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

Optimal. Leaf size=67 \[ \frac {2 (a+b x)^{7/2} (A b-2 a B)}{7 b^3}-\frac {2 a (a+b x)^{5/2} (A b-a B)}{5 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \begin {gather*} \frac {2 (a+b x)^{7/2} (A b-2 a B)}{7 b^3}-\frac {2 a (a+b x)^{5/2} (A b-a B)}{5 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^(3/2)*(A + B*x),x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 49, normalized size = 0.73 \begin {gather*} \frac {2 (a+b x)^{5/2} \left (8 a^2 B-2 a b (9 A+10 B x)+5 b^2 x (9 A+7 B x)\right )}{315 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*(8*a^2*B + 5*b^2*x*(9*A + 7*B*x) - 2*a*b*(9*A + 10*B*x)))/(315*b^3)

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IntegrateAlgebraic [A]  time = 0.03, size = 56, normalized size = 0.84 \begin {gather*} \frac {2 (a+b x)^{5/2} \left (63 a^2 B+45 A b (a+b x)-63 a A b-90 a B (a+b x)+35 B (a+b x)^2\right )}{315 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*(-63*a*A*b + 63*a^2*B + 45*A*b*(a + b*x) - 90*a*B*(a + b*x) + 35*B*(a + b*x)^2))/(315*b^3)

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fricas [A]  time = 1.32, size = 95, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (35 \, B b^{4} x^{4} + 8 \, B a^{4} - 18 \, A a^{3} b + 5 \, {\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{3} + 3 \, {\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{2} - {\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x + a}}{315 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*b^4*x^4 + 8*B*a^4 - 18*A*a^3*b + 5*(10*B*a*b^3 + 9*A*b^4)*x^3 + 3*(B*a^2*b^2 + 24*A*a*b^3)*x^2 - (
4*B*a^3*b - 9*A*a^2*b^2)*x)*sqrt(b*x + a)/b^3

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giac [B]  time = 1.21, size = 275, normalized size = 4.10 \begin {gather*} \frac {2 \, {\left (\frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A a^{2}}{b} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B a^{2}}{b^{2}} + \frac {42 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A a}{b} + \frac {18 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B a}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A}{b} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B}{b^{2}}\right )}}{315 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A*a^2/b + 21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*s
qrt(b*x + a)*a^2)*B*a^2/b^2 + 42*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*A*a/b + 18*
(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B*a/b^2 + 9*(5*(b*x
+ a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*A/b + (35*(b*x + a)^(9/2)
- 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*B/b^2)/b

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maple [A]  time = 0.00, size = 47, normalized size = 0.70 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-35 B \,b^{2} x^{2}-45 A \,b^{2} x +20 B a b x +18 A a b -8 B \,a^{2}\right )}{315 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(b*x+a)^(5/2)*(-35*B*b^2*x^2-45*A*b^2*x+20*B*a*b*x+18*A*a*b-8*B*a^2)/b^3

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maxima [A]  time = 0.90, size = 54, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} B - 45 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 63 \, {\left (B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{315 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(b*x + a)^(9/2)*B - 45*(2*B*a - A*b)*(b*x + a)^(7/2) + 63*(B*a^2 - A*a*b)*(b*x + a)^(5/2))/b^3

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mupad [B]  time = 0.06, size = 52, normalized size = 0.78 \begin {gather*} \frac {2\,{\left (a+b\,x\right )}^{5/2}\,\left (63\,B\,a^2+35\,B\,{\left (a+b\,x\right )}^2-63\,A\,a\,b+45\,A\,b\,\left (a+b\,x\right )-90\,B\,a\,\left (a+b\,x\right )\right )}{315\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(A + B*x)*(a + b*x)^(3/2),x)

[Out]

(2*(a + b*x)^(5/2)*(63*B*a^2 + 35*B*(a + b*x)^2 - 63*A*a*b + 45*A*b*(a + b*x) - 90*B*a*(a + b*x)))/(315*b^3)

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sympy [B]  time = 8.46, size = 178, normalized size = 2.66 \begin {gather*} \frac {2 A a \left (- \frac {a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x\right )^{\frac {5}{2}}}{5}\right )}{b^{2}} + \frac {2 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^{2}} + \frac {2 B 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 B \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*(-a*(a + b*x)**(3/2)/3 + (a + b*x)**(5/2)/5)/b**2 + 2*A*(a**2*(a + b*x)**(3/2)/3 - 2*a*(a + b*x)**(5/2)/
5 + (a + b*x)**(7/2)/7)/b**2 + 2*B*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*B*(-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

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