3.748 \(\int x^{5/2} (A+B x) (a^2+2 a b x+b^2 x^2)^3 \, dx\)

Optimal. Leaf size=159 \[ \frac{2}{3} a^2 b^3 x^{15/2} (4 a B+3 A b)+\frac{10}{13} a^3 b^2 x^{13/2} (3 a B+4 A b)+\frac{6}{11} a^4 b x^{11/2} (2 a B+5 A b)+\frac{2}{9} a^5 x^{9/2} (a B+6 A b)+\frac{2}{7} a^6 A x^{7/2}+\frac{2}{19} b^5 x^{19/2} (6 a B+A b)+\frac{6}{17} a b^4 x^{17/2} (5 a B+2 A b)+\frac{2}{21} b^6 B x^{21/2} \]

[Out]

(2*a^6*A*x^(7/2))/7 + (2*a^5*(6*A*b + a*B)*x^(9/2))/9 + (6*a^4*b*(5*A*b + 2*a*B)*x^(11/2))/11 + (10*a^3*b^2*(4
*A*b + 3*a*B)*x^(13/2))/13 + (2*a^2*b^3*(3*A*b + 4*a*B)*x^(15/2))/3 + (6*a*b^4*(2*A*b + 5*a*B)*x^(17/2))/17 +
(2*b^5*(A*b + 6*a*B)*x^(19/2))/19 + (2*b^6*B*x^(21/2))/21

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Rubi [A]  time = 0.0795575, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {27, 76} \[ \frac{2}{3} a^2 b^3 x^{15/2} (4 a B+3 A b)+\frac{10}{13} a^3 b^2 x^{13/2} (3 a B+4 A b)+\frac{6}{11} a^4 b x^{11/2} (2 a B+5 A b)+\frac{2}{9} a^5 x^{9/2} (a B+6 A b)+\frac{2}{7} a^6 A x^{7/2}+\frac{2}{19} b^5 x^{19/2} (6 a B+A b)+\frac{6}{17} a b^4 x^{17/2} (5 a B+2 A b)+\frac{2}{21} b^6 B x^{21/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(2*a^6*A*x^(7/2))/7 + (2*a^5*(6*A*b + a*B)*x^(9/2))/9 + (6*a^4*b*(5*A*b + 2*a*B)*x^(11/2))/11 + (10*a^3*b^2*(4
*A*b + 3*a*B)*x^(13/2))/13 + (2*a^2*b^3*(3*A*b + 4*a*B)*x^(15/2))/3 + (6*a*b^4*(2*A*b + 5*a*B)*x^(17/2))/17 +
(2*b^5*(A*b + 6*a*B)*x^(19/2))/19 + (2*b^6*B*x^(21/2))/21

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 76

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*} \int x^{5/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx &=\int x^{5/2} (a+b x)^6 (A+B x) \, dx\\ &=\int \left (a^6 A x^{5/2}+a^5 (6 A b+a B) x^{7/2}+3 a^4 b (5 A b+2 a B) x^{9/2}+5 a^3 b^2 (4 A b+3 a B) x^{11/2}+5 a^2 b^3 (3 A b+4 a B) x^{13/2}+3 a b^4 (2 A b+5 a B) x^{15/2}+b^5 (A b+6 a B) x^{17/2}+b^6 B x^{19/2}\right ) \, dx\\ &=\frac{2}{7} a^6 A x^{7/2}+\frac{2}{9} a^5 (6 A b+a B) x^{9/2}+\frac{6}{11} a^4 b (5 A b+2 a B) x^{11/2}+\frac{10}{13} a^3 b^2 (4 A b+3 a B) x^{13/2}+\frac{2}{3} a^2 b^3 (3 A b+4 a B) x^{15/2}+\frac{6}{17} a b^4 (2 A b+5 a B) x^{17/2}+\frac{2}{19} b^5 (A b+6 a B) x^{19/2}+\frac{2}{21} b^6 B x^{21/2}\\ \end{align*}

Mathematica [A]  time = 0.0880796, size = 128, normalized size = 0.81 \[ \frac{2 \left (B x^{7/2} (a+b x)^7-\frac{7}{2} \left (2 a^2 b^4 x^{15/2}+\frac{40}{13} a^3 b^3 x^{13/2}+\frac{30}{11} a^4 b^2 x^{11/2}+\frac{4}{3} a^5 b x^{9/2}+\frac{2}{7} a^6 x^{7/2}+\frac{12}{17} a b^5 x^{17/2}+\frac{2}{19} b^6 x^{19/2}\right ) (a B-3 A b)\right )}{21 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(2*(B*x^(7/2)*(a + b*x)^7 - (7*(-3*A*b + a*B)*((2*a^6*x^(7/2))/7 + (4*a^5*b*x^(9/2))/3 + (30*a^4*b^2*x^(11/2))
/11 + (40*a^3*b^3*x^(13/2))/13 + 2*a^2*b^4*x^(15/2) + (12*a*b^5*x^(17/2))/17 + (2*b^6*x^(19/2))/19))/2))/(21*b
)

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Maple [A]  time = 0.009, size = 148, normalized size = 0.9 \begin{align*}{\frac{277134\,B{b}^{6}{x}^{7}+306306\,A{b}^{6}{x}^{6}+1837836\,B{x}^{6}a{b}^{5}+2054052\,aA{b}^{5}{x}^{5}+5135130\,B{x}^{5}{a}^{2}{b}^{4}+5819814\,{a}^{2}A{b}^{4}{x}^{4}+7759752\,B{x}^{4}{a}^{3}{b}^{3}+8953560\,{a}^{3}A{b}^{3}{x}^{3}+6715170\,B{x}^{3}{a}^{4}{b}^{2}+7936110\,{a}^{4}A{b}^{2}{x}^{2}+3174444\,B{x}^{2}{a}^{5}b+3879876\,{a}^{5}Abx+646646\,B{a}^{6}x+831402\,A{a}^{6}}{2909907}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2/2909907*x^(7/2)*(138567*B*b^6*x^7+153153*A*b^6*x^6+918918*B*a*b^5*x^6+1027026*A*a*b^5*x^5+2567565*B*a^2*b^4*
x^5+2909907*A*a^2*b^4*x^4+3879876*B*a^3*b^3*x^4+4476780*A*a^3*b^3*x^3+3357585*B*a^4*b^2*x^3+3968055*A*a^4*b^2*
x^2+1587222*B*a^5*b*x^2+1939938*A*a^5*b*x+323323*B*a^6*x+415701*A*a^6)

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Maxima [A]  time = 1.0118, size = 198, normalized size = 1.25 \begin{align*} \frac{2}{21} \, B b^{6} x^{\frac{21}{2}} + \frac{2}{7} \, A a^{6} x^{\frac{7}{2}} + \frac{2}{19} \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac{19}{2}} + \frac{6}{17} \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac{17}{2}} + \frac{2}{3} \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac{15}{2}} + \frac{10}{13} \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac{13}{2}} + \frac{6}{11} \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac{11}{2}} + \frac{2}{9} \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x^{\frac{9}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

2/21*B*b^6*x^(21/2) + 2/7*A*a^6*x^(7/2) + 2/19*(6*B*a*b^5 + A*b^6)*x^(19/2) + 6/17*(5*B*a^2*b^4 + 2*A*a*b^5)*x
^(17/2) + 2/3*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^(15/2) + 10/13*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^(13/2) + 6/11*(2*B*a^
5*b + 5*A*a^4*b^2)*x^(11/2) + 2/9*(B*a^6 + 6*A*a^5*b)*x^(9/2)

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Fricas [A]  time = 1.56087, size = 385, normalized size = 2.42 \begin{align*} \frac{2}{2909907} \,{\left (138567 \, B b^{6} x^{10} + 415701 \, A a^{6} x^{3} + 153153 \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{9} + 513513 \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{8} + 969969 \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{7} + 1119195 \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{6} + 793611 \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{5} + 323323 \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x^{4}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

2/2909907*(138567*B*b^6*x^10 + 415701*A*a^6*x^3 + 153153*(6*B*a*b^5 + A*b^6)*x^9 + 513513*(5*B*a^2*b^4 + 2*A*a
*b^5)*x^8 + 969969*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^7 + 1119195*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^6 + 793611*(2*B*a^5
*b + 5*A*a^4*b^2)*x^5 + 323323*(B*a^6 + 6*A*a^5*b)*x^4)*sqrt(x)

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Sympy [A]  time = 18.4307, size = 214, normalized size = 1.35 \begin{align*} \frac{2 A a^{6} x^{\frac{7}{2}}}{7} + \frac{4 A a^{5} b x^{\frac{9}{2}}}{3} + \frac{30 A a^{4} b^{2} x^{\frac{11}{2}}}{11} + \frac{40 A a^{3} b^{3} x^{\frac{13}{2}}}{13} + 2 A a^{2} b^{4} x^{\frac{15}{2}} + \frac{12 A a b^{5} x^{\frac{17}{2}}}{17} + \frac{2 A b^{6} x^{\frac{19}{2}}}{19} + \frac{2 B a^{6} x^{\frac{9}{2}}}{9} + \frac{12 B a^{5} b x^{\frac{11}{2}}}{11} + \frac{30 B a^{4} b^{2} x^{\frac{13}{2}}}{13} + \frac{8 B a^{3} b^{3} x^{\frac{15}{2}}}{3} + \frac{30 B a^{2} b^{4} x^{\frac{17}{2}}}{17} + \frac{12 B a b^{5} x^{\frac{19}{2}}}{19} + \frac{2 B b^{6} x^{\frac{21}{2}}}{21} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

2*A*a**6*x**(7/2)/7 + 4*A*a**5*b*x**(9/2)/3 + 30*A*a**4*b**2*x**(11/2)/11 + 40*A*a**3*b**3*x**(13/2)/13 + 2*A*
a**2*b**4*x**(15/2) + 12*A*a*b**5*x**(17/2)/17 + 2*A*b**6*x**(19/2)/19 + 2*B*a**6*x**(9/2)/9 + 12*B*a**5*b*x**
(11/2)/11 + 30*B*a**4*b**2*x**(13/2)/13 + 8*B*a**3*b**3*x**(15/2)/3 + 30*B*a**2*b**4*x**(17/2)/17 + 12*B*a*b**
5*x**(19/2)/19 + 2*B*b**6*x**(21/2)/21

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Giac [A]  time = 1.13989, size = 201, normalized size = 1.26 \begin{align*} \frac{2}{21} \, B b^{6} x^{\frac{21}{2}} + \frac{12}{19} \, B a b^{5} x^{\frac{19}{2}} + \frac{2}{19} \, A b^{6} x^{\frac{19}{2}} + \frac{30}{17} \, B a^{2} b^{4} x^{\frac{17}{2}} + \frac{12}{17} \, A a b^{5} x^{\frac{17}{2}} + \frac{8}{3} \, B a^{3} b^{3} x^{\frac{15}{2}} + 2 \, A a^{2} b^{4} x^{\frac{15}{2}} + \frac{30}{13} \, B a^{4} b^{2} x^{\frac{13}{2}} + \frac{40}{13} \, A a^{3} b^{3} x^{\frac{13}{2}} + \frac{12}{11} \, B a^{5} b x^{\frac{11}{2}} + \frac{30}{11} \, A a^{4} b^{2} x^{\frac{11}{2}} + \frac{2}{9} \, B a^{6} x^{\frac{9}{2}} + \frac{4}{3} \, A a^{5} b x^{\frac{9}{2}} + \frac{2}{7} \, A a^{6} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

2/21*B*b^6*x^(21/2) + 12/19*B*a*b^5*x^(19/2) + 2/19*A*b^6*x^(19/2) + 30/17*B*a^2*b^4*x^(17/2) + 12/17*A*a*b^5*
x^(17/2) + 8/3*B*a^3*b^3*x^(15/2) + 2*A*a^2*b^4*x^(15/2) + 30/13*B*a^4*b^2*x^(13/2) + 40/13*A*a^3*b^3*x^(13/2)
+ 12/11*B*a^5*b*x^(11/2) + 30/11*A*a^4*b^2*x^(11/2) + 2/9*B*a^6*x^(9/2) + 4/3*A*a^5*b*x^(9/2) + 2/7*A*a^6*x^(
7/2)