\(\int x^{5/2} (A+B x) (a^2+2 a b x+b^2 x^2)^2 \, dx\) [739]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{7} a^4 A x^{7/2}+\frac {2}{9} a^3 (4 A b+a B) x^{9/2}+\frac {4}{11} a^2 b (3 A b+2 a B) x^{11/2}+\frac {4}{13} a b^2 (2 A b+3 a B) x^{13/2}+\frac {2}{15} b^3 (A b+4 a B) x^{15/2}+\frac {2}{17} b^4 B x^{17/2} \]

[Out]

2/7*a^4*A*x^(7/2)+2/9*a^3*(4*A*b+B*a)*x^(9/2)+4/11*a^2*b*(3*A*b+2*B*a)*x^(11/2)+4/13*a*b^2*(2*A*b+3*B*a)*x^(13
/2)+2/15*b^3*(A*b+4*B*a)*x^(15/2)+2/17*b^4*B*x^(17/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 77} \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{7} a^4 A x^{7/2}+\frac {2}{9} a^3 x^{9/2} (a B+4 A b)+\frac {4}{11} a^2 b x^{11/2} (2 a B+3 A b)+\frac {2}{15} b^3 x^{15/2} (4 a B+A b)+\frac {4}{13} a b^2 x^{13/2} (3 a B+2 A b)+\frac {2}{17} b^4 B x^{17/2} \]

[In]

Int[x^(5/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*a^4*A*x^(7/2))/7 + (2*a^3*(4*A*b + a*B)*x^(9/2))/9 + (4*a^2*b*(3*A*b + 2*a*B)*x^(11/2))/11 + (4*a*b^2*(2*A*
b + 3*a*B)*x^(13/2))/13 + (2*b^3*(A*b + 4*a*B)*x^(15/2))/15 + (2*b^4*B*x^(17/2))/17

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int x^{5/2} (a+b x)^4 (A+B x) \, dx \\ & = \int \left (a^4 A x^{5/2}+a^3 (4 A b+a B) x^{7/2}+2 a^2 b (3 A b+2 a B) x^{9/2}+2 a b^2 (2 A b+3 a B) x^{11/2}+b^3 (A b+4 a B) x^{13/2}+b^4 B x^{15/2}\right ) \, dx \\ & = \frac {2}{7} a^4 A x^{7/2}+\frac {2}{9} a^3 (4 A b+a B) x^{9/2}+\frac {4}{11} a^2 b (3 A b+2 a B) x^{11/2}+\frac {4}{13} a b^2 (2 A b+3 a B) x^{13/2}+\frac {2}{15} b^3 (A b+4 a B) x^{15/2}+\frac {2}{17} b^4 B x^{17/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 x^{7/2} \left (12155 a^4 (9 A+7 B x)+30940 a^3 b x (11 A+9 B x)+32130 a^2 b^2 x^2 (13 A+11 B x)+15708 a b^3 x^3 (15 A+13 B x)+3003 b^4 x^4 (17 A+15 B x)\right )}{765765} \]

[In]

Integrate[x^(5/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*x^(7/2)*(12155*a^4*(9*A + 7*B*x) + 30940*a^3*b*x*(11*A + 9*B*x) + 32130*a^2*b^2*x^2*(13*A + 11*B*x) + 15708
*a*b^3*x^3*(15*A + 13*B*x) + 3003*b^4*x^4*(17*A + 15*B*x)))/765765

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90

method	result	size
gosper	\(\frac {2 x^{\frac {7}{2}} \left (45045 b^{4} B \,x^{5}+51051 A \,b^{4} x^{4}+204204 x^{4} B \,b^{3} a +235620 a A \,b^{3} x^{3}+353430 x^{3} B \,a^{2} b^{2}+417690 a^{2} A \,b^{2} x^{2}+278460 x^{2} B \,a^{3} b +340340 a^{3} A b x +85085 a^{4} B x +109395 A \,a^{4}\right )}{765765}\)	\(100\)
derivativedivides	\(\frac {2 b^{4} B \,x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,b^{4}+4 B \,b^{3} a \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (4 A \,b^{3} a +6 B \,a^{2} b^{2}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (4 A \,a^{3} b +B \,a^{4}\right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{4} A \,x^{\frac {7}{2}}}{7}\)	\(100\)
default	\(\frac {2 b^{4} B \,x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,b^{4}+4 B \,b^{3} a \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (4 A \,b^{3} a +6 B \,a^{2} b^{2}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (4 A \,a^{3} b +B \,a^{4}\right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{4} A \,x^{\frac {7}{2}}}{7}\)	\(100\)
trager	\(\frac {2 x^{\frac {7}{2}} \left (45045 b^{4} B \,x^{5}+51051 A \,b^{4} x^{4}+204204 x^{4} B \,b^{3} a +235620 a A \,b^{3} x^{3}+353430 x^{3} B \,a^{2} b^{2}+417690 a^{2} A \,b^{2} x^{2}+278460 x^{2} B \,a^{3} b +340340 a^{3} A b x +85085 a^{4} B x +109395 A \,a^{4}\right )}{765765}\)	\(100\)
risch	\(\frac {2 x^{\frac {7}{2}} \left (45045 b^{4} B \,x^{5}+51051 A \,b^{4} x^{4}+204204 x^{4} B \,b^{3} a +235620 a A \,b^{3} x^{3}+353430 x^{3} B \,a^{2} b^{2}+417690 a^{2} A \,b^{2} x^{2}+278460 x^{2} B \,a^{3} b +340340 a^{3} A b x +85085 a^{4} B x +109395 A \,a^{4}\right )}{765765}\)	\(100\)

[In]

int(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

2/765765*x^(7/2)*(45045*B*b^4*x^5+51051*A*b^4*x^4+204204*B*a*b^3*x^4+235620*A*a*b^3*x^3+353430*B*a^2*b^2*x^3+4
17690*A*a^2*b^2*x^2+278460*B*a^3*b*x^2+340340*A*a^3*b*x+85085*B*a^4*x+109395*A*a^4)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{765765} \, {\left (45045 \, B b^{4} x^{8} + 109395 \, A a^{4} x^{3} + 51051 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{7} + 117810 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{6} + 139230 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{5} + 85085 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{4}\right )} \sqrt {x} \]

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

2/765765*(45045*B*b^4*x^8 + 109395*A*a^4*x^3 + 51051*(4*B*a*b^3 + A*b^4)*x^7 + 117810*(3*B*a^2*b^2 + 2*A*a*b^3
)*x^6 + 139230*(2*B*a^3*b + 3*A*a^2*b^2)*x^5 + 85085*(B*a^4 + 4*A*a^3*b)*x^4)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.33 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 A a^{4} x^{\frac {7}{2}}}{7} + \frac {8 A a^{3} b x^{\frac {9}{2}}}{9} + \frac {12 A a^{2} b^{2} x^{\frac {11}{2}}}{11} + \frac {8 A a b^{3} x^{\frac {13}{2}}}{13} + \frac {2 A b^{4} x^{\frac {15}{2}}}{15} + \frac {2 B a^{4} x^{\frac {9}{2}}}{9} + \frac {8 B a^{3} b x^{\frac {11}{2}}}{11} + \frac {12 B a^{2} b^{2} x^{\frac {13}{2}}}{13} + \frac {8 B a b^{3} x^{\frac {15}{2}}}{15} + \frac {2 B b^{4} x^{\frac {17}{2}}}{17} \]

[In]

integrate(x**(5/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

2*A*a**4*x**(7/2)/7 + 8*A*a**3*b*x**(9/2)/9 + 12*A*a**2*b**2*x**(11/2)/11 + 8*A*a*b**3*x**(13/2)/13 + 2*A*b**4
*x**(15/2)/15 + 2*B*a**4*x**(9/2)/9 + 8*B*a**3*b*x**(11/2)/11 + 12*B*a**2*b**2*x**(13/2)/13 + 8*B*a*b**3*x**(1
5/2)/15 + 2*B*b**4*x**(17/2)/17

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.89 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{17} \, B b^{4} x^{\frac {17}{2}} + \frac {2}{7} \, A a^{4} x^{\frac {7}{2}} + \frac {2}{15} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac {15}{2}} + \frac {4}{13} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac {13}{2}} + \frac {4}{11} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{\frac {9}{2}} \]

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

2/17*B*b^4*x^(17/2) + 2/7*A*a^4*x^(7/2) + 2/15*(4*B*a*b^3 + A*b^4)*x^(15/2) + 4/13*(3*B*a^2*b^2 + 2*A*a*b^3)*x
^(13/2) + 4/11*(2*B*a^3*b + 3*A*a^2*b^2)*x^(11/2) + 2/9*(B*a^4 + 4*A*a^3*b)*x^(9/2)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{17} \, B b^{4} x^{\frac {17}{2}} + \frac {8}{15} \, B a b^{3} x^{\frac {15}{2}} + \frac {2}{15} \, A b^{4} x^{\frac {15}{2}} + \frac {12}{13} \, B a^{2} b^{2} x^{\frac {13}{2}} + \frac {8}{13} \, A a b^{3} x^{\frac {13}{2}} + \frac {8}{11} \, B a^{3} b x^{\frac {11}{2}} + \frac {12}{11} \, A a^{2} b^{2} x^{\frac {11}{2}} + \frac {2}{9} \, B a^{4} x^{\frac {9}{2}} + \frac {8}{9} \, A a^{3} b x^{\frac {9}{2}} + \frac {2}{7} \, A a^{4} x^{\frac {7}{2}} \]

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

2/17*B*b^4*x^(17/2) + 8/15*B*a*b^3*x^(15/2) + 2/15*A*b^4*x^(15/2) + 12/13*B*a^2*b^2*x^(13/2) + 8/13*A*a*b^3*x^
(13/2) + 8/11*B*a^3*b*x^(11/2) + 12/11*A*a^2*b^2*x^(11/2) + 2/9*B*a^4*x^(9/2) + 8/9*A*a^3*b*x^(9/2) + 2/7*A*a^
4*x^(7/2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^{9/2}\,\left (\frac {2\,B\,a^4}{9}+\frac {8\,A\,b\,a^3}{9}\right )+x^{15/2}\,\left (\frac {2\,A\,b^4}{15}+\frac {8\,B\,a\,b^3}{15}\right )+\frac {2\,A\,a^4\,x^{7/2}}{7}+\frac {2\,B\,b^4\,x^{17/2}}{17}+\frac {4\,a^2\,b\,x^{11/2}\,\left (3\,A\,b+2\,B\,a\right )}{11}+\frac {4\,a\,b^2\,x^{13/2}\,\left (2\,A\,b+3\,B\,a\right )}{13} \]

[In]

int(x^(5/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

x^(9/2)*((2*B*a^4)/9 + (8*A*a^3*b)/9) + x^(15/2)*((2*A*b^4)/15 + (8*B*a*b^3)/15) + (2*A*a^4*x^(7/2))/7 + (2*B*
b^4*x^(17/2))/17 + (4*a^2*b*x^(11/2)*(3*A*b + 2*B*a))/11 + (4*a*b^2*x^(13/2)*(2*A*b + 3*B*a))/13