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

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

[Out]

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

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Rubi [A]  time = 0.084969, 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} \[ \frac{10}{13} a^2 b^3 x^{13/2} (4 a B+3 A b)+\frac{10}{11} a^3 b^2 x^{11/2} (3 a B+4 A b)+\frac{2}{3} a^4 b x^{9/2} (2 a B+5 A b)+\frac{2}{7} a^5 x^{7/2} (a B+6 A b)+\frac{2}{5} a^6 A x^{5/2}+\frac{2}{17} b^5 x^{17/2} (6 a B+A b)+\frac{2}{5} a b^4 x^{15/2} (5 a B+2 A b)+\frac{2}{19} b^6 B x^{19/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0876669, size = 103, normalized size = 0.65 \[ \frac{2 \left (\frac{x^{5/2} \left (425425 a^4 b^2 x^2+464100 a^3 b^3 x^3+294525 a^2 b^4 x^4+218790 a^5 b x+51051 a^6+102102 a b^5 x^5+15015 b^6 x^6\right ) (19 A b-5 a B)}{255255}+B x^{5/2} (a+b x)^7\right )}{19 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*x^(5/2)*(a + b*x)^7 + ((19*A*b - 5*a*B)*x^(5/2)*(51051*a^6 + 218790*a^5*b*x + 425425*a^4*b^2*x^2 + 46410
0*a^3*b^3*x^3 + 294525*a^2*b^4*x^4 + 102102*a*b^5*x^5 + 15015*b^6*x^6))/255255))/(19*b)

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Maple [A]  time = 0.007, size = 148, normalized size = 0.9 \begin{align*}{\frac{510510\,B{b}^{6}{x}^{7}+570570\,A{b}^{6}{x}^{6}+3423420\,B{x}^{6}a{b}^{5}+3879876\,aA{b}^{5}{x}^{5}+9699690\,B{x}^{5}{a}^{2}{b}^{4}[phone]\,{a}^{2}A{b}^{4}{x}^{4}[phone]\,B{x}^{4}{a}^{3}{b}^{3}[phone]\,{a}^{3}A{b}^{3}{x}^{3}[phone]\,B{x}^{3}{a}^{4}{b}^{2}[phone]\,{a}^{4}A{b}^{2}{x}^{2}+6466460\,B{x}^{2}{a}^{5}b+8314020\,{a}^{5}Abx+1385670\,B{a}^{6}x+1939938\,A{a}^{6}}{4849845}{x}^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/4849845*x^(5/2)*(255255*B*b^6*x^7+285285*A*b^6*x^6+1711710*B*a*b^5*x^6+1939938*A*a*b^5*x^5+4849845*B*a^2*b^4
*x^5+5595975*A*a^2*b^4*x^4+7461300*B*a^3*b^3*x^4+8817900*A*a^3*b^3*x^3+6613425*B*a^4*b^2*x^3+8083075*A*a^4*b^2
*x^2+3233230*B*a^5*b*x^2+4157010*A*a^5*b*x+692835*B*a^6*x+969969*A*a^6)

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Maxima [A]  time = 1.02491, size = 198, normalized size = 1.25 \begin{align*} \frac{2}{19} \, B b^{6} x^{\frac{19}{2}} + \frac{2}{5} \, A a^{6} x^{\frac{5}{2}} + \frac{2}{17} \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac{17}{2}} + \frac{2}{5} \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac{15}{2}} + \frac{10}{13} \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac{13}{2}} + \frac{10}{11} \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac{11}{2}} + \frac{2}{3} \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac{9}{2}} + \frac{2}{7} \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46344, size = 386, normalized size = 2.43 \begin{align*} \frac{2}{4849845} \,{\left (255255 \, B b^{6} x^{9} + 969969 \, A a^{6} x^{2} + 285285 \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{8} + 969969 \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{7} + 1865325 \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{6} + 2204475 \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{5} + 1616615 \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{4} + 692835 \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/4849845*(255255*B*b^6*x^9 + 969969*A*a^6*x^2 + 285285*(6*B*a*b^5 + A*b^6)*x^8 + 969969*(5*B*a^2*b^4 + 2*A*a*
b^5)*x^7 + 1865325*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^6 + 2204475*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^5 + 1616615*(2*B*a^
5*b + 5*A*a^4*b^2)*x^4 + 692835*(B*a^6 + 6*A*a^5*b)*x^3)*sqrt(x)

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Sympy [A]  time = 10.5436, size = 214, normalized size = 1.35 \begin{align*} \frac{2 A a^{6} x^{\frac{5}{2}}}{5} + \frac{12 A a^{5} b x^{\frac{7}{2}}}{7} + \frac{10 A a^{4} b^{2} x^{\frac{9}{2}}}{3} + \frac{40 A a^{3} b^{3} x^{\frac{11}{2}}}{11} + \frac{30 A a^{2} b^{4} x^{\frac{13}{2}}}{13} + \frac{4 A a b^{5} x^{\frac{15}{2}}}{5} + \frac{2 A b^{6} x^{\frac{17}{2}}}{17} + \frac{2 B a^{6} x^{\frac{7}{2}}}{7} + \frac{4 B a^{5} b x^{\frac{9}{2}}}{3} + \frac{30 B a^{4} b^{2} x^{\frac{11}{2}}}{11} + \frac{40 B a^{3} b^{3} x^{\frac{13}{2}}}{13} + 2 B a^{2} b^{4} x^{\frac{15}{2}} + \frac{12 B a b^{5} x^{\frac{17}{2}}}{17} + \frac{2 B b^{6} x^{\frac{19}{2}}}{19} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**6*x**(5/2)/5 + 12*A*a**5*b*x**(7/2)/7 + 10*A*a**4*b**2*x**(9/2)/3 + 40*A*a**3*b**3*x**(11/2)/11 + 30*A*
a**2*b**4*x**(13/2)/13 + 4*A*a*b**5*x**(15/2)/5 + 2*A*b**6*x**(17/2)/17 + 2*B*a**6*x**(7/2)/7 + 4*B*a**5*b*x**
(9/2)/3 + 30*B*a**4*b**2*x**(11/2)/11 + 40*B*a**3*b**3*x**(13/2)/13 + 2*B*a**2*b**4*x**(15/2) + 12*B*a*b**5*x*
*(17/2)/17 + 2*B*b**6*x**(19/2)/19

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Giac [A]  time = 1.10948, size = 201, normalized size = 1.26 \begin{align*} \frac{2}{19} \, B b^{6} x^{\frac{19}{2}} + \frac{12}{17} \, B a b^{5} x^{\frac{17}{2}} + \frac{2}{17} \, A b^{6} x^{\frac{17}{2}} + 2 \, B a^{2} b^{4} x^{\frac{15}{2}} + \frac{4}{5} \, A a b^{5} x^{\frac{15}{2}} + \frac{40}{13} \, B a^{3} b^{3} x^{\frac{13}{2}} + \frac{30}{13} \, A a^{2} b^{4} x^{\frac{13}{2}} + \frac{30}{11} \, B a^{4} b^{2} x^{\frac{11}{2}} + \frac{40}{11} \, A a^{3} b^{3} x^{\frac{11}{2}} + \frac{4}{3} \, B a^{5} b x^{\frac{9}{2}} + \frac{10}{3} \, A a^{4} b^{2} x^{\frac{9}{2}} + \frac{2}{7} \, B a^{6} x^{\frac{7}{2}} + \frac{12}{7} \, A a^{5} b x^{\frac{7}{2}} + \frac{2}{5} \, A a^{6} x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*B*b^6*x^(19/2) + 12/17*B*a*b^5*x^(17/2) + 2/17*A*b^6*x^(17/2) + 2*B*a^2*b^4*x^(15/2) + 4/5*A*a*b^5*x^(15/
2) + 40/13*B*a^3*b^3*x^(13/2) + 30/13*A*a^2*b^4*x^(13/2) + 30/11*B*a^4*b^2*x^(11/2) + 40/11*A*a^3*b^3*x^(11/2)
+ 4/3*B*a^5*b*x^(9/2) + 10/3*A*a^4*b^2*x^(9/2) + 2/7*B*a^6*x^(7/2) + 12/7*A*a^5*b*x^(7/2) + 2/5*A*a^6*x^(5/2)