\(\int (c+d x)^3 \Gamma (-1,a+b x) \, dx\) [137]

Optimal result

Integrand size = 15, antiderivative size = 139 \[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=-\frac {3 d (b c-a d)^2 e^{-a-b x}}{2 b^4}-\frac {(b c-a d)^4 \Gamma (-1,a+b x)}{4 b^4 d}+\frac {(c+d x)^4 \Gamma (-1,a+b x)}{4 d}-\frac {(b c-a d)^3 \Gamma (0,a+b x)}{b^4}-\frac {d^2 (b c-a d) \Gamma (2,a+b x)}{b^4}-\frac {d^3 \Gamma (3,a+b x)}{4 b^4} \] Output:

-3/2*d*(-a*d+b*c)^2*exp(-b*x-a)/b^4-1/4*(-a*d+b*c)^4/(b*x+a)*Ei(2,b*x+a)/b 
^4/d+1/4*(d*x+c)^4/(b*x+a)*Ei(2,b*x+a)/d-(-a*d+b*c)^3*Ei(1,b*x+a)/b^4-d^2* 
(-a*d+b*c)*exp(-b*x-a)*(b*x+a+1)/b^4-1/2*d^3*exp(-b*x-a)*(1+b*x+a+1/2*(b*x 
+a)^2)/b^4
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(282\) vs. \(2(139)=278\).

Time = 0.33 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.03 \[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\frac {\left (4 (1+a) b^3 c^3-6 a (2+a) b^2 c^2 d+4 a^2 (3+a) b c d^2-a^3 (4+a) d^3\right ) \operatorname {ExpIntegralEi}(-a-b x)+\frac {e^{-a-b x} \left (a^3 (4 b c-3 d) d^2-a^4 d^3+a^2 d \left (-6 b^2 c^2+2 d^2+b d (8 c-d x)\right )+a \left (4 b^3 c^3-4 b c d^2-2 d^3+b^2 d \left (-6 c^2+4 c d x+d^2 x^2\right )\right )-b d x \left (2 d^2+2 b d (2 c+d x)+b^2 \left (6 c^2+4 c d x+d^2 x^2\right )\right )+b^4 e^{a+b x} x (a+b x) \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \Gamma (-1,a+b x)\right )}{a+b x}}{4 b^4} \] Input:

Integrate[(c + d*x)^3*Gamma[-1, a + b*x],x]
 

Output:

((4*(1 + a)*b^3*c^3 - 6*a*(2 + a)*b^2*c^2*d + 4*a^2*(3 + a)*b*c*d^2 - a^3* 
(4 + a)*d^3)*ExpIntegralEi[-a - b*x] + (E^(-a - b*x)*(a^3*(4*b*c - 3*d)*d^ 
2 - a^4*d^3 + a^2*d*(-6*b^2*c^2 + 2*d^2 + b*d*(8*c - d*x)) + a*(4*b^3*c^3 
- 4*b*c*d^2 - 2*d^3 + b^2*d*(-6*c^2 + 4*c*d*x + d^2*x^2)) - b*d*x*(2*d^2 + 
 2*b*d*(2*c + d*x) + b^2*(6*c^2 + 4*c*d*x + d^2*x^2)) + b^4*E^(a + b*x)*x* 
(a + b*x)*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Gamma[-1, a + b*x])) 
/(a + b*x))/(4*b^4)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7119, 2629, 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.

\(\displaystyle \int (c+d x)^3 \Gamma (-1,a+b x) \, dx\)

\(\Big \downarrow \) 7119

\(\displaystyle \frac {b \int \frac {e^{-a-b x} (c+d x)^4}{(a+b x)^2}dx}{4 d}+\frac {(c+d x)^4 \Gamma (-1,a+b x)}{4 d}\)

\(\Big \downarrow \) 2629

\(\displaystyle \frac {b \int \left (\frac {e^{-a-b x} (a+b x)^2 d^4}{b^4}+\frac {4 (b c-a d) e^{-a-b x} (a+b x) d^3}{b^4}+\frac {6 (b c-a d)^2 e^{-a-b x} d^2}{b^4}+\frac {4 (b c-a d)^3 e^{-a-b x} d}{b^4 (a+b x)}+\frac {(b c-a d)^4 e^{-a-b x}}{b^4 (a+b x)^2}\right )dx}{4 d}+\frac {(c+d x)^4 \Gamma (-1,a+b x)}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (-\frac {4 d^3 e^{-a-b x} (b c-a d)}{b^5}-\frac {4 d^3 e^{-a-b x} (a+b x) (b c-a d)}{b^5}-\frac {6 d^2 e^{-a-b x} (b c-a d)^2}{b^5}+\frac {4 d (b c-a d)^3 \operatorname {ExpIntegralEi}(-a-b x)}{b^5}-\frac {(b c-a d)^4 \operatorname {ExpIntegralEi}(-a-b x)}{b^5}-\frac {e^{-a-b x} (b c-a d)^4}{b^5 (a+b x)}-\frac {2 d^4 e^{-a-b x}}{b^5}-\frac {d^4 e^{-a-b x} (a+b x)^2}{b^5}-\frac {2 d^4 e^{-a-b x} (a+b x)}{b^5}\right )}{4 d}+\frac {(c+d x)^4 \Gamma (-1,a+b x)}{4 d}\)

Input:

Int[(c + d*x)^3*Gamma[-1, a + b*x],x]
 

Output:

(b*((-2*d^4*E^(-a - b*x))/b^5 - (4*d^3*(b*c - a*d)*E^(-a - b*x))/b^5 - (6* 
d^2*(b*c - a*d)^2*E^(-a - b*x))/b^5 - ((b*c - a*d)^4*E^(-a - b*x))/(b^5*(a 
 + b*x)) - (2*d^4*E^(-a - b*x)*(a + b*x))/b^5 - (4*d^3*(b*c - a*d)*E^(-a - 
 b*x)*(a + b*x))/b^5 - (d^4*E^(-a - b*x)*(a + b*x)^2)/b^5 + (4*d*(b*c - a* 
d)^3*ExpIntegralEi[-a - b*x])/b^5 - ((b*c - a*d)^4*ExpIntegralEi[-a - b*x] 
)/b^5))/(4*d) + ((c + d*x)^4*Gamma[-1, a + b*x])/(4*d)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2629

Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte 
grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ 
Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

rule 7119

Int[Gamma[n_, (a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> 
Block[{$UseGamma = True}, Simp[(c + d*x)^(m + 1)*(Gamma[n, a + b*x]/(d*(m + 
 1))), x] + Simp[b/(d*(m + 1))   Int[(c + d*x)^(m + 1)*((a + b*x)^(n - 1)/E 
^(a + b*x)), x], x]] /; FreeQ[{a, b, c, d, m, n}, x] && (IGtQ[m, 0] || IGtQ 
[n, 0] || IntegersQ[m, n]) && NeQ[m, -1]
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{3} \operatorname {expIntegral}_{2}\left (b x +a \right )}{b x +a}d x\]

Input:

int((d*x+c)^3/(b*x+a)*Ei(2,b*x+a),x)
 

Output:

int((d*x+c)^3/(b*x+a)*Ei(2,b*x+a),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (130) = 260\).

Time = 0.10 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.59 \[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=-\frac {{\left (b^{3} d^{3} x^{3} + 4 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d + 4 \, {\left (a^{2} + a\right )} b c d^{2} - {\left (a^{3} + 2 \, a^{2} - 2 \, a\right )} d^{3} + {\left (4 \, b^{3} c d^{2} - {\left (a - 2\right )} b^{2} d^{3}\right )} x^{2} + {\left (6 \, b^{3} c^{2} d - 4 \, {\left (a - 1\right )} b^{2} c d^{2} + {\left (a^{2} + 2\right )} b d^{3}\right )} x\right )} e^{\left (-b x - a\right )} - {\left (b^{5} d^{3} x^{5} + 4 \, {\left (a^{2} + a\right )} b^{3} c^{3} - 6 \, {\left (a^{3} + 2 \, a^{2}\right )} b^{2} c^{2} d + 4 \, {\left (a^{4} + 3 \, a^{3}\right )} b c d^{2} + {\left (4 \, b^{5} c d^{2} + a b^{4} d^{3}\right )} x^{4} - {\left (a^{5} + 4 \, a^{4}\right )} d^{3} + 2 \, {\left (3 \, b^{5} c^{2} d + 2 \, a b^{4} c d^{2}\right )} x^{3} + 2 \, {\left (2 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d\right )} x^{2} + {\left (4 \, {\left (2 \, a + 1\right )} b^{4} c^{3} - 6 \, {\left (a^{2} + 2 \, a\right )} b^{3} c^{2} d + 4 \, {\left (a^{3} + 3 \, a^{2}\right )} b^{2} c d^{2} - {\left (a^{4} + 4 \, a^{3}\right )} b d^{3}\right )} x\right )} \Gamma \left (-1, b x + a\right )}{4 \, {\left (b^{5} x + a b^{4}\right )}} \] Input:

integrate((d*x+c)^3*gamma(-1,b*x+a),x, algorithm="fricas")
 

Output:

-1/4*((b^3*d^3*x^3 + 4*b^3*c^3 - 6*a*b^2*c^2*d + 4*(a^2 + a)*b*c*d^2 - (a^ 
3 + 2*a^2 - 2*a)*d^3 + (4*b^3*c*d^2 - (a - 2)*b^2*d^3)*x^2 + (6*b^3*c^2*d 
- 4*(a - 1)*b^2*c*d^2 + (a^2 + 2)*b*d^3)*x)*e^(-b*x - a) - (b^5*d^3*x^5 + 
4*(a^2 + a)*b^3*c^3 - 6*(a^3 + 2*a^2)*b^2*c^2*d + 4*(a^4 + 3*a^3)*b*c*d^2 
+ (4*b^5*c*d^2 + a*b^4*d^3)*x^4 - (a^5 + 4*a^4)*d^3 + 2*(3*b^5*c^2*d + 2*a 
*b^4*c*d^2)*x^3 + 2*(2*b^5*c^3 + 3*a*b^4*c^2*d)*x^2 + (4*(2*a + 1)*b^4*c^3 
 - 6*(a^2 + 2*a)*b^3*c^2*d + 4*(a^3 + 3*a^2)*b^2*c*d^2 - (a^4 + 4*a^3)*b*d 
^3)*x)*gamma(-1, b*x + a))/(b^5*x + a*b^4)
 

Sympy [F]

\[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\int \frac {\left (c + d x\right )^{3} \operatorname {E}_{2}\left (a + b x\right )}{a + b x}\, dx \] Input:

integrate((d*x+c)**3*uppergamma(-1,b*x+a),x)
 

Output:

Integral((c + d*x)**3*expint(2, a + b*x)/(a + b*x), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \Gamma \left (-1, b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^3*gamma(-1,b*x+a),x, algorithm="maxima")
 

Output:

((b*x + a)*gamma(-1, b*x + a) + Ei(-b*x - a))*c^3/b + integrate(d^3*x^3*ga 
mma(-1, b*x + a) + 3*c*d^2*x^2*gamma(-1, b*x + a) + 3*c^2*d*x*gamma(-1, b* 
x + a), x)
 

Giac [F]

\[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \Gamma \left (-1, b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^3*gamma(-1,b*x+a),x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*gamma(-1, b*x + a), x)
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.17 \[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\int \frac {\mathrm {expint}\left (2,a+b\,x\right )\,{\left (c+d\,x\right )}^3}{a+b\,x} \,d x \] Input:

int((expint(2, a + b*x)*(c + d*x)^3)/(a + b*x),x)
 

Output:

int((expint(2, a + b*x)*(c + d*x)^3)/(a + b*x), x)
 

Reduce [F]

\[ \int (c+d x)^3 \Gamma (-1,a+b x) \, dx=\left (\int \frac {\mathit {ei} \left (2, b x +a \right )}{b x +a}d x \right ) c^{3}+\left (\int \frac {\mathit {ei} \left (2, b x +a \right ) x^{3}}{b x +a}d x \right ) d^{3}+3 \left (\int \frac {\mathit {ei} \left (2, b x +a \right ) x^{2}}{b x +a}d x \right ) c \,d^{2}+3 \left (\int \frac {\mathit {ei} \left (2, b x +a \right ) x}{b x +a}d x \right ) c^{2} d \] Input:

int((d*x+c)^3/(b*x+a)*Ei(2,b*x+a),x)
 

Output:

int(ei(2,a + b*x)/(a + b*x),x)*c**3 + int((ei(2,a + b*x)*x**3)/(a + b*x),x 
)*d**3 + 3*int((ei(2,a + b*x)*x**2)/(a + b*x),x)*c*d**2 + 3*int((ei(2,a + 
b*x)*x)/(a + b*x),x)*c**2*d