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

Optimal. Leaf size=111 \[ \frac{4}{9} a^2 b x^{9/2} (2 a B+3 A b)+\frac{2}{7} a^3 x^{7/2} (a B+4 A b)+\frac{2}{5} a^4 A x^{5/2}+\frac{2}{13} b^3 x^{13/2} (4 a B+A b)+\frac{4}{11} a b^2 x^{11/2} (3 a B+2 A b)+\frac{2}{15} b^4 B x^{15/2} \]

[Out]

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

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Rubi [A]  time = 0.0522332, antiderivative size = 111, 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} \[ \frac{4}{9} a^2 b x^{9/2} (2 a B+3 A b)+\frac{2}{7} a^3 x^{7/2} (a B+4 A b)+\frac{2}{5} a^4 A x^{5/2}+\frac{2}{13} b^3 x^{13/2} (4 a B+A b)+\frac{4}{11} a b^2 x^{11/2} (3 a B+2 A b)+\frac{2}{15} b^4 B x^{15/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0598668, size = 81, normalized size = 0.73 \[ \frac{2 \left (\frac{x^{5/2} \left (10010 a^2 b^2 x^2+8580 a^3 b x+3003 a^4+5460 a b^3 x^3+1155 b^4 x^4\right ) (3 A b-a B)}{3003}+B x^{5/2} (a+b x)^5\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*x^(5/2)*(a + b*x)^5 + ((3*A*b - a*B)*x^(5/2)*(3003*a^4 + 8580*a^3*b*x + 10010*a^2*b^2*x^2 + 5460*a*b^3*x
^3 + 1155*b^4*x^4))/3003))/(15*b)

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Maple [A]  time = 0.006, size = 100, normalized size = 0.9 \begin{align*}{\frac{6006\,{b}^{4}B{x}^{5}+6930\,A{b}^{4}{x}^{4}+27720\,B{x}^{4}a{b}^{3}+32760\,aA{b}^{3}{x}^{3}+49140\,B{x}^{3}{a}^{2}{b}^{2}+60060\,{a}^{2}A{b}^{2}{x}^{2}+40040\,B{x}^{2}{a}^{3}b+51480\,{a}^{3}Abx+12870\,{a}^{4}Bx+18018\,A{a}^{4}}{45045}{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)^2,x)

[Out]

2/45045*x^(5/2)*(3003*B*b^4*x^5+3465*A*b^4*x^4+13860*B*a*b^3*x^4+16380*A*a*b^3*x^3+24570*B*a^2*b^2*x^3+30030*A
*a^2*b^2*x^2+20020*B*a^3*b*x^2+25740*A*a^3*b*x+6435*B*a^4*x+9009*A*a^4)

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Maxima [A]  time = 1.02669, size = 134, normalized size = 1.21 \begin{align*} \frac{2}{15} \, B b^{4} x^{\frac{15}{2}} + \frac{2}{5} \, A a^{4} x^{\frac{5}{2}} + \frac{2}{13} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac{13}{2}} + \frac{4}{11} \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac{11}{2}} + \frac{4}{9} \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac{9}{2}} + \frac{2}{7} \,{\left (B a^{4} + 4 \, A a^{3} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56019, size = 254, normalized size = 2.29 \begin{align*} \frac{2}{45045} \,{\left (3003 \, B b^{4} x^{7} + 9009 \, A a^{4} x^{2} + 3465 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{6} + 8190 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{5} + 10010 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 6435 \,{\left (B a^{4} + 4 \, A a^{3} 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)^2,x, algorithm="fricas")

[Out]

2/45045*(3003*B*b^4*x^7 + 9009*A*a^4*x^2 + 3465*(4*B*a*b^3 + A*b^4)*x^6 + 8190*(3*B*a^2*b^2 + 2*A*a*b^3)*x^5 +
10010*(2*B*a^3*b + 3*A*a^2*b^2)*x^4 + 6435*(B*a^4 + 4*A*a^3*b)*x^3)*sqrt(x)

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Sympy [A]  time = 4.4561, size = 148, normalized size = 1.33 \begin{align*} \frac{2 A a^{4} x^{\frac{5}{2}}}{5} + \frac{8 A a^{3} b x^{\frac{7}{2}}}{7} + \frac{4 A a^{2} b^{2} x^{\frac{9}{2}}}{3} + \frac{8 A a b^{3} x^{\frac{11}{2}}}{11} + \frac{2 A b^{4} x^{\frac{13}{2}}}{13} + \frac{2 B a^{4} x^{\frac{7}{2}}}{7} + \frac{8 B a^{3} b x^{\frac{9}{2}}}{9} + \frac{12 B a^{2} b^{2} x^{\frac{11}{2}}}{11} + \frac{8 B a b^{3} x^{\frac{13}{2}}}{13} + \frac{2 B b^{4} x^{\frac{15}{2}}}{15} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.15417, size = 136, normalized size = 1.23 \begin{align*} \frac{2}{15} \, B b^{4} x^{\frac{15}{2}} + \frac{8}{13} \, B a b^{3} x^{\frac{13}{2}} + \frac{2}{13} \, A b^{4} x^{\frac{13}{2}} + \frac{12}{11} \, B a^{2} b^{2} x^{\frac{11}{2}} + \frac{8}{11} \, A a b^{3} x^{\frac{11}{2}} + \frac{8}{9} \, B a^{3} b x^{\frac{9}{2}} + \frac{4}{3} \, A a^{2} b^{2} x^{\frac{9}{2}} + \frac{2}{7} \, B a^{4} x^{\frac{7}{2}} + \frac{8}{7} \, A a^{3} b x^{\frac{7}{2}} + \frac{2}{5} \, A a^{4} 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)^2,x, algorithm="giac")

[Out]

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