Integrand size = 15, antiderivative size = 181 \[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=-\frac {b^3 e^{-a+\frac {b c}{d}} \Gamma \left (-2,\frac {b (c+d x)}{d}\right )}{3 d^3 (b c-a d)}-\frac {b^3 e^{-a+\frac {b c}{d}} \Gamma \left (-1,\frac {b (c+d x)}{d}\right )}{3 d^2 (b c-a d)^2}+\frac {b^3 \Gamma (0,a+b x)}{3 d (b c-a d)^3}-\frac {\Gamma (0,a+b x)}{3 d (c+d x)^3}-\frac {b^3 e^{-a+\frac {b c}{d}} \Gamma \left (0,\frac {b (c+d x)}{d}\right )}{3 d (b c-a d)^3} \] Output:
-1/3*b*exp(-a+b*c/d)/(d*x+c)^2/d*Ei(3,b*(d*x+c)/d)/(-a*d+b*c)-1/3*b^2*exp( -a+b*c/d)/(d*x+c)/d*Ei(2,b*(d*x+c)/d)/(-a*d+b*c)^2+1/3*b^3*Ei(1,b*x+a)/d/( -a*d+b*c)^3-1/3*Ei(1,b*x+a)/d/(d*x+c)^3-1/3*b^3*exp(-a+b*c/d)*Ei(1,b*(d*x+ c)/d)/d/(-a*d+b*c)^3
Time = 0.46 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.99 \[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\frac {\frac {2 b^3 d^2 \operatorname {ExpIntegralEi}(-a-b x)}{(-b c+a d)^3}+\frac {b^3 \left (b^2 c^2-2 (1+a) b c d+\left (2+2 a+a^2\right ) d^2\right ) e^{-a+\frac {b c}{d}} \operatorname {ExpIntegralEi}\left (-\frac {b (c+d x)}{d}\right )}{(b c-a d)^3}+\frac {d \left (\frac {b e^{-a-b x} (c+d x) \left (a d^2+b^2 c (c+d x)-b d ((3+a) c+(2+a) d x)\right )}{(b c-a d)^2}-2 d \Gamma (0,a+b x)\right )}{(c+d x)^3}}{6 d^3} \] Input:
Integrate[Gamma[0, a + b*x]/(c + d*x)^4,x]
Output:
((2*b^3*d^2*ExpIntegralEi[-a - b*x])/(-(b*c) + a*d)^3 + (b^3*(b^2*c^2 - 2* (1 + a)*b*c*d + (2 + 2*a + a^2)*d^2)*E^(-a + (b*c)/d)*ExpIntegralEi[-((b*( c + d*x))/d)])/(b*c - a*d)^3 + (d*((b*E^(-a - b*x)*(c + d*x)*(a*d^2 + b^2* c*(c + d*x) - b*d*((3 + a)*c + (2 + a)*d*x)))/(b*c - a*d)^2 - 2*d*Gamma[0, a + b*x]))/(c + d*x)^3)/(6*d^3)
Time = 0.87 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7119, 7293, 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.
\(\displaystyle \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle -\frac {b \int \frac {e^{-a-b x}}{(a+b x) (c+d x)^3}dx}{3 d}-\frac {\Gamma (0,a+b x)}{3 d (c+d x)^3}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {b \int \left (\frac {e^{-a-b x} b^3}{(b c-a d)^3 (a+b x)}-\frac {d e^{-a-b x} b^2}{(b c-a d)^3 (c+d x)}-\frac {d e^{-a-b x} b}{(b c-a d)^2 (c+d x)^2}-\frac {d e^{-a-b x}}{(b c-a d) (c+d x)^3}\right )dx}{3 d}-\frac {\Gamma (0,a+b x)}{3 d (c+d x)^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (-\frac {b^2 e^{\frac {b c}{d}-a} \operatorname {ExpIntegralEi}\left (-\frac {b (c+d x)}{d}\right )}{2 d^2 (b c-a d)}+\frac {b^2 \operatorname {ExpIntegralEi}(-a-b x)}{(b c-a d)^3}+\frac {b^2 e^{\frac {b c}{d}-a} \operatorname {ExpIntegralEi}\left (-\frac {b (c+d x)}{d}\right )}{d (b c-a d)^2}-\frac {b^2 e^{\frac {b c}{d}-a} \operatorname {ExpIntegralEi}\left (-\frac {b (c+d x)}{d}\right )}{(b c-a d)^3}-\frac {b e^{-a-b x}}{2 d (c+d x) (b c-a d)}+\frac {b e^{-a-b x}}{(c+d x) (b c-a d)^2}+\frac {e^{-a-b x}}{2 (c+d x)^2 (b c-a d)}\right )}{3 d}-\frac {\Gamma (0,a+b x)}{3 d (c+d x)^3}\) |
Input:
Int[Gamma[0, a + b*x]/(c + d*x)^4,x]
Output:
-1/3*(b*(E^(-a - b*x)/(2*(b*c - a*d)*(c + d*x)^2) + (b*E^(-a - b*x))/((b*c - a*d)^2*(c + d*x)) - (b*E^(-a - b*x))/(2*d*(b*c - a*d)*(c + d*x)) + (b^2 *ExpIntegralEi[-a - b*x])/(b*c - a*d)^3 - (b^2*E^(-a + (b*c)/d)*ExpIntegra lEi[-((b*(c + d*x))/d)])/(b*c - a*d)^3 + (b^2*E^(-a + (b*c)/d)*ExpIntegral Ei[-((b*(c + d*x))/d)])/(d*(b*c - a*d)^2) - (b^2*E^(-a + (b*c)/d)*ExpInteg ralEi[-((b*(c + d*x))/d)])/(2*d^2*(b*c - a*d))))/d - Gamma[0, a + b*x]/(3* d*(c + d*x)^3)
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]
Time = 3.49 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.68
method | result | size |
parts | \(-\frac {\operatorname {expIntegral}_{1}\left (b x +a \right )}{3 d \left (d x +c \right )^{3}}-\frac {\frac {b^{3} \operatorname {expIntegral}_{1}\left (b x +a \right )}{\left (a d -c b \right )^{3}}-\frac {b^{3} {\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{\left (a d -c b \right )^{3}}+\frac {b^{3} \left (-\frac {{\mathrm e}^{-b x -a}}{-b x -a +\frac {a d -c b}{d}}-{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )\right )}{\left (a d -c b \right )^{2} d}+\frac {b^{3} \left (-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )^{2}}-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )}-\frac {{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{2}\right )}{\left (a d -c b \right ) d^{2}}}{3 d}\) | \(304\) |
derivativedivides | \(\frac {\frac {b^{4} \operatorname {expIntegral}_{1}\left (b x +a \right )}{3 \left (a d -c b -d \left (b x +a \right )\right )^{3} d}+\frac {b^{4} \left (-\frac {-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )^{2}}-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )}-\frac {{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{2}}{\left (a d -c b \right ) d^{2}}+\frac {{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{\left (a d -c b \right )^{3}}-\frac {\operatorname {expIntegral}_{1}\left (b x +a \right )}{\left (a d -c b \right )^{3}}-\frac {-\frac {{\mathrm e}^{-b x -a}}{-b x -a +\frac {a d -c b}{d}}-{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{\left (a d -c b \right )^{2} d}\right )}{3 d}}{b}\) | \(315\) |
default | \(\frac {\frac {b^{4} \operatorname {expIntegral}_{1}\left (b x +a \right )}{3 \left (a d -c b -d \left (b x +a \right )\right )^{3} d}+\frac {b^{4} \left (-\frac {-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )^{2}}-\frac {{\mathrm e}^{-b x -a}}{2 \left (-b x -a +\frac {a d -c b}{d}\right )}-\frac {{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{2}}{\left (a d -c b \right ) d^{2}}+\frac {{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{\left (a d -c b \right )^{3}}-\frac {\operatorname {expIntegral}_{1}\left (b x +a \right )}{\left (a d -c b \right )^{3}}-\frac {-\frac {{\mathrm e}^{-b x -a}}{-b x -a +\frac {a d -c b}{d}}-{\mathrm e}^{-\frac {a d -c b}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {a d -c b}{d}\right )}{\left (a d -c b \right )^{2} d}\right )}{3 d}}{b}\) | \(315\) |
Input:
int(Ei(1,b*x+a)/(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/3*Ei(1,b*x+a)/d/(d*x+c)^3-1/3/d*(b^3/(a*d-b*c)^3*Ei(1,b*x+a)-b^3/(a*d-b *c)^3*exp(-(a*d-b*c)/d)*Ei(1,b*x+a-(a*d-b*c)/d)+b^3/(a*d-b*c)^2/d*(-exp(-b *x-a)/(-b*x-a+(a*d-b*c)/d)-exp(-(a*d-b*c)/d)*Ei(1,b*x+a-(a*d-b*c)/d))+b^3/ (a*d-b*c)/d^2*(-1/2*exp(-b*x-a)/(-b*x-a+(a*d-b*c)/d)^2-1/2*exp(-b*x-a)/(-b *x-a+(a*d-b*c)/d)-1/2*exp(-(a*d-b*c)/d)*Ei(1,b*x+a-(a*d-b*c)/d)))
Exception generated. \[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(exp_integral_e(1,b*x+a)/(d*x+c)^4,x, algorithm="fricas")
Output:
Exception raised: TypeError >> An error occurred when FriCAS evaluated ((( (d)*(x))+(c))^(((-4)::EXPR INT)))*(exp_integral_e(((1)::EXPR INT),((b)*(x) )+(a))): There are no library operations named exp_integral_e Use HyperDo
Timed out. \[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\text {Timed out} \] Input:
integrate(expint(1,b*x+a)/(d*x+c)**4,x)
Output:
Timed out
\[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\int { \frac {E_{1}\left (b x + a\right )}{{\left (d x + c\right )}^{4}} \,d x } \] Input:
integrate(exp_integral_e(1,b*x+a)/(d*x+c)^4,x, algorithm="maxima")
Output:
integrate(exp_integral_e(1, b*x + a)/(d*x + c)^4, x)
\[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\int { \frac {E_{1}\left (b x + a\right )}{{\left (d x + c\right )}^{4}} \,d x } \] Input:
integrate(exp_integral_e(1,b*x+a)/(d*x+c)^4,x, algorithm="giac")
Output:
integrate(exp_integral_e(1, b*x + a)/(d*x + c)^4, x)
Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.09 \[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\int \frac {\mathrm {expint}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \] Input:
int(expint(a + b*x)/(c + d*x)^4,x)
Output:
int(expint(a + b*x)/(c + d*x)^4, x)
\[ \int \frac {\Gamma (0,a+b x)}{(c+d x)^4} \, dx=\int \frac {\mathit {ei} \left (1, b x +a \right )}{d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}}d x \] Input:
int(Ei(1,b*x+a)/(d*x+c)^4,x)
Output:
int(ei(1,a + b*x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4),x)