Integrand size = 13, antiderivative size = 68 \[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=-\frac {3 a e^{-a} \sqrt [3]{x} \Gamma \left (\frac {4}{3},b x\right )}{\sqrt [3]{b x}}+3 \sqrt [3]{x} \Gamma (2,a+b x)-\frac {3 e^{-a} \sqrt [3]{x} \Gamma \left (\frac {7}{3},b x\right )}{\sqrt [3]{b x}} \] Output:
-3*a*x^(1/3)*GAMMA(4/3,b*x)/exp(a)/(b*x)^(1/3)+3*x^(1/3)*exp(-b*x-a)*(b*x+ a+1)-3*x^(1/3)*GAMMA(7/3,b*x)/exp(a)/(b*x)^(1/3)
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=\frac {1}{3} e^{-a} \sqrt [3]{x} \left (-\frac {(4+3 a) \Gamma \left (\frac {1}{3},b x\right )}{\sqrt [3]{b x}}-3 e^{-b x} \left (4+3 a+3 b x-3 e^{a+b x} \Gamma (2,a+b x)\right )\right ) \] Input:
Integrate[Gamma[2, a + b*x]/x^(2/3),x]
Output:
(x^(1/3)*(-(((4 + 3*a)*Gamma[1/3, b*x])/(b*x)^(1/3)) - (3*(4 + 3*a + 3*b*x - 3*E^(a + b*x)*Gamma[2, a + b*x]))/E^(b*x)))/(3*E^a)
Time = 0.50 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15, 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.
\(\displaystyle \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle 3 b \int e^{-a-b x} \sqrt [3]{x} (a+b x)dx+3 \sqrt [3]{x} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle 3 b \int \left (b e^{-a-b x} x^{4/3}+a e^{-a-b x} \sqrt [3]{x}\right )dx+3 \sqrt [3]{x} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \sqrt [3]{x} \Gamma (2,a+b x)+3 b \left (-\frac {a e^{-a} \sqrt [3]{x} \Gamma \left (\frac {4}{3},b x\right )}{b \sqrt [3]{b x}}-\frac {e^{-a} \sqrt [3]{x} \Gamma \left (\frac {7}{3},b x\right )}{b \sqrt [3]{b x}}\right )\) |
Input:
Int[Gamma[2, a + b*x]/x^(2/3),x]
Output:
3*x^(1/3)*Gamma[2, a + b*x] + 3*b*(-((a*x^(1/3)*Gamma[4/3, b*x])/(b*E^a*(b *x)^(1/3))) - (x^(1/3)*Gamma[7/3, b*x])/(b*E^a*(b*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 = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.09
method | result | size |
meijerg | \(\frac {3 \,{\mathrm e}^{-a -\frac {b x}{2}} x^{\frac {1}{3}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{4 \left (b x \right )^{\frac {1}{6}}}+\frac {{\mathrm e}^{-a} a \left (\frac {9 x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{4 \left (b x \right )^{\frac {1}{6}}}+\frac {3 \,{\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, b x \right )}{x^{\frac {2}{3}} b^{\frac {2}{3}} \left (b x \right )^{\frac {1}{6}}}\right )}{b^{\frac {1}{3}}}+\frac {{\mathrm e}^{-a} \left (\frac {9 x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{4 \left (b x \right )^{\frac {1}{6}}}+\frac {3 \,{\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, b x \right )}{x^{\frac {2}{3}} b^{\frac {2}{3}} \left (b x \right )^{\frac {1}{6}}}\right )}{b^{\frac {1}{3}}}\) | \(142\) |
Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(2/3),x,method=_RETURNVERBOSE)
Output:
3/4*exp(-a-1/2*b*x)*x^(1/3)/(b*x)^(1/6)*WhittakerM(1/6,2/3,b*x)+exp(-a)*a/ b^(1/3)*(9/4*x^(1/3)*b^(1/3)/(b*x)^(1/6)*exp(-1/2*b*x)*WhittakerM(1/6,2/3, b*x)+3/x^(2/3)/b^(2/3)/(b*x)^(1/6)*exp(-1/2*b*x)*WhittakerM(7/6,2/3,b*x))+ 1/b^(1/3)*exp(-a)*(9/4*x^(1/3)*b^(1/3)/(b*x)^(1/6)*exp(-1/2*b*x)*Whittaker M(1/6,2/3,b*x)+3/x^(2/3)/b^(2/3)/(b*x)^(1/6)*exp(-1/2*b*x)*WhittakerM(7/6, 2/3,b*x))
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=-\frac {{\left (3 \, a + 4\right )} b^{\frac {2}{3}} e^{\left (-a\right )} \Gamma \left (\frac {1}{3}, b x\right ) + 3 \, {\left ({\left (3 \, b^{2} x + {\left (3 \, a + 4\right )} b\right )} e^{\left (-b x - a\right )} - 3 \, b \Gamma \left (2, b x + a\right )\right )} x^{\frac {1}{3}}}{3 \, b} \] Input:
integrate(gamma(2,b*x+a)/x^(2/3),x, algorithm="fricas")
Output:
-1/3*((3*a + 4)*b^(2/3)*e^(-a)*gamma(1/3, b*x) + 3*((3*b^2*x + (3*a + 4)*b )*e^(-b*x - a) - 3*b*gamma(2, b*x + a))*x^(1/3))/b
Time = 4.82 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=\left (- \frac {a \left (b x\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, b x\right )}{b x^{\frac {2}{3}}} - \frac {\sqrt [3]{x} \Gamma \left (\frac {4}{3}, b x\right )}{\sqrt [3]{b x}} - \frac {\left (b x\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, b x\right )}{b x^{\frac {2}{3}}}\right ) e^{- a} \] Input:
integrate(uppergamma(2,b*x+a)/x**(2/3),x)
Output:
(-a*(b*x)**(2/3)*uppergamma(1/3, b*x)/(b*x**(2/3)) - x**(1/3)*uppergamma(4 /3, b*x)/(b*x)**(1/3) - (b*x)**(2/3)*uppergamma(1/3, b*x)/(b*x**(2/3)))*ex p(-a)
\[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {2}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(2/3),x, algorithm="maxima")
Output:
integrate(gamma(2, b*x + a)/x^(2/3), x)
\[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=\int { \frac {\Gamma \left (2, b x + a\right )}{x^{\frac {2}{3}}} \,d x } \] Input:
integrate(gamma(2,b*x+a)/x^(2/3),x, algorithm="giac")
Output:
integrate(gamma(2, b*x + a)/x^(2/3), x)
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=-x^{1/3}\,{\mathrm {e}}^{-a-b\,x}-\frac {x^{1/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {1}{3},b\,x\right )}{3\,{\left (b\,x\right )}^{1/3}}-\frac {{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{2/3}\,\Gamma \left (\frac {1}{3},b\,x\right )}{b\,x^{2/3}}-\frac {a\,{\mathrm {e}}^{-a}\,{\left (b\,x\right )}^{2/3}\,\Gamma \left (\frac {1}{3},b\,x\right )}{b\,x^{2/3}} \] Input:
int((exp(- a - b*x)*(a + b*x + 1))/x^(2/3),x)
Output:
- x^(1/3)*exp(- a - b*x) - (x^(1/3)*exp(-a)*igamma(1/3, b*x))/(3*(b*x)^(1/ 3)) - (exp(-a)*(b*x)^(2/3)*igamma(1/3, b*x))/(b*x^(2/3)) - (a*exp(-a)*(b*x )^(2/3)*igamma(1/3, b*x))/(b*x^(2/3))
\[ \int \frac {\Gamma (2,a+b x)}{x^{2/3}} \, dx=\frac {-6 x^{\frac {2}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {5}{3}} e^{b x}}d x \right ) a -8 x^{\frac {2}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {5}{3}} e^{b x}}d x \right )-9 a -9 b x -12}{9 x^{\frac {2}{3}} e^{b x +a} b} \] Input:
int(exp(-b*x-a)*(b*x+a+1)/x^(2/3),x)
Output:
( - 6*x**(2/3)*e**(b*x)*int(1/(x**(2/3)*e**(b*x)*x),x)*a - 8*x**(2/3)*e**( b*x)*int(1/(x**(2/3)*e**(b*x)*x),x) - 9*a - 9*b*x - 12)/(9*x**(2/3)*e**(a + b*x)*b)