Integrand size = 13, antiderivative size = 80 \[ \int x^{4/3} \Gamma (2,a+b x) \, dx=\frac {3}{7} x^{7/3} \Gamma (2,a+b x)-\frac {3 a e^{-a} \sqrt [3]{x} \Gamma \left (\frac {10}{3},b x\right )}{7 b^2 \sqrt [3]{b x}}-\frac {3 e^{-a} \sqrt [3]{x} \Gamma \left (\frac {13}{3},b x\right )}{7 b^2 \sqrt [3]{b x}} \] Output:
3/7*x^(7/3)*exp(-b*x-a)*(b*x+a+1)-3/7*a*x^(1/3)*GAMMA(10/3,b*x)/b^2/exp(a) /(b*x)^(1/3)-3/7*x^(1/3)*GAMMA(13/3,b*x)/b^2/exp(a)/(b*x)^(1/3)
Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.49 \[ \int x^{4/3} \Gamma (2,a+b x) \, dx=-\frac {e^{-a-b x} \sqrt [3]{x} \left (28 (10+3 a) e^{b x} \Gamma \left (\frac {1}{3},b x\right )+3 \sqrt [3]{b x} \left (280+210 b x+90 b^2 x^2+27 b^3 x^3+3 a \left (28+21 b x+9 b^2 x^2\right )-27 b^2 e^{a+b x} x^2 \Gamma (2,a+b x)\right )\right )}{189 b^2 \sqrt [3]{b x}} \] Input:
Integrate[x^(4/3)*Gamma[2, a + b*x],x]
Output:
-1/189*(E^(-a - b*x)*x^(1/3)*(28*(10 + 3*a)*E^(b*x)*Gamma[1/3, b*x] + 3*(b *x)^(1/3)*(280 + 210*b*x + 90*b^2*x^2 + 27*b^3*x^3 + 3*a*(28 + 21*b*x + 9* b^2*x^2) - 27*b^2*E^(a + b*x)*x^2*Gamma[2, a + b*x])))/(b^2*(b*x)^(1/3))
Time = 0.33 (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.
\(\displaystyle \int x^{4/3} \Gamma (2,a+b x) \, dx\) |
\(\Big \downarrow \) 7119 |
\(\displaystyle \frac {3}{7} b \int e^{-a-b x} x^{7/3} (a+b x)dx+\frac {3}{7} x^{7/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \frac {3}{7} b \int \left (b e^{-a-b x} x^{10/3}+a e^{-a-b x} x^{7/3}\right )dx+\frac {3}{7} x^{7/3} \Gamma (2,a+b x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{7} b \left (-\frac {a e^{-a} \sqrt [3]{x} \Gamma \left (\frac {10}{3},b x\right )}{b^3 \sqrt [3]{b x}}-\frac {e^{-a} \sqrt [3]{x} \Gamma \left (\frac {13}{3},b x\right )}{b^3 \sqrt [3]{b x}}\right )+\frac {3}{7} x^{7/3} \Gamma (2,a+b x)\) |
Input:
Int[x^(4/3)*Gamma[2, a + b*x],x]
Output:
(3*x^(7/3)*Gamma[2, a + b*x])/7 + (3*b*(-((a*x^(1/3)*Gamma[10/3, b*x])/(b^ 3*E^a*(b*x)^(1/3))) - (x^(1/3)*Gamma[13/3, b*x])/(b^3*E^a*(b*x)^(1/3))))/7
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 = 178, normalized size of antiderivative = 2.22
method | result | size |
meijerg | \(\frac {{\mathrm e}^{-a} \left (\frac {7 x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{3 \left (b x \right )^{\frac {1}{6}}}-\frac {x^{\frac {1}{3}} b^{\frac {1}{3}} \left (30 b x +70\right ) {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, b x \right )}{30 \left (b x \right )^{\frac {1}{6}}}\right )}{b^{\frac {7}{3}}}+\frac {{\mathrm e}^{-a} a \left (\frac {x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{\left (b x \right )^{\frac {1}{6}}}-\frac {x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, b x \right )}{\left (b x \right )^{\frac {1}{6}}}\right )}{b^{\frac {7}{3}}}+\frac {{\mathrm e}^{-a} \left (\frac {x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, b x \right )}{\left (b x \right )^{\frac {1}{6}}}-\frac {x^{\frac {1}{3}} b^{\frac {1}{3}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, b x \right )}{\left (b x \right )^{\frac {1}{6}}}\right )}{b^{\frac {7}{3}}}\) | \(178\) |
Input:
int(x^(4/3)*exp(-b*x-a)*(b*x+a+1),x,method=_RETURNVERBOSE)
Output:
1/b^(7/3)*exp(-a)*(7/3*x^(1/3)*b^(1/3)/(b*x)^(1/6)*exp(-1/2*b*x)*Whittaker M(1/6,2/3,b*x)-1/30*x^(1/3)*b^(1/3)*(30*b*x+70)/(b*x)^(1/6)*exp(-1/2*b*x)* WhittakerM(7/6,2/3,b*x))+1/b^(7/3)*exp(-a)*a*(x^(1/3)*b^(1/3)/(b*x)^(1/6)* exp(-1/2*b*x)*WhittakerM(1/6,2/3,b*x)-x^(1/3)*b^(1/3)/(b*x)^(1/6)*exp(-1/2 *b*x)*WhittakerM(7/6,2/3,b*x))+1/b^(7/3)*exp(-a)*(x^(1/3)*b^(1/3)/(b*x)^(1 /6)*exp(-1/2*b*x)*WhittakerM(1/6,2/3,b*x)-x^(1/3)*b^(1/3)/(b*x)^(1/6)*exp( -1/2*b*x)*WhittakerM(7/6,2/3,b*x))
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22 \[ \int x^{4/3} \Gamma (2,a+b x) \, dx=-\frac {28 \, {\left (3 \, a + 10\right )} b^{\frac {2}{3}} e^{\left (-a\right )} \Gamma \left (\frac {1}{3}, b x\right ) - 3 \, {\left (27 \, b^{3} x^{2} \Gamma \left (2, b x + a\right ) - {\left (27 \, b^{4} x^{3} + 9 \, {\left (3 \, a + 10\right )} b^{3} x^{2} + 21 \, {\left (3 \, a + 10\right )} b^{2} x + 28 \, {\left (3 \, a + 10\right )} b\right )} e^{\left (-b x - a\right )}\right )} x^{\frac {1}{3}}}{189 \, b^{3}} \] Input:
integrate(x^(4/3)*gamma(2,b*x+a),x, algorithm="fricas")
Output:
-1/189*(28*(3*a + 10)*b^(2/3)*e^(-a)*gamma(1/3, b*x) - 3*(27*b^3*x^2*gamma (2, b*x + a) - (27*b^4*x^3 + 9*(3*a + 10)*b^3*x^2 + 21*(3*a + 10)*b^2*x + 28*(3*a + 10)*b)*e^(-b*x - a))*x^(1/3))/b^3
Timed out. \[ \int x^{4/3} \Gamma (2,a+b x) \, dx=\text {Timed out} \] Input:
integrate(x**(4/3)*uppergamma(2,b*x+a),x)
Output:
Timed out
\[ \int x^{4/3} \Gamma (2,a+b x) \, dx=\int { x^{\frac {4}{3}} \Gamma \left (2, b x + a\right ) \,d x } \] Input:
integrate(x^(4/3)*gamma(2,b*x+a),x, algorithm="maxima")
Output:
integrate(x^(4/3)*gamma(2, b*x + a), x)
\[ \int x^{4/3} \Gamma (2,a+b x) \, dx=\int { x^{\frac {4}{3}} \Gamma \left (2, b x + a\right ) \,d x } \] Input:
integrate(x^(4/3)*gamma(2,b*x+a),x, algorithm="giac")
Output:
integrate(x^(4/3)*gamma(2, b*x + a), x)
Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55 \[ \int x^{4/3} \Gamma (2,a+b x) \, dx=-x^{7/3}\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}-\frac {40\,x^{7/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {1}{3},b\,x\right )}{27\,{\left (b\,x\right )}^{7/3}}-\frac {40\,x^{1/3}\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{9\,b^2}-\frac {10\,x^{4/3}\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{3\,b}-\frac {4\,a\,x^{1/3}\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{3\,b^2}-\frac {a\,x^{4/3}\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{b}-\frac {4\,a\,x^{7/3}\,{\mathrm {e}}^{-a}\,\Gamma \left (\frac {1}{3},b\,x\right )}{9\,{\left (b\,x\right )}^{7/3}} \] Input:
int(x^(4/3)*exp(- a - b*x)*(a + b*x + 1),x)
Output:
- x^(7/3)*exp(-a)*exp(-b*x) - (40*x^(7/3)*exp(-a)*igamma(1/3, b*x))/(27*(b *x)^(7/3)) - (40*x^(1/3)*exp(-a)*exp(-b*x))/(9*b^2) - (10*x^(4/3)*exp(-a)* exp(-b*x))/(3*b) - (4*a*x^(1/3)*exp(-a)*exp(-b*x))/(3*b^2) - (a*x^(4/3)*ex p(-a)*exp(-b*x))/b - (4*a*x^(7/3)*exp(-a)*igamma(1/3, b*x))/(9*(b*x)^(7/3) )
\[ \int x^{4/3} \Gamma (2,a+b x) \, dx=\frac {-24 x^{\frac {2}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {5}{3}} e^{b x}}d x \right ) a -80 x^{\frac {2}{3}} e^{b x} \left (\int \frac {1}{x^{\frac {5}{3}} e^{b x}}d x \right )-81 a \,b^{2} x^{2}-108 a b x -36 a -81 b^{3} x^{3}-270 b^{2} x^{2}-360 b x -120}{81 x^{\frac {2}{3}} e^{b x +a} b^{3}} \] Input:
int(x^(4/3)*exp(-b*x-a)*(b*x+a+1),x)
Output:
( - 24*x**(2/3)*e**(b*x)*int(1/(x**(2/3)*e**(b*x)*x),x)*a - 80*x**(2/3)*e* *(b*x)*int(1/(x**(2/3)*e**(b*x)*x),x) - 81*a*b**2*x**2 - 108*a*b*x - 36*a - 81*b**3*x**3 - 270*b**2*x**2 - 360*b*x - 120)/(81*x**(2/3)*e**(a + b*x)* b**3)