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

Optimal. Leaf size=95 \[ \frac{2 a^2 (a+b x)^{5/2} (A b-a B)}{5 b^4}+\frac{2 (a+b x)^{9/2} (A b-3 a B)}{9 b^4}-\frac{2 a (a+b x)^{7/2} (2 A b-3 a B)}{7 b^4}+\frac{2 B (a+b x)^{11/2}}{11 b^4} \]

[Out]

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

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Rubi [A]  time = 0.0349589, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {77} \[ \frac{2 a^2 (a+b x)^{5/2} (A b-a B)}{5 b^4}+\frac{2 (a+b x)^{9/2} (A b-3 a B)}{9 b^4}-\frac{2 a (a+b x)^{7/2} (2 A b-3 a B)}{7 b^4}+\frac{2 B (a+b x)^{11/2}}{11 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^(3/2)*(A + B*x),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0505772, size = 68, normalized size = 0.72 \[ \frac{2 (a+b x)^{5/2} \left (8 a^2 b (11 A+15 B x)-48 a^3 B-10 a b^2 x (22 A+21 B x)+35 b^3 x^2 (11 A+9 B x)\right )}{3465 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*(-48*a^3*B + 35*b^3*x^2*(11*A + 9*B*x) + 8*a^2*b*(11*A + 15*B*x) - 10*a*b^2*x*(22*A + 21*B*
x)))/(3465*b^4)

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Maple [A]  time = 0.004, size = 71, normalized size = 0.8 \begin{align*}{\frac{630\,{b}^{3}B{x}^{3}+770\,A{x}^{2}{b}^{3}-420\,B{x}^{2}a{b}^{2}-440\,a{b}^{2}Ax+240\,{a}^{2}bBx+176\,A{a}^{2}b-96\,B{a}^{3}}{3465\,{b}^{4}} \left ( bx+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(b*x+a)^(5/2)*(315*B*b^3*x^3+385*A*b^3*x^2-210*B*a*b^2*x^2-220*A*a*b^2*x+120*B*a^2*b*x+88*A*a^2*b-48*B*
a^3)/b^4

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Maxima [A]  time = 1.14303, size = 104, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} B - 385 \,{\left (3 \, B a - A b\right )}{\left (b x + a\right )}^{\frac{9}{2}} + 495 \,{\left (3 \, B a^{2} - 2 \, A a b\right )}{\left (b x + a\right )}^{\frac{7}{2}} - 693 \,{\left (B a^{3} - A a^{2} b\right )}{\left (b x + a\right )}^{\frac{5}{2}}\right )}}{3465 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*(b*x + a)^(11/2)*B - 385*(3*B*a - A*b)*(b*x + a)^(9/2) + 495*(3*B*a^2 - 2*A*a*b)*(b*x + a)^(7/2) -
693*(B*a^3 - A*a^2*b)*(b*x + a)^(5/2))/b^4

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Fricas [A]  time = 2.26345, size = 277, normalized size = 2.92 \begin{align*} \frac{2 \,{\left (315 \, B b^{5} x^{5} - 48 \, B a^{5} + 88 \, A a^{4} b + 35 \,{\left (12 \, B a b^{4} + 11 \, A b^{5}\right )} x^{4} + 5 \,{\left (3 \, B a^{2} b^{3} + 110 \, A a b^{4}\right )} x^{3} - 3 \,{\left (6 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{2} + 4 \,{\left (6 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x\right )} \sqrt{b x + a}}{3465 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*B*b^5*x^5 - 48*B*a^5 + 88*A*a^4*b + 35*(12*B*a*b^4 + 11*A*b^5)*x^4 + 5*(3*B*a^2*b^3 + 110*A*a*b^4)
*x^3 - 3*(6*B*a^3*b^2 - 11*A*a^2*b^3)*x^2 + 4*(6*B*a^4*b - 11*A*a^3*b^2)*x)*sqrt(b*x + a)/b^4

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Sympy [B]  time = 9.66342, size = 240, normalized size = 2.53 \begin{align*} \frac{2 A a \left (\frac{a^{2} \left (a + b x\right )^{\frac{3}{2}}}{3} - \frac{2 a \left (a + b x\right )^{\frac{5}{2}}}{5} + \frac{\left (a + b x\right )^{\frac{7}{2}}}{7}\right )}{b^{3}} + \frac{2 A \left (- \frac{a^{3} \left (a + b x\right )^{\frac{3}{2}}}{3} + \frac{3 a^{2} \left (a + b x\right )^{\frac{5}{2}}}{5} - \frac{3 a \left (a + b x\right )^{\frac{7}{2}}}{7} + \frac{\left (a + b x\right )^{\frac{9}{2}}}{9}\right )}{b^{3}} + \frac{2 B a \left (- \frac{a^{3} \left (a + b x\right )^{\frac{3}{2}}}{3} + \frac{3 a^{2} \left (a + b x\right )^{\frac{5}{2}}}{5} - \frac{3 a \left (a + b x\right )^{\frac{7}{2}}}{7} + \frac{\left (a + b x\right )^{\frac{9}{2}}}{9}\right )}{b^{4}} + \frac{2 B \left (\frac{a^{4} \left (a + b x\right )^{\frac{3}{2}}}{3} - \frac{4 a^{3} \left (a + b x\right )^{\frac{5}{2}}}{5} + \frac{6 a^{2} \left (a + b x\right )^{\frac{7}{2}}}{7} - \frac{4 a \left (a + b x\right )^{\frac{9}{2}}}{9} + \frac{\left (a + b x\right )^{\frac{11}{2}}}{11}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*(a**2*(a + b*x)**(3/2)/3 - 2*a*(a + b*x)**(5/2)/5 + (a + b*x)**(7/2)/7)/b**3 + 2*A*(-a**3*(a + b*x)**(3/
2)/3 + 3*a**2*(a + b*x)**(5/2)/5 - 3*a*(a + b*x)**(7/2)/7 + (a + b*x)**(9/2)/9)/b**3 + 2*B*a*(-a**3*(a + b*x)*
*(3/2)/3 + 3*a**2*(a + b*x)**(5/2)/5 - 3*a*(a + b*x)**(7/2)/7 + (a + b*x)**(9/2)/9)/b**4 + 2*B*(a**4*(a + b*x)
**(3/2)/3 - 4*a**3*(a + b*x)**(5/2)/5 + 6*a**2*(a + b*x)**(7/2)/7 - 4*a*(a + b*x)**(9/2)/9 + (a + b*x)**(11/2)
/11)/b**4

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Giac [B]  time = 1.18442, size = 279, normalized size = 2.94 \begin{align*} \frac{2 \,{\left (\frac{33 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} A a}{b^{2}} + \frac{11 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} B a}{b^{3}} + \frac{11 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} A}{b^{2}} + \frac{{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}\right )} B}{b^{3}}\right )}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(33*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*A*a/b^2 + 11*(35*(b*x + a)^(9/
2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)*B*a/b^3 + 11*(35*(b*x + a)^(9/
2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)*A/b^2 + (315*(b*x + a)^(11/2)
- 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(7/2)*a^2 - 2772*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4)*B/b
^3)/b