Integrand size = 18, antiderivative size = 95 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {2 a^2 (A b-a B) (a+b x)^{7/2}}{7 b^4}-\frac {2 a (2 A b-3 a B) (a+b x)^{9/2}}{9 b^4}+\frac {2 (A b-3 a B) (a+b x)^{11/2}}{11 b^4}+\frac {2 B (a+b x)^{13/2}}{13 b^4} \] Output:
2/7*a^2*(A*b-B*a)*(b*x+a)^(7/2)/b^4-2/9*a*(2*A*b-3*B*a)*(b*x+a)^(9/2)/b^4+ 2/11*(A*b-3*B*a)*(b*x+a)^(11/2)/b^4+2/13*B*(b*x+a)^(13/2)/b^4
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {2 (a+b x)^{7/2} \left (-48 a^3 B+63 b^3 x^2 (13 A+11 B x)+8 a^2 b (13 A+21 B x)-14 a b^2 x (26 A+27 B x)\right )}{9009 b^4} \] Input:
Integrate[x^2*(a + b*x)^(5/2)*(A + B*x),x]
Output:
(2*(a + b*x)^(7/2)*(-48*a^3*B + 63*b^3*x^2*(13*A + 11*B*x) + 8*a^2*b*(13*A + 21*B*x) - 14*a*b^2*x*(26*A + 27*B*x)))/(9009*b^4)
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {86, 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.
\(\displaystyle \int x^2 (a+b x)^{5/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {a^2 (a+b x)^{5/2} (a B-A b)}{b^3}+\frac {(a+b x)^{9/2} (A b-3 a B)}{b^3}+\frac {a (a+b x)^{7/2} (3 a B-2 A b)}{b^3}+\frac {B (a+b x)^{11/2}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2 (a+b x)^{7/2} (A b-a B)}{7 b^4}+\frac {2 (a+b x)^{11/2} (A b-3 a B)}{11 b^4}-\frac {2 a (a+b x)^{9/2} (2 A b-3 a B)}{9 b^4}+\frac {2 B (a+b x)^{13/2}}{13 b^4}\) |
Input:
Int[x^2*(a + b*x)^(5/2)*(A + B*x),x]
Output:
(2*a^2*(A*b - a*B)*(a + b*x)^(7/2))/(7*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b* x)^(9/2))/(9*b^4) + (2*(A*b - 3*a*B)*(a + b*x)^(11/2))/(11*b^4) + (2*B*(a + b*x)^(13/2))/(13*b^4)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.38 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {16 \left (\frac {63 \left (\frac {11 B x}{13}+A \right ) x^{2} b^{3}}{8}-\frac {7 a \left (\frac {27 B x}{26}+A \right ) x \,b^{2}}{2}+a^{2} \left (\frac {21 B x}{13}+A \right ) b -\frac {6 a^{3} B}{13}\right ) \left (b x +a \right )^{\frac {7}{2}}}{693 b^{4}}\) | \(58\) |
gosper | \(\frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (693 b^{3} B \,x^{3}+819 A \,x^{2} b^{3}-378 B \,x^{2} a \,b^{2}-364 A x a \,b^{2}+168 B x \,a^{2} b +104 a^{2} b A -48 a^{3} B \right )}{9009 b^{4}}\) | \(71\) |
orering | \(\frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (693 b^{3} B \,x^{3}+819 A \,x^{2} b^{3}-378 B \,x^{2} a \,b^{2}-364 A x a \,b^{2}+168 B x \,a^{2} b +104 a^{2} b A -48 a^{3} B \right )}{9009 b^{4}}\) | \(71\) |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {13}{2}}}{13}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}}{b^{4}}\) | \(80\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {13}{2}}}{13}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}}{b^{4}}\) | \(80\) |
trager | \(\frac {2 \left (693 B \,b^{6} x^{6}+819 A \,b^{6} x^{5}+1701 B a \,b^{5} x^{5}+2093 A a \,b^{5} x^{4}+1113 B \,a^{2} b^{4} x^{4}+1469 A \,a^{2} b^{4} x^{3}+15 B \,a^{3} b^{3} x^{3}+39 A \,a^{3} b^{3} x^{2}-18 B \,a^{4} b^{2} x^{2}-52 A \,a^{4} b^{2} x +24 B \,a^{5} b x +104 A \,a^{5} b -48 B \,a^{6}\right ) \sqrt {b x +a}}{9009 b^{4}}\) | \(143\) |
risch | \(\frac {2 \left (693 B \,b^{6} x^{6}+819 A \,b^{6} x^{5}+1701 B a \,b^{5} x^{5}+2093 A a \,b^{5} x^{4}+1113 B \,a^{2} b^{4} x^{4}+1469 A \,a^{2} b^{4} x^{3}+15 B \,a^{3} b^{3} x^{3}+39 A \,a^{3} b^{3} x^{2}-18 B \,a^{4} b^{2} x^{2}-52 A \,a^{4} b^{2} x +24 B \,a^{5} b x +104 A \,a^{5} b -48 B \,a^{6}\right ) \sqrt {b x +a}}{9009 b^{4}}\) | \(143\) |
Input:
int(x^2*(b*x+a)^(5/2)*(B*x+A),x,method=_RETURNVERBOSE)
Output:
16/693*(63/8*(11/13*B*x+A)*x^2*b^3-7/2*a*(27/26*B*x+A)*x*b^2+a^2*(21/13*B* x+A)*b-6/13*a^3*B)*(b*x+a)^(7/2)/b^4
Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.51 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {2 \, {\left (693 \, B b^{6} x^{6} - 48 \, B a^{6} + 104 \, A a^{5} b + 63 \, {\left (27 \, B a b^{5} + 13 \, A b^{6}\right )} x^{5} + 7 \, {\left (159 \, B a^{2} b^{4} + 299 \, A a b^{5}\right )} x^{4} + {\left (15 \, B a^{3} b^{3} + 1469 \, A a^{2} b^{4}\right )} x^{3} - 3 \, {\left (6 \, B a^{4} b^{2} - 13 \, A a^{3} b^{3}\right )} x^{2} + 4 \, {\left (6 \, B a^{5} b - 13 \, A a^{4} b^{2}\right )} x\right )} \sqrt {b x + a}}{9009 \, b^{4}} \] Input:
integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="fricas")
Output:
2/9009*(693*B*b^6*x^6 - 48*B*a^6 + 104*A*a^5*b + 63*(27*B*a*b^5 + 13*A*b^6 )*x^5 + 7*(159*B*a^2*b^4 + 299*A*a*b^5)*x^4 + (15*B*a^3*b^3 + 1469*A*a^2*b ^4)*x^3 - 3*(6*B*a^4*b^2 - 13*A*a^3*b^3)*x^2 + 4*(6*B*a^5*b - 13*A*a^4*b^2 )*x)*sqrt(b*x + a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (94) = 188\).
Time = 0.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.07 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\begin {cases} \frac {16 A a^{5} \sqrt {a + b x}}{693 b^{3}} - \frac {8 A a^{4} x \sqrt {a + b x}}{693 b^{2}} + \frac {2 A a^{3} x^{2} \sqrt {a + b x}}{231 b} + \frac {226 A a^{2} x^{3} \sqrt {a + b x}}{693} + \frac {46 A a b x^{4} \sqrt {a + b x}}{99} + \frac {2 A b^{2} x^{5} \sqrt {a + b x}}{11} - \frac {32 B a^{6} \sqrt {a + b x}}{3003 b^{4}} + \frac {16 B a^{5} x \sqrt {a + b x}}{3003 b^{3}} - \frac {4 B a^{4} x^{2} \sqrt {a + b x}}{1001 b^{2}} + \frac {10 B a^{3} x^{3} \sqrt {a + b x}}{3003 b} + \frac {106 B a^{2} x^{4} \sqrt {a + b x}}{429} + \frac {54 B a b x^{5} \sqrt {a + b x}}{143} + \frac {2 B b^{2} x^{6} \sqrt {a + b x}}{13} & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(b*x+a)**(5/2)*(B*x+A),x)
Output:
Piecewise((16*A*a**5*sqrt(a + b*x)/(693*b**3) - 8*A*a**4*x*sqrt(a + b*x)/( 693*b**2) + 2*A*a**3*x**2*sqrt(a + b*x)/(231*b) + 226*A*a**2*x**3*sqrt(a + b*x)/693 + 46*A*a*b*x**4*sqrt(a + b*x)/99 + 2*A*b**2*x**5*sqrt(a + b*x)/1 1 - 32*B*a**6*sqrt(a + b*x)/(3003*b**4) + 16*B*a**5*x*sqrt(a + b*x)/(3003* b**3) - 4*B*a**4*x**2*sqrt(a + b*x)/(1001*b**2) + 10*B*a**3*x**3*sqrt(a + b*x)/(3003*b) + 106*B*a**2*x**4*sqrt(a + b*x)/429 + 54*B*a*b*x**5*sqrt(a + b*x)/143 + 2*B*b**2*x**6*sqrt(a + b*x)/13, Ne(b, 0)), (a**(5/2)*(A*x**3/3 + B*x**4/4), True))
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {2 \, {\left (693 \, {\left (b x + a\right )}^{\frac {13}{2}} B - 819 \, {\left (3 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {11}{2}} + 1001 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} {\left (b x + a\right )}^{\frac {9}{2}} - 1287 \, {\left (B a^{3} - A a^{2} b\right )} {\left (b x + a\right )}^{\frac {7}{2}}\right )}}{9009 \, b^{4}} \] Input:
integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="maxima")
Output:
2/9009*(693*(b*x + a)^(13/2)*B - 819*(3*B*a - A*b)*(b*x + a)^(11/2) + 1001 *(3*B*a^2 - 2*A*a*b)*(b*x + a)^(9/2) - 1287*(B*a^3 - A*a^2*b)*(b*x + a)^(7 /2))/b^4
Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (80) = 160\).
Time = 0.12 (sec) , antiderivative size = 516, normalized size of antiderivative = 5.43 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx =\text {Too large to display} \] Input:
integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="giac")
Output:
2/45045*(3003*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a) *a^2)*A*a^3/b^2 + 1287*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B*a^3/b^3 + 3861*(5*(b*x + a)^(7/2 ) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)* A*a^2/b^2 + 429*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a )^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*B*a^2/b^3 + 429*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*A*a/b^2 + 195*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*B*a/b^ 3 + 65*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)* a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*A/b^2 + 15*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005 *(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)*B/b^3)/b
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {\left (6\,B\,a^2-4\,A\,a\,b\right )\,{\left (a+b\,x\right )}^{9/2}}{9\,b^4}+\frac {2\,B\,{\left (a+b\,x\right )}^{13/2}}{13\,b^4}+\frac {\left (2\,A\,b-6\,B\,a\right )\,{\left (a+b\,x\right )}^{11/2}}{11\,b^4}-\frac {\left (2\,B\,a^3-2\,A\,a^2\,b\right )\,{\left (a+b\,x\right )}^{7/2}}{7\,b^4} \] Input:
int(x^2*(A + B*x)*(a + b*x)^(5/2),x)
Output:
((6*B*a^2 - 4*A*a*b)*(a + b*x)^(9/2))/(9*b^4) + (2*B*(a + b*x)^(13/2))/(13 *b^4) + ((2*A*b - 6*B*a)*(a + b*x)^(11/2))/(11*b^4) - ((2*B*a^3 - 2*A*a^2* b)*(a + b*x)^(7/2))/(7*b^4)
Time = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.78 \[ \int x^2 (a+b x)^{5/2} (A+B x) \, dx=\frac {2 \sqrt {b x +a}\, \left (99 b^{6} x^{6}+360 a \,b^{5} x^{5}+458 a^{2} b^{4} x^{4}+212 a^{3} b^{3} x^{3}+3 a^{4} b^{2} x^{2}-4 a^{5} b x +8 a^{6}\right )}{1287 b^{3}} \] Input:
int(x^2*(b*x+a)^(5/2)*(B*x+A),x)
Output:
(2*sqrt(a + b*x)*(8*a**6 - 4*a**5*b*x + 3*a**4*b**2*x**2 + 212*a**3*b**3*x **3 + 458*a**2*b**4*x**4 + 360*a*b**5*x**5 + 99*b**6*x**6))/(1287*b**3)