 + (3*d*h + 3*b*d*g*Log[F] - 2*b*c*h*Log[F])/(9*d^2* Log[F])),x]
Output:
(F^(a + x*(b + x*(c + d*x)))*(-2*c*h + 3*d*(g + h*x)))/(9*d^2*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.
\(\displaystyle \int F^{a+b x+c x^2+d x^3} \left (\frac {x \left (3 b d h-4 c^2 h+6 c d g\right )}{9 d^2}+\frac {-2 b c h \log (F)+3 b d g \log (F)+3 d h}{9 d^2 \log (F)}+g x^2+h x^3\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int F^{a+b x+c x^2+d x^3} \left (\frac {x \left (3 b d h-4 c^2 h+6 c d g\right )}{9 d^2}+\frac {b \log (F) (3 d g-2 c h)+3 d h}{9 d^2 \log (F)}+g x^2+h x^3\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (3 b d h-4 c^2 h+6 c d g\right ) F^{a+b x+c x^2+d x^3}}{9 d^2}+\frac {F^{a+b x+c x^2+d x^3} (b \log (F) (3 d g-2 c h)+3 d h)}{9 d^2 \log (F)}+g x^2 F^{a+b x+c x^2+d x^3}+h x^3 F^{a+b x+c x^2+d x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (3 b d h-4 c^2 h+6 c d g\right ) \int F^{d x^3+c x^2+b x+a} xdx}{9 d^2}+\frac {(b \log (F) (3 d g-2 c h)+3 d h) \int F^{d x^3+c x^2+b x+a}dx}{9 d^2 \log (F)}+g \int F^{d x^3+c x^2+b x+a} x^2dx+h \int F^{d x^3+c x^2+b x+a} x^3dx\) |
Input:
Int[F^(a + b*x + c*x^2 + d*x^3)*(((6*c*d*g - 4*c^2*h + 3*b*d*h)*x)/(9*d^2) + g*x^2 + h*x^3 + (3*d*h + 3*b*d*g*Log[F] - 2*b*c*h*Log[F])/(9*d^2*Log[F] )),x]
Output:
$Aborted
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98
method | result | size |
gosper | \(-\frac {F^{d \,x^{3}+c \,x^{2}+b x +a} \left (-3 d h x +2 c h -3 d g \right )}{9 d^{2} \ln \left (F \right )}\) | \(41\) |
risch | \(-\frac {F^{d \,x^{3}+c \,x^{2}+b x +a} \left (-3 d h x +2 c h -3 d g \right )}{9 d^{2} \ln \left (F \right )}\) | \(41\) |
norman | \(\frac {-\frac {\left (2 c h -3 d g \right ) {\mathrm e}^{\left (d \,x^{3}+c \,x^{2}+b x +a \right ) \ln \left (F \right )}}{9 d \ln \left (F \right )}+\frac {h x \,{\mathrm e}^{\left (d \,x^{3}+c \,x^{2}+b x +a \right ) \ln \left (F \right )}}{3 \ln \left (F \right )}}{d}\) | \(70\) |
parallelrisch | \(\frac {3 d h \,F^{d \,x^{3}+c \,x^{2}+b x +a} x -2 F^{d \,x^{3}+c \,x^{2}+b x +a} c h +3 F^{d \,x^{3}+c \,x^{2}+b x +a} d g}{9 d^{2} \ln \left (F \right )}\) | \(75\) |
Input:
int(F^(d*x^3+c*x^2+b*x+a)*(1/9*(3*b*d*h-4*c^2*h+6*c*d*g)*x/d^2+g*x^2+h*x^3 +1/9*(3*d*h+3*b*d*g*ln(F)-2*b*c*h*ln(F))/d^2/ln(F)),x,method=_RETURNVERBOS E)
Output:
-1/9*F^(d*x^3+c*x^2+b*x+a)*(-3*d*h*x+2*c*h-3*d*g)/d^2/ln(F)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {{\left (3 \, d h x + 3 \, d g - 2 \, c h\right )} F^{d x^{3} + c x^{2} + b x + a}}{9 \, d^{2} \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(1/9*(3*b*d*h-4*c^2*h+6*c*d*g)*x/d^2+g*x^2 +h*x^3+1/9*(3*d*h+3*b*d*g*log(F)-2*b*c*h*log(F))/d^2/log(F)),x, algorithm= "fricas")
Output:
1/9*(3*d*h*x + 3*d*g - 2*c*h)*F^(d*x^3 + c*x^2 + b*x + a)/(d^2*log(F))
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (41) = 82\).
Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.95 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\begin {cases} \frac {F^{a + b x + c x^{2} + d x^{3}} \left (- 2 c h + 3 d g + 3 d h x\right )}{9 d^{2} \log {\left (F \right )}} & \text {for}\: d^{2} \log {\left (F \right )} \neq 0 \\\frac {g x^{3}}{3} + \frac {h x^{4}}{4} + \frac {x^{2} \cdot \left (3 b d h - 4 c^{2} h + 6 c d g\right )}{18 d^{2}} + \frac {x \left (- 2 b c h \log {\left (F \right )} + 3 b d g \log {\left (F \right )} + 3 d h\right )}{9 d^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(d*x**3+c*x**2+b*x+a)*(1/9*(3*b*d*h-4*c**2*h+6*c*d*g)*x/d**2+ g*x**2+h*x**3+1/9*(3*d*h+3*b*d*g*ln(F)-2*b*c*h*ln(F))/d**2/ln(F)),x)
Output:
Piecewise((F**(a + b*x + c*x**2 + d*x**3)*(-2*c*h + 3*d*g + 3*d*h*x)/(9*d* *2*log(F)), Ne(d**2*log(F), 0)), (g*x**3/3 + h*x**4/4 + x**2*(3*b*d*h - 4* c**2*h + 6*c*d*g)/(18*d**2) + x*(-2*b*c*h*log(F) + 3*b*d*g*log(F) + 3*d*h) /(9*d**2*log(F)), True))
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {{\left (3 \, F^{a} d h x + {\left (3 \, d g - 2 \, c h\right )} F^{a}\right )} e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )}}{9 \, d^{2} \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(1/9*(3*b*d*h-4*c^2*h+6*c*d*g)*x/d^2+g*x^2 +h*x^3+1/9*(3*d*h+3*b*d*g*log(F)-2*b*c*h*log(F))/d^2/log(F)),x, algorithm= "maxima")
Output:
1/9*(3*F^a*d*h*x + (3*d*g - 2*c*h)*F^a)*e^(d*x^3*log(F) + c*x^2*log(F) + b *x*log(F))/(d^2*log(F))
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (40) = 80\).
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.26 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {3 \, F^{a} d h x e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )} + 3 \, F^{a} d g e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )} - 2 \, F^{a} c h e^{\left (d x^{3} \log \left (F\right ) + c x^{2} \log \left (F\right ) + b x \log \left (F\right )\right )}}{9 \, d^{2} \log \left (F\right )} \] Input:
integrate(F^(d*x^3+c*x^2+b*x+a)*(1/9*(3*b*d*h-4*c^2*h+6*c*d*g)*x/d^2+g*x^2 +h*x^3+1/9*(3*d*h+3*b*d*g*log(F)-2*b*c*h*log(F))/d^2/log(F)),x, algorithm= "giac")
Output:
1/9*(3*F^a*d*h*x*e^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)) + 3*F^a*d*g* e^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)) - 2*F^a*c*h*e^(d*x^3*log(F) + c*x^2*log(F) + b*x*log(F)))/(d^2*log(F))
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.64 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=-\frac {F^a\,F^{c\,x^2}\,F^{d\,x^3}\,F^{b\,x}\,\left (2\,c\,h-3\,d\,g\right )-3\,F^a\,F^{c\,x^2}\,F^{d\,x^3}\,F^{b\,x}\,d\,h\,x}{9\,d^2\,\ln \left (F\right )} \] Input:
int(F^(a + b*x + c*x^2 + d*x^3)*(g*x^2 + h*x^3 + ((d*h)/3 - (2*b*c*h*log(F ))/9 + (b*d*g*log(F))/3)/(d^2*log(F)) + (x*(3*b*d*h - 4*c^2*h + 6*c*d*g))/ (9*d^2)),x)
Output:
-(F^a*F^(c*x^2)*F^(d*x^3)*F^(b*x)*(2*c*h - 3*d*g) - 3*F^a*F^(c*x^2)*F^(d*x ^3)*F^(b*x)*d*h*x)/(9*d^2*log(F))
Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int F^{a+b x+c x^2+d x^3} \left (\frac {\left (6 c d g-4 c^2 h+3 b d h\right ) x}{9 d^2}+g x^2+h x^3+\frac {3 d h+3 b d g \log (F)-2 b c h \log (F)}{9 d^2 \log (F)}\right ) \, dx=\frac {f^{d \,x^{3}+c \,x^{2}+b x +a} \left (3 d h x -2 c h +3 d g \right )}{9 \,\mathrm {log}\left (f \right ) d^{2}} \] Input:
int(F^(d*x^3+c*x^2+b*x+a)*(1/9*(3*b*d*h-4*c^2*h+6*c*d*g)*x/d^2+g*x^2+h*x^3 +1/9*(3*d*h+3*b*d*g*log(F)-2*b*c*h*log(F))/d^2/log(F)),x)
Output:
(f**(a + b*x + c*x**2 + d*x**3)*( - 2*c*h + 3*d*g + 3*d*h*x))/(9*log(f)*d* *2)