Integrand size = 13, antiderivative size = 74 \[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=-\frac {3 a e^{-a} x^{2/3} \Gamma \left (\frac {5}{3},b x\right )}{2 (b x)^{2/3}}+\frac {3}{2} x^{2/3} \Gamma (2,a+b x)-\frac {3 e^{-a} x^{2/3} \Gamma \left (\frac {8}{3},b x\right )}{2 (b x)^{2/3}} \] Output:
-3/2*a*x^(2/3)*GAMMA(5/3,b*x)/exp(a)/(b*x)^(2/3)+3/2*x^(2/3)*exp(-b*x-a)*( b*x+a+1)-3/2*x^(2/3)*GAMMA(8/3,b*x)/exp(a)/(b*x)^(2/3)
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=-\frac {e^{-a-b x} x^{2/3} \left (2 (5+3 a) e^{b x} \Gamma \left (\frac {2}{3},b x\right )+3 (b x)^{2/3} \left (5+3 a+3 b x-3 e^{a+b x} \Gamma (2,a+b x)\right )\right )}{6 (b x)^{2/3}} \] Input:
Integrate[Gamma[2, a + b*x]/x^(1/3),x]
Output:
-1/6*(E^(-a - b*x)*x^(2/3)*(2*(5 + 3*a)*E^(b*x)*Gamma[2/3, b*x] + 3*(b*x)^ (2/3)*(5 + 3*a + 3*b*x - 3*E^(a + b*x)*Gamma[2, a + b*x])))/(b*x)^(2/3)
Time = 0.53 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle \frac {3}{2} b \int e^{-a-b x} x^{2/3} (a+b x)dx+\frac {3}{2} x^{2/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \frac {3}{2} b \int \left (b e^{-a-b x} x^{5/3}+a e^{-a-b x} x^{2/3}\right )dx+\frac {3}{2} x^{2/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{2} x^{2/3} \Gamma (2,a+b x)+\frac {3}{2} b \left (-\frac {a e^{-a} x^{2/3} \Gamma \left (\frac {5}{3},b x\right )}{b (b x)^{2/3}}-\frac {e^{-a} x^{2/3} \Gamma \left (\frac {8}{3},b x\right )}{b (b x)^{2/3}}\right )\) |
Input:
Int[Gamma[2, a + b*x]/x^(1/3),x]
Output:
(3*x^(2/3)*Gamma[2, a + b*x])/2 + (3*b*(-((a*x^(2/3)*Gamma[5/3, b*x])/(b*E ^a*(b*x)^(2/3))) - (x^(2/3)*Gamma[8/3, b*x])/(b*E^a*(b*x)^(2/3))))/2
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 = 0.42 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.92
method | result | size |
meijerg | \(\frac {3 \,{\mathrm e}^{-a -\frac {b x}{2}} x^{\frac {2}{3}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{5 \left (b x \right )^{\frac {1}{3}}}+\frac {{\mathrm e}^{-a} a \left (\frac {9 x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{10 \left (b x \right )^{\frac {1}{3}}}+\frac {3 \,{\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{2 x^{\frac {1}{3}} b^{\frac {1}{3}} \left (b x \right )^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}}+\frac {{\mathrm e}^{-a} \left (\frac {9 x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{10 \left (b x \right )^{\frac {1}{3}}}+\frac {3 \,{\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{2 x^{\frac {1}{3}} b^{\frac {1}{3}} \left (b x \right )^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}}\) | \(142\) |
Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(1/3),x,method=_RETURNVERBOSE)
Output:
3/5*exp(-a-1/2*b*x)*x^(2/3)/(b*x)^(1/3)*WhittakerM(1/3,5/6,b*x)+exp(-a)*a/ b^(2/3)*(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 ))+1/b^(2/3)*exp(-a)*(9/10*x^(2/3)*b^(2/3)/(b*x)^(1/3)*exp(-1/2*b*x)*Whitt akerM(1/3,5/6,b*x)+3/2/x^(1/3)/b^(1/3)/(b*x)^(1/3)*exp(-1/2*b*x)*Whittaker M(4/3,5/6,b*x))
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=-\frac {2 \, {\left (3 \, a + 5\right )} b^{\frac {1}{3}} e^{\left (-a\right )} \Gamma \left (\frac {2}{3}, b x\right ) + 3 \, {\left ({\left (3 \, b^{2} x + {\left (3 \, a + 5\right )} b\right )} e^{\left (-b x - a\right )} - 3 \, b \Gamma \left (2, b x + a\right )\right )} x^{\frac {2}{3}}}{6 \, b} \] Input:
integrate(gamma(2,b*x+a)/x^(1/3),x, algorithm="fricas")
Output:
-1/6*(2*(3*a + 5)*b^(1/3)*e^(-a)*gamma(2/3, b*x) + 3*((3*b^2*x + (3*a + 5) *b)*e^(-b*x - a) - 3*b*gamma(2, b*x + a))*x^(2/3))/b
Time = 6.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=\left (- \frac {a \sqrt [3]{b x} \Gamma \left (\frac {2}{3}, b x\right )}{b \sqrt [3]{x}} - \frac {x^{\frac {2}{3}} \Gamma \left (\frac {5}{3}, b x\right )}{\left (b x\right )^{\frac {2}{3}}} - \frac {\sqrt [3]{b x} \Gamma \left (\frac {2}{3}, b x\right )}{b \sqrt [3]{x}}\right ) e^{- a} \] Input:
integrate(uppergamma(2,b*x+a)/x**(1/3),x)
Output:
(-a*(b*x)**(1/3)*uppergamma(2/3, b*x)/(b*x**(1/3)) - x**(2/3)*uppergamma(5 /3, b*x)/(b*x)**(2/3) - (b*x)**(1/3)*uppergamma(2/3, b*x)/(b*x**(1/3)))*ex p(-a)
\[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {1}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(1/3),x, algorithm="maxima")
Output:
integrate(gamma(2, b*x + a)/x^(1/3), x)
\[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {1}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(1/3),x, algorithm="giac")
Output:
integrate(gamma(2, b*x + a)/x^(1/3), x)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=-x^{2/3}\,{\mathrm {e}}^{-a-b\,x}-\frac {2\,x^{2/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {2}{3},b\,x\right )}{3\,{\left (b\,x\right )}^{2/3}}-\frac {{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{1/3}\,\Gamma \left (\frac {2}{3},b\,x\right )}{b\,x^{1/3}}-\frac {a\,{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{1/3}\,\Gamma \left (\frac {2}{3},b\,x\right )}{b\,x^{1/3}} \] Input:
int((exp(- a - b*x)*(a + b*x + 1))/x^(1/3),x)
Output:
- x^(2/3)*exp(- a - b*x) - (2*x^(2/3)*exp(-a)*igamma(2/3, b*x))/(3*(b*x)^( 2/3)) - (exp(-a)*(b*x)^(1/3)*igamma(2/3, b*x))/(b*x^(1/3)) - (a*exp(-a)*(b *x)^(1/3)*igamma(2/3, b*x))/(b*x^(1/3))
\[ \int \frac {\Gamma (2,a+b x)}{\sqrt [3]{x}} \, dx=\frac {-3 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right ) a -5 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right )-9 a -9 b x -15}{9 x^{\frac {1}{3}} e^{b x +a} b} \] Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(1/3),x)
Output:
( - 3*x**(1/3)*e**(b*x)*int(1/(x**(1/3)*e**(b*x)*x),x)*a - 5*x**(1/3)*e**( b*x)*int(1/(x**(1/3)*e**(b*x)*x),x) - 9*a - 9*b*x - 15)/(9*x**(1/3)*e**(a + b*x)*b)