Integrand size = 13, antiderivative size = 80 \[ \int x^{2/3} \Gamma (2,a+b x) \, dx=\frac {3}{5} x^{5/3} \Gamma (2,a+b x)-\frac {3 a e^{-a} x^{2/3} \Gamma \left (\frac {8}{3},b x\right )}{5 b (b x)^{2/3}}-\frac {3 e^{-a} x^{2/3} \Gamma \left (\frac {11}{3},b x\right )}{5 b (b x)^{2/3}} \] Output:
3/5*x^(5/3)*exp(-b*x-a)*(b*x+a+1)-3/5*a*x^(2/3)*GAMMA(8/3,b*x)/b/exp(a)/(b *x)^(2/3)-3/5*x^(2/3)*GAMMA(11/3,b*x)/b/exp(a)/(b*x)^(2/3)
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.19 \[ \int x^{2/3} \Gamma (2,a+b x) \, dx=-\frac {e^{-a-b x} x^{5/3} \left (10 (8+3 a) e^{b x} \Gamma \left (\frac {2}{3},b x\right )+3 (b x)^{2/3} \left (40+15 a+24 b x+9 a b x+9 b^2 x^2-9 b e^{a+b x} x \Gamma (2,a+b x)\right )\right )}{45 (b x)^{5/3}} \] Input:
Integrate[x^(2/3)*Gamma[2, a + b*x],x]
Output:
-1/45*(E^(-a - b*x)*x^(5/3)*(10*(8 + 3*a)*E^(b*x)*Gamma[2/3, b*x] + 3*(b*x )^(2/3)*(40 + 15*a + 24*b*x + 9*a*b*x + 9*b^2*x^2 - 9*b*E^(a + b*x)*x*Gamm a[2, a + b*x])))/(b*x)^(5/3)
Time = 0.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int x^{2/3} \Gamma (2,a+b x) \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle \frac {3}{5} b \int e^{-a-b x} x^{5/3} (a+b x)dx+\frac {3}{5} x^{5/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \frac {3}{5} b \int \left (b e^{-a-b x} x^{8/3}+a e^{-a-b x} x^{5/3}\right )dx+\frac {3}{5} x^{5/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{5} b \left (-\frac {a e^{-a} x^{2/3} \Gamma \left (\frac {8}{3},b x\right )}{b^2 (b x)^{2/3}}-\frac {e^{-a} x^{2/3} \Gamma \left (\frac {11}{3},b x\right )}{b^2 (b x)^{2/3}}\right )+\frac {3}{5} x^{5/3} \Gamma (2,a+b x)\) |
Input:
Int[x^(2/3)*Gamma[2, a + b*x],x]
Output:
(3*x^(5/3)*Gamma[2, a + b*x])/5 + (3*b*(-((a*x^(2/3)*Gamma[8/3, b*x])/(b^2 *E^a*(b*x)^(2/3))) - (x^(2/3)*Gamma[11/3, b*x])/(b^2*E^a*(b*x)^(2/3))))/5
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 = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.44
method | result | size |
meijerg | \(\frac {{\mathrm e}^{-a} \left (\frac {x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{\left (b x \right )^{\frac {1}{3}}}-\frac {x^{\frac {2}{3}} b^{\frac {2}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, b x \right )}{\left (b x \right )^{\frac {1}{3}}}\right )}{b^{\frac {5}{3}}}+\frac {3 \,{\mathrm e}^{-a -\frac {b x}{2}} a \,x^{\frac {2}{3}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{5 b \left (b x \right )^{\frac {1}{3}}}+\frac {3 \,{\mathrm e}^{-a -\frac {b x}{2}} x^{\frac {2}{3}} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, b x \right )}{5 b \left (b x \right )^{\frac {1}{3}}}\) | \(115\) |
Input:
int(x^(2/3)*exp(-b*x-a)*(b*x+a+1),x,method=_RETURNVERBOSE)
Output:
1/b^(5/3)*exp(-a)*(x^(2/3)*b^(2/3)/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM(1/ 3,5/6,b*x)-x^(2/3)*b^(2/3)/(b*x)^(1/3)*exp(-1/2*b*x)*WhittakerM(4/3,5/6,b* x))+3/5/b*exp(-a-1/2*b*x)*a*x^(2/3)/(b*x)^(1/3)*WhittakerM(1/3,5/6,b*x)+3/ 5/b*exp(-a-1/2*b*x)*x^(2/3)/(b*x)^(1/3)*WhittakerM(1/3,5/6,b*x)
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int x^{2/3} \Gamma (2,a+b x) \, dx=-\frac {10 \, {\left (3 \, a + 8\right )} b^{\frac {1}{3}} e^{\left (-a\right )} \Gamma \left (\frac {2}{3}, b x\right ) - 3 \, {\left (9 \, b^{2} x \Gamma \left (2, b x + a\right ) - {\left (9 \, b^{3} x^{2} + 3 \, {\left (3 \, a + 8\right )} b^{2} x + 5 \, {\left (3 \, a + 8\right )} b\right )} e^{\left (-b x - a\right )}\right )} x^{\frac {2}{3}}}{45 \, b^{2}} \] Input:
integrate(x^(2/3)*gamma(2,b*x+a),x, algorithm="fricas")
Output:
-1/45*(10*(3*a + 8)*b^(1/3)*e^(-a)*gamma(2/3, b*x) - 3*(9*b^2*x*gamma(2, b *x + a) - (9*b^3*x^2 + 3*(3*a + 8)*b^2*x + 5*(3*a + 8)*b)*e^(-b*x - a))*x^ (2/3))/b^2
Time = 28.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int x^{2/3} \Gamma (2,a+b x) \, dx=\left (- \frac {a x^{\frac {2}{3}} \Gamma \left (\frac {5}{3}, b x\right )}{b \left (b x\right )^{\frac {2}{3}}} - \frac {x^{\frac {5}{3}} \Gamma \left (\frac {8}{3}, b x\right )}{\left (b x\right )^{\frac {5}{3}}} - \frac {x^{\frac {2}{3}} \Gamma \left (\frac {5}{3}, b x\right )}{b \left (b x\right )^{\frac {2}{3}}}\right ) e^{- a} \] Input:
integrate(x**(2/3)*uppergamma(2,b*x+a),x)
Output:
(-a*x**(2/3)*uppergamma(5/3, b*x)/(b*(b*x)**(2/3)) - x**(5/3)*uppergamma(8 /3, b*x)/(b*x)**(5/3) - x**(2/3)*uppergamma(5/3, b*x)/(b*(b*x)**(2/3)))*ex p(-a)
\[ \int x^{2/3} \Gamma (2,a+b x) \, dx=\int { x^{\frac {2}{3}} \Gamma \left (2, b x + a\right ) \,d x } \] Input:
integrate(x^(2/3)*gamma(2,b*x+a),x, algorithm="maxima")
Output:
integrate(x^(2/3)*gamma(2, b*x + a), x)
\[ \int x^{2/3} \Gamma (2,a+b x) \, dx=\int { x^{\frac {2}{3}} \Gamma \left (2, b x + a\right ) \,d x } \] Input:
integrate(x^(2/3)*gamma(2,b*x+a),x, algorithm="giac")
Output:
integrate(x^(2/3)*gamma(2, b*x + a), x)
Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int x^{2/3} \Gamma (2,a+b x) \, dx=-x^{5/3}\,{\mathrm {e}}^{-a-b\,x}-\frac {8\,x^{2/3}\,{\mathrm {e}}^{-a-b\,x}}{3\,b}-\frac {a\,x^{2/3}\,{\mathrm {e}}^{-a-b\,x}}{b}-\frac {10\,x^{5/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {2}{3},b\,x\right )}{9\,{\left (b\,x\right )}^{5/3}}-\frac {2\,x^{2/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {2}{3},b\,x\right )}{3\,b\,{\left (b\,x\right )}^{2/3}}-\frac {2\,a\,x^{2/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {2}{3},b\,x\right )}{3\,b\,{\left (b\,x\right )}^{2/3}} \] Input:
int(x^(2/3)*exp(- a - b*x)*(a + b*x + 1),x)
Output:
- x^(5/3)*exp(- a - b*x) - (8*x^(2/3)*exp(- a - b*x))/(3*b) - (a*x^(2/3)*e xp(- a - b*x))/b - (10*x^(5/3)*exp(-a)*igamma(2/3, b*x))/(9*(b*x)^(5/3)) - (2*x^(2/3)*exp(-a)*igamma(2/3, b*x))/(3*b*(b*x)^(2/3)) - (2*a*x^(2/3)*exp (-a)*igamma(2/3, b*x))/(3*b*(b*x)^(2/3))
\[ \int x^{2/3} \Gamma (2,a+b x) \, dx=\frac {-6 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right ) a -16 x^{\frac {1}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {4}{3}} e^{b x}}d x \right )-27 a b x -18 a -27 b^{2} x^{2}-72 b x -48}{27 x^{\frac {1}{3}} e^{b x +a} b^{2}} \] Input:
int(x^(2/3)*exp(-b*x-a)*(b*x+a+1),x)
Output:
( - 6*x**(1/3)*e**(b*x)*int(1/(x**(1/3)*e**(b*x)*x),x)*a - 16*x**(1/3)*e** (b*x)*int(1/(x**(1/3)*e**(b*x)*x),x) - 27*a*b*x - 18*a - 27*b**2*x**2 - 72 *b*x - 48)/(27*x**(1/3)*e**(a + b*x)*b**2)