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

Optimal. Leaf size=63 \[ \frac{2}{5} a^2 A x^{5/2}+\frac{2}{9} b x^{9/2} (2 a B+A b)+\frac{2}{7} a x^{7/2} (a B+2 A b)+\frac{2}{11} b^2 B x^{11/2} \]

[Out]

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

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Rubi [A]  time = 0.0276435, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {27, 76} \[ \frac{2}{5} a^2 A x^{5/2}+\frac{2}{9} b x^{9/2} (2 a B+A b)+\frac{2}{7} a x^{7/2} (a B+2 A b)+\frac{2}{11} b^2 B x^{11/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0169175, size = 52, normalized size = 0.83 \[ \frac{2 x^{5/2} \left (99 a^2 (7 A+5 B x)+110 a b x (9 A+7 B x)+35 b^2 x^2 (11 A+9 B x)\right )}{3465} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(99*a^2*(7*A + 5*B*x) + 110*a*b*x*(9*A + 7*B*x) + 35*b^2*x^2*(11*A + 9*B*x)))/3465

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Maple [A]  time = 0.007, size = 52, normalized size = 0.8 \begin{align*}{\frac{630\,{b}^{2}B{x}^{3}+770\,A{b}^{2}{x}^{2}+1540\,B{x}^{2}ab+1980\,aAbx+990\,{a}^{2}Bx+1386\,A{a}^{2}}{3465}{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),x)

[Out]

2/3465*x^(5/2)*(315*B*b^2*x^3+385*A*b^2*x^2+770*B*a*b*x^2+990*A*a*b*x+495*B*a^2*x+693*A*a^2)

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Maxima [A]  time = 1.00835, size = 69, normalized size = 1.1 \begin{align*} \frac{2}{11} \, B b^{2} x^{\frac{11}{2}} + \frac{2}{5} \, A a^{2} x^{\frac{5}{2}} + \frac{2}{9} \,{\left (2 \, B a b + A b^{2}\right )} x^{\frac{9}{2}} + \frac{2}{7} \,{\left (B a^{2} + 2 \, A a 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53849, size = 140, normalized size = 2.22 \begin{align*} \frac{2}{3465} \,{\left (315 \, B b^{2} x^{5} + 693 \, A a^{2} x^{2} + 385 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} + 495 \,{\left (B a^{2} + 2 \, A a 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),x, algorithm="fricas")

[Out]

2/3465*(315*B*b^2*x^5 + 693*A*a^2*x^2 + 385*(2*B*a*b + A*b^2)*x^4 + 495*(B*a^2 + 2*A*a*b)*x^3)*sqrt(x)

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Sympy [A]  time = 1.83519, size = 80, normalized size = 1.27 \begin{align*} \frac{2 A a^{2} x^{\frac{5}{2}}}{5} + \frac{4 A a b x^{\frac{7}{2}}}{7} + \frac{2 A b^{2} x^{\frac{9}{2}}}{9} + \frac{2 B a^{2} x^{\frac{7}{2}}}{7} + \frac{4 B a b x^{\frac{9}{2}}}{9} + \frac{2 B b^{2} x^{\frac{11}{2}}}{11} \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),x)

[Out]

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

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Giac [A]  time = 1.18957, size = 72, normalized size = 1.14 \begin{align*} \frac{2}{11} \, B b^{2} x^{\frac{11}{2}} + \frac{4}{9} \, B a b x^{\frac{9}{2}} + \frac{2}{9} \, A b^{2} x^{\frac{9}{2}} + \frac{2}{7} \, B a^{2} x^{\frac{7}{2}} + \frac{4}{7} \, A a b x^{\frac{7}{2}} + \frac{2}{5} \, A a^{2} 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),x, algorithm="giac")

[Out]

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