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

Optimal. Leaf size=122 \[ \frac{2 a^2 (a+b x)^{7/2} (3 A b-4 a B)}{7 b^5}-\frac{2 a^3 (a+b x)^{5/2} (A b-a B)}{5 b^5}+\frac{2 (a+b x)^{11/2} (A b-4 a B)}{11 b^5}-\frac{2 a (a+b x)^{9/2} (A b-2 a B)}{3 b^5}+\frac{2 B (a+b x)^{13/2}}{13 b^5} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0479089, antiderivative size = 122, 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} \[ \frac{2 a^2 (a+b x)^{7/2} (3 A b-4 a B)}{7 b^5}-\frac{2 a^3 (a+b x)^{5/2} (A b-a B)}{5 b^5}+\frac{2 (a+b x)^{11/2} (A b-4 a B)}{11 b^5}-\frac{2 a (a+b x)^{9/2} (A b-2 a B)}{3 b^5}+\frac{2 B (a+b x)^{13/2}}{13 b^5} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x)^(3/2)*(A + B*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0651178, size = 87, normalized size = 0.71 \[ \frac{2 (a+b x)^{5/2} \left (40 a^2 b^2 x (13 A+14 B x)-16 a^3 b (13 A+20 B x)+128 a^4 B-70 a b^3 x^2 (13 A+12 B x)+105 b^4 x^3 (13 A+11 B x)\right )}{15015 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*(128*a^4*B + 105*b^4*x^3*(13*A + 11*B*x) - 70*a*b^3*x^2*(13*A + 12*B*x) + 40*a^2*b^2*x*(13*
A + 14*B*x) - 16*a^3*b*(13*A + 20*B*x)))/(15015*b^5)

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 95, normalized size = 0.8 \begin{align*} -{\frac{-2310\,B{x}^{4}{b}^{4}-2730\,A{b}^{4}{x}^{3}+1680\,Ba{b}^{3}{x}^{3}+1820\,Aa{b}^{3}{x}^{2}-1120\,B{a}^{2}{b}^{2}{x}^{2}-1040\,A{a}^{2}{b}^{2}x+640\,B{a}^{3}bx+416\,A{a}^{3}b-256\,B{a}^{4}}{15015\,{b}^{5}} \left ( bx+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15015*(b*x+a)^(5/2)*(-1155*B*b^4*x^4-1365*A*b^4*x^3+840*B*a*b^3*x^3+910*A*a*b^3*x^2-560*B*a^2*b^2*x^2-520*A
*a^2*b^2*x+320*B*a^3*b*x+208*A*a^3*b-128*B*a^4)/b^5

________________________________________________________________________________________

Maxima [A]  time = 1.11627, size = 135, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (1155 \,{\left (b x + a\right )}^{\frac{13}{2}} B - 1365 \,{\left (4 \, B a - A b\right )}{\left (b x + a\right )}^{\frac{11}{2}} + 5005 \,{\left (2 \, B a^{2} - A a b\right )}{\left (b x + a\right )}^{\frac{9}{2}} - 2145 \,{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )}{\left (b x + a\right )}^{\frac{7}{2}} + 3003 \,{\left (B a^{4} - A a^{3} b\right )}{\left (b x + a\right )}^{\frac{5}{2}}\right )}}{15015 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(1155*(b*x + a)^(13/2)*B - 1365*(4*B*a - A*b)*(b*x + a)^(11/2) + 5005*(2*B*a^2 - A*a*b)*(b*x + a)^(9/2
) - 2145*(4*B*a^3 - 3*A*a^2*b)*(b*x + a)^(7/2) + 3003*(B*a^4 - A*a^3*b)*(b*x + a)^(5/2))/b^5

________________________________________________________________________________________

Fricas [A]  time = 2.35458, size = 331, normalized size = 2.71 \begin{align*} \frac{2 \,{\left (1155 \, B b^{6} x^{6} + 128 \, B a^{6} - 208 \, A a^{5} b + 105 \,{\left (14 \, B a b^{5} + 13 \, A b^{6}\right )} x^{5} + 35 \,{\left (B a^{2} b^{4} + 52 \, A a b^{5}\right )} x^{4} - 5 \,{\left (8 \, B a^{3} b^{3} - 13 \, A a^{2} b^{4}\right )} x^{3} + 6 \,{\left (8 \, B a^{4} b^{2} - 13 \, A a^{3} b^{3}\right )} x^{2} - 8 \,{\left (8 \, B a^{5} b - 13 \, A a^{4} b^{2}\right )} x\right )} \sqrt{b x + a}}{15015 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(1155*B*b^6*x^6 + 128*B*a^6 - 208*A*a^5*b + 105*(14*B*a*b^5 + 13*A*b^6)*x^5 + 35*(B*a^2*b^4 + 52*A*a*b
^5)*x^4 - 5*(8*B*a^3*b^3 - 13*A*a^2*b^4)*x^3 + 6*(8*B*a^4*b^2 - 13*A*a^3*b^3)*x^2 - 8*(8*B*a^5*b - 13*A*a^4*b^
2)*x)*sqrt(b*x + a)/b^5

________________________________________________________________________________________

Sympy [B]  time = 11.6963, size = 298, normalized size = 2.44 \begin{align*} \frac{2 A 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 A \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}} + \frac{2 B a \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^{5}} + \frac{2 B \left (- \frac{a^{5} \left (a + b x\right )^{\frac{3}{2}}}{3} + a^{4} \left (a + b x\right )^{\frac{5}{2}} - \frac{10 a^{3} \left (a + b x\right )^{\frac{7}{2}}}{7} + \frac{10 a^{2} \left (a + b x\right )^{\frac{9}{2}}}{9} - \frac{5 a \left (a + b x\right )^{\frac{11}{2}}}{11} + \frac{\left (a + b x\right )^{\frac{13}{2}}}{13}\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*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*A*(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 + 2*B*a*(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**5 + 2*B*(-a**5*(a + b*x)**(3/2)/3 + a**4*(a +
b*x)**(5/2) - 10*a**3*(a + b*x)**(7/2)/7 + 10*a**2*(a + b*x)**(9/2)/9 - 5*a*(a + b*x)**(11/2)/11 + (a + b*x)**
(13/2)/13)/b**5

________________________________________________________________________________________

Giac [B]  time = 1.16257, size = 346, normalized size = 2.84 \begin{align*} \frac{2 \,{\left (\frac{143 \,{\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 a}{b^{3}} + \frac{13 \,{\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 a}{b^{4}} + \frac{13 \,{\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 )} A}{b^{3}} + \frac{5 \,{\left (693 \,{\left (b x + a\right )}^{\frac{13}{2}} - 4095 \,{\left (b x + a\right )}^{\frac{11}{2}} a + 10010 \,{\left (b x + a\right )}^{\frac{9}{2}} a^{2} - 12870 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{3} + 9009 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{4} - 3003 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{5}\right )} B}{b^{4}}\right )}}{45045 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(143*(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*a/b^3 + 13*(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*a/b^4 + 13*(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)*A/b^3 + 5*(693*(b*x + a)^(13/2) - 4095*(b*x +
a)^(11/2)*a + 10010*(b*x + a)^(9/2)*a^2 - 12870*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 3003*(b*x + a
)^(3/2)*a^5)*B/b^4)/b