Integrand size = 13, antiderivative size = 69 \[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\frac {3 a e^{-a} \sqrt [3]{b x} \Gamma \left (\frac {2}{3},b x\right )}{\sqrt [3]{x}}+\frac {3 b e^{-a} x^{2/3} \Gamma \left (\frac {5}{3},b x\right )}{(b x)^{2/3}}-\frac {3 \Gamma (2,a+b x)}{\sqrt [3]{x}} \] Output:
3*a*(b*x)^(1/3)*GAMMA(2/3,b*x)/exp(a)/x^(1/3)+3*b*x^(2/3)*GAMMA(5/3,b*x)/e xp(a)/(b*x)^(2/3)-3*exp(-b*x-a)*(b*x+a+1)/x^(1/3)
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\frac {e^{-a} \left (3 b e^{-b x} x+(2+3 a) \sqrt [3]{b x} \Gamma \left (\frac {2}{3},b x\right )-3 e^a \Gamma (2,a+b x)\right )}{\sqrt [3]{x}} \] Input:
Integrate[Gamma[2, a + b*x]/x^(4/3),x]
Output:
((3*b*x)/E^(b*x) + (2 + 3*a)*(b*x)^(1/3)*Gamma[2/3, b*x] - 3*E^a*Gamma[2, a + b*x])/(E^a*x^(1/3))
Time = 0.55 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.
\(\displaystyle \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle -3 b \int \frac {e^{-a-b x} (a+b x)}{\sqrt [3]{x}}dx-\frac {3 \Gamma (2,a+b x)}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle -3 b \int \left (\frac {e^{-a-b x} a}{\sqrt [3]{x}}+b e^{-a-b x} x^{2/3}\right )dx-\frac {3 \Gamma (2,a+b x)}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 b \left (-\frac {e^{-a} x^{2/3} \Gamma \left (\frac {5}{3},b x\right )}{(b x)^{2/3}}-\frac {a e^{-a} \sqrt [3]{b x} \Gamma \left (\frac {2}{3},b x\right )}{b \sqrt [3]{x}}\right )-\frac {3 \Gamma (2,a+b x)}{\sqrt [3]{x}}\) |
Input:
Int[Gamma[2, a + b*x]/x^(4/3),x]
Output:
-3*b*(-((a*(b*x)^(1/3)*Gamma[2/3, b*x])/(b*E^a*x^(1/3))) - (x^(2/3)*Gamma[ 5/3, b*x])/(E^a*(b*x)^(2/3))) - (3*Gamma[2, a + b*x])/x^(1/3)
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[$UseGamma]
Int[Gamma[n_, (a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Block[{$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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.45 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.70
method | result | size |
meijerg | \(b^{\frac {1}{3}} {\mathrm e}^{-a} \left (\frac {9 x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{10 \left (b x \right )^{\frac {1}{3}}}+\frac {3 \,{\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{2 x^{\frac {1}{3}} b^{\frac {1}{3}} \left (b x \right )^{\frac {1}{3}}}\right )+{\mathrm e}^{-a} a \,b^{\frac {1}{3}} \left (-\frac {27 x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{10 \left (b x \right )^{\frac {1}{3}}}-\frac {3 \left (3 b x +2\right ) {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{2 x^{\frac {4}{3}} b^{\frac {4}{3}} \left (b x \right )^{\frac {1}{3}}}\right )+b^{\frac {1}{3}} {\mathrm e}^{-a} \left (-\frac {27 x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{10 \left (b x \right )^{\frac {1}{3}}}-\frac {3 \left (3 b x +2\right ) {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{2 x^{\frac {4}{3}} b^{\frac {4}{3}} \left (b x \right )^{\frac {1}{3}}}\right )\) | \(186\) |
Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(4/3),x,method=_RETURNVERBOSE)
Output:
b^(1/3)*exp(-a)*(9/10*x^(2/3)*b^(2/3)/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM (1/3,5/6,b*x)+3/2/x^(1/3)/b^(1/3)/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM(4/3 ,5/6,b*x))+exp(-a)*a*b^(1/3)*(-27/10*x^(2/3)*b^(2/3)/(b*x)^(1/3)*exp(-1/2* b*x)*WhittakerM(1/3,5/6,b*x)-3/2/x^(4/3)/b^(4/3)*(3*b*x+2)/(b*x)^(1/3)*exp (-1/2*b*x)*WhittakerM(4/3,5/6,b*x))+b^(1/3)*exp(-a)*(-27/10*x^(2/3)*b^(2/3 )/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM(1/3,5/6,b*x)-3/2/x^(4/3)/b^(4/3)*(3 *b*x+2)/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM(4/3,5/6,b*x))
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.74 \[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\frac {{\left (3 \, a + 2\right )} b^{\frac {1}{3}} x e^{\left (-a\right )} \Gamma \left (\frac {2}{3}, b x\right ) + 3 \, {\left (b x e^{\left (-b x - a\right )} - \Gamma \left (2, b x + a\right )\right )} x^{\frac {2}{3}}}{x} \] Input:
integrate(gamma(2,b*x+a)/x^(4/3),x, algorithm="fricas")
Output:
((3*a + 2)*b^(1/3)*x*e^(-a)*gamma(2/3, b*x) + 3*(b*x*e^(-b*x - a) - gamma( 2, b*x + a))*x^(2/3))/x
Time = 4.82 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\left (- \frac {a \left (b x\right )^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}, b x\right )}{b x^{\frac {4}{3}}} - \frac {\sqrt [3]{b x} \Gamma \left (\frac {2}{3}, b x\right )}{\sqrt [3]{x}} - \frac {\left (b x\right )^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}, b x\right )}{b x^{\frac {4}{3}}}\right ) e^{- a} \] Input:
integrate(uppergamma(2,b*x+a)/x**(4/3),x)
Output:
(-a*(b*x)**(4/3)*uppergamma(-1/3, b*x)/(b*x**(4/3)) - (b*x)**(1/3)*upperga mma(2/3, b*x)/x**(1/3) - (b*x)**(4/3)*uppergamma(-1/3, b*x)/(b*x**(4/3)))* exp(-a)
\[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {4}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(4/3),x, algorithm="maxima")
Output:
integrate(gamma(2, b*x + a)/x^(4/3), x)
\[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {4}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(4/3),x, algorithm="giac")
Output:
integrate(gamma(2, b*x + a)/x^(4/3), x)
Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\frac {2\,{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{1/3}\,\Gamma \left (\frac {2}{3},b\,x\right )}{x^{1/3}}-\frac {3\,a\,{\mathrm {e}}^{-a-b\,x}}{x^{1/3}}-\frac {3\,{\mathrm {e}}^{-a-b\,x}}{x^{1/3}}+\frac {3\,a\,{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{1/3}\,\Gamma \left (\frac {2}{3},b\,x\right )}{x^{1/3}} \] Input:
int((exp(- a - b*x)*(a + b*x + 1))/x^(4/3),x)
Output:
(2*exp(-a)*(b*x)^(1/3)*igamma(2/3, b*x))/x^(1/3) - (3*a*exp(- a - b*x))/x^ (1/3) - (3*exp(- a - b*x))/x^(1/3) + (3*a*exp(-a)*(b*x)^(1/3)*igamma(2/3, b*x))/x^(1/3)
\[ \int \frac {\Gamma (2,a+b x)}{x^{4/3}} \, dx=\frac {3 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right ) a +2 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right )-3}{3 x^{\frac {1}{3}} e^{b x +a}} \] Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(4/3),x)
Output:
(3*x**(1/3)*e**(b*x)*int(1/(x**(1/3)*e**(b*x)*x),x)*a + 2*x**(1/3)*e**(b*x )*int(1/(x**(1/3)*e**(b*x)*x),x) - 3)/(3*x**(1/3)*e**(a + b*x))