Integrand size = 18, antiderivative size = 80 \[ \int e^{-a-b x} (a+b x)^3 \, dx=-\frac {6 e^{-a-b x}}{b}-\frac {6 e^{-a-b x} (a+b x)}{b}-\frac {3 e^{-a-b x} (a+b x)^2}{b}-\frac {e^{-a-b x} (a+b x)^3}{b} \] Output:
-6*exp(-b*x-a)/b-6*exp(-b*x-a)*(b*x+a)/b-3*exp(-b*x-a)*(b*x+a)^2/b-exp(-b* x-a)*(b*x+a)^3/b
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.51 \[ \int e^{-a-b x} (a+b x)^3 \, dx=\frac {e^{-a-b x} \left (-6-6 (a+b x)-3 (a+b x)^2-(a+b x)^3\right )}{b} \] Input:
Integrate[E^(-a - b*x)*(a + b*x)^3,x]
Output:
(E^(-a - b*x)*(-6 - 6*(a + b*x) - 3*(a + b*x)^2 - (a + b*x)^3))/b
Time = 0.52 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2607, 2607, 2607, 2624}
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.
\(\displaystyle \int e^{-a-b x} (a+b x)^3 \, dx\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle 3 \int e^{-a-b x} (a+b x)^2dx-\frac {e^{-a-b x} (a+b x)^3}{b}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle 3 \left (2 \int e^{-a-b x} (a+b x)dx-\frac {e^{-a-b x} (a+b x)^2}{b}\right )-\frac {e^{-a-b x} (a+b x)^3}{b}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle 3 \left (2 \left (\int e^{-a-b x}dx-\frac {e^{-a-b x} (a+b x)}{b}\right )-\frac {e^{-a-b x} (a+b x)^2}{b}\right )-\frac {e^{-a-b x} (a+b x)^3}{b}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle 3 \left (2 \left (-\frac {e^{-a-b x} (a+b x)}{b}-\frac {e^{-a-b x}}{b}\right )-\frac {e^{-a-b x} (a+b x)^2}{b}\right )-\frac {e^{-a-b x} (a+b x)^3}{b}\) |
Input:
Int[E^(-a - b*x)*(a + b*x)^3,x]
Output:
-((E^(-a - b*x)*(a + b*x)^3)/b) + 3*(-((E^(-a - b*x)*(a + b*x)^2)/b) + 2*( -(E^(-a - b*x)/b) - (E^(-a - b*x)*(a + b*x))/b))
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Simp[(c + d*x)^m*((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Simp[d*(m/(f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x)))^ n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2* m] && !TrueQ[$UseGamma]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(-\frac {\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +3 b^{2} x^{2}+a^{3}+6 a b x +3 a^{2}+6 b x +6 a +6\right ) {\mathrm e}^{-b x -a}}{b}\) | \(68\) |
risch | \(-\frac {\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +3 b^{2} x^{2}+a^{3}+6 a b x +3 a^{2}+6 b x +6 a +6\right ) {\mathrm e}^{-b x -a}}{b}\) | \(68\) |
orering | \(-\frac {\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +3 b^{2} x^{2}+a^{3}+6 a b x +3 a^{2}+6 b x +6 a +6\right ) {\mathrm e}^{-b x -a}}{b}\) | \(68\) |
derivativedivides | \(\frac {{\mathrm e}^{-b x -a} \left (-b x -a \right )^{3}-3 \left (-b x -a \right )^{2} {\mathrm e}^{-b x -a}+6 \left (-b x -a \right ) {\mathrm e}^{-b x -a}-6 \,{\mathrm e}^{-b x -a}}{b}\) | \(77\) |
default | \(\frac {{\mathrm e}^{-b x -a} \left (-b x -a \right )^{3}-3 \left (-b x -a \right )^{2} {\mathrm e}^{-b x -a}+6 \left (-b x -a \right ) {\mathrm e}^{-b x -a}-6 \,{\mathrm e}^{-b x -a}}{b}\) | \(77\) |
norman | \(\left (-3 a b -3 b \right ) x^{2} {\mathrm e}^{-b x -a}+\left (-3 a^{2}-6 a -6\right ) x \,{\mathrm e}^{-b x -a}-b^{2} x^{3} {\mathrm e}^{-b x -a}-\frac {\left (a^{3}+3 a^{2}+6 a +6\right ) {\mathrm e}^{-b x -a}}{b}\) | \(88\) |
meijerg | \(\frac {{\mathrm e}^{-a} \left (6-\frac {\left (4 b^{3} x^{3}+12 b^{2} x^{2}+24 b x +24\right ) {\mathrm e}^{-b x}}{4}\right )}{b}+\frac {3 \,{\mathrm e}^{-a} a \left (2-\frac {\left (3 b^{2} x^{2}+6 b x +6\right ) {\mathrm e}^{-b x}}{3}\right )}{b}+\frac {3 \,{\mathrm e}^{-a} a^{2} \left (1-\frac {\left (2 b x +2\right ) {\mathrm e}^{-b x}}{2}\right )}{b}+\frac {{\mathrm e}^{-a} a^{3} \left (1-{\mathrm e}^{-b x}\right )}{b}\) | \(121\) |
parts | \(-b^{2} x^{3} {\mathrm e}^{-b x -a}-3 \,{\mathrm e}^{-b x -a} b a \,x^{2}-3 \,{\mathrm e}^{-b x -a} a^{2} x -\frac {{\mathrm e}^{-b x -a} a^{3}}{b}-\frac {3 \left (\left (-b x -a \right )^{2} {\mathrm e}^{-b x -a}-2 \left (-b x -a \right ) {\mathrm e}^{-b x -a}+2 \,{\mathrm e}^{-b x -a}\right )}{b}\) | \(123\) |
parallelrisch | \(-\frac {{\mathrm e}^{-b x -a} x^{3} b^{3}+3 x^{2} {\mathrm e}^{-b x -a} a \,b^{2}+3 b^{2} {\mathrm e}^{-b x -a} x^{2}+3 x \,{\mathrm e}^{-b x -a} a^{2} b +6 a b \,{\mathrm e}^{-b x -a} x +{\mathrm e}^{-b x -a} a^{3}+6 b \,{\mathrm e}^{-b x -a} x +3 a^{2} {\mathrm e}^{-b x -a}+6 a \,{\mathrm e}^{-b x -a}+6 \,{\mathrm e}^{-b x -a}}{b}\) | \(151\) |
Input:
int(exp(-b*x-a)*(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
-(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+3*b^2*x^2+a^3+6*a*b*x+3*a^2+6*b*x+6*a+6)*e xp(-b*x-a)/b
Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int e^{-a-b x} (a+b x)^3 \, dx=-\frac {{\left (b^{3} x^{3} + 3 \, {\left (a + 1\right )} b^{2} x^{2} + a^{3} + 3 \, {\left (a^{2} + 2 \, a + 2\right )} b x + 3 \, a^{2} + 6 \, a + 6\right )} e^{\left (-b x - a\right )}}{b} \] Input:
integrate(exp(-b*x-a)*(b*x+a)^3,x, algorithm="fricas")
Output:
-(b^3*x^3 + 3*(a + 1)*b^2*x^2 + a^3 + 3*(a^2 + 2*a + 2)*b*x + 3*a^2 + 6*a + 6)*e^(-b*x - a)/b
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30 \[ \int e^{-a-b x} (a+b x)^3 \, dx=\begin {cases} \frac {\left (- a^{3} - 3 a^{2} b x - 3 a^{2} - 3 a b^{2} x^{2} - 6 a b x - 6 a - b^{3} x^{3} - 3 b^{2} x^{2} - 6 b x - 6\right ) e^{- a - b x}}{b} & \text {for}\: b \neq 0 \\a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(exp(-b*x-a)*(b*x+a)**3,x)
Output:
Piecewise(((-a**3 - 3*a**2*b*x - 3*a**2 - 3*a*b**2*x**2 - 6*a*b*x - 6*a - b**3*x**3 - 3*b**2*x**2 - 6*b*x - 6)*exp(-a - b*x)/b, Ne(b, 0)), (a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4, True))
Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29 \[ \int e^{-a-b x} (a+b x)^3 \, dx=-\frac {3 \, {\left (b x + 1\right )} a^{2} e^{\left (-b x - a\right )}}{b} - \frac {a^{3} e^{\left (-b x - a\right )}}{b} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, b x + 2\right )} a e^{\left (-b x - a\right )}}{b} - \frac {{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{b} \] Input:
integrate(exp(-b*x-a)*(b*x+a)^3,x, algorithm="maxima")
Output:
-3*(b*x + 1)*a^2*e^(-b*x - a)/b - a^3*e^(-b*x - a)/b - 3*(b^2*x^2 + 2*b*x + 2)*a*e^(-b*x - a)/b - (b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09 \[ \int e^{-a-b x} (a+b x)^3 \, dx=-\frac {{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + 3 \, b^{5} x^{2} + a^{3} b^{3} + 6 \, a b^{4} x + 3 \, a^{2} b^{3} + 6 \, b^{4} x + 6 \, a b^{3} + 6 \, b^{3}\right )} e^{\left (-b x - a\right )}}{b^{4}} \] Input:
integrate(exp(-b*x-a)*(b*x+a)^3,x, algorithm="giac")
Output:
-(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + 3*b^5*x^2 + a^3*b^3 + 6*a*b^4*x + 3*a^2*b^3 + 6*b^4*x + 6*a*b^3 + 6*b^3)*e^(-b*x - a)/b^4
Time = 22.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int e^{-a-b x} (a+b x)^3 \, dx=-x\,{\mathrm {e}}^{-a-b\,x}\,\left (3\,a^2+3\,a\,b\,x+6\,a+b^2\,x^2+3\,b\,x+6\right )-\frac {{\mathrm {e}}^{-a-b\,x}\,\left (a^3+3\,a^2+6\,a+6\right )}{b} \] Input:
int(exp(- a - b*x)*(a + b*x)^3,x)
Output:
- x*exp(- a - b*x)*(6*a + 3*b*x + 3*a^2 + b^2*x^2 + 3*a*b*x + 6) - (exp(- a - b*x)*(6*a + 3*a^2 + a^3 + 6))/b
Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86 \[ \int e^{-a-b x} (a+b x)^3 \, dx=\frac {-b^{3} x^{3}-3 a \,b^{2} x^{2}-3 a^{2} b x -3 b^{2} x^{2}-a^{3}-6 a b x -3 a^{2}-6 b x -6 a -6}{e^{b x +a} b} \] Input:
int(exp(-b*x-a)*(b*x+a)^3,x)
Output:
( - a**3 - 3*a**2*b*x - 3*a**2 - 3*a*b**2*x**2 - 6*a*b*x - 6*a - b**3*x**3 - 3*b**2*x**2 - 6*b*x - 6)/(e**(a + b*x)*b)