3.412 \(\int x^2 (a+b x)^{5/2} (A+B x) \, dx\)

Optimal. Leaf size=95 \[ \frac{2 a^2 (a+b x)^{7/2} (A b-a B)}{7 b^4}+\frac{2 (a+b x)^{11/2} (A b-3 a B)}{11 b^4}-\frac{2 a (a+b x)^{9/2} (2 A b-3 a B)}{9 b^4}+\frac{2 B (a+b x)^{13/2}}{13 b^4} \]

[Out]

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

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Rubi [A]  time = 0.0336559, 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} \[ \frac{2 a^2 (a+b x)^{7/2} (A b-a B)}{7 b^4}+\frac{2 (a+b x)^{11/2} (A b-3 a B)}{11 b^4}-\frac{2 a (a+b x)^{9/2} (2 A b-3 a B)}{9 b^4}+\frac{2 B (a+b x)^{13/2}}{13 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^(5/2)*(A + B*x),x]

[Out]

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

Mathematica [A]  time = 0.0545299, size = 68, normalized size = 0.72 \[ \frac{2 (a+b x)^{7/2} \left (8 a^2 b (13 A+21 B x)-48 a^3 B-14 a b^2 x (26 A+27 B x)+63 b^3 x^2 (13 A+11 B x)\right )}{9009 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x)^(5/2)*(A + B*x),x]

[Out]

(2*(a + b*x)^(7/2)*(-48*a^3*B + 63*b^3*x^2*(13*A + 11*B*x) + 8*a^2*b*(13*A + 21*B*x) - 14*a*b^2*x*(26*A + 27*B
*x)))/(9009*b^4)

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Maple [A]  time = 0.004, size = 71, normalized size = 0.8 \begin{align*}{\frac{1386\,{b}^{3}B{x}^{3}+1638\,A{x}^{2}{b}^{3}-756\,B{x}^{2}a{b}^{2}-728\,a{b}^{2}Ax+336\,{a}^{2}bBx+208\,A{a}^{2}b-96\,B{a}^{3}}{9009\,{b}^{4}} \left ( bx+a \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^(5/2)*(B*x+A),x)

[Out]

2/9009*(b*x+a)^(7/2)*(693*B*b^3*x^3+819*A*b^3*x^2-378*B*a*b^2*x^2-364*A*a*b^2*x+168*B*a^2*b*x+104*A*a^2*b-48*B
*a^3)/b^4

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Maxima [A]  time = 1.32472, size = 104, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (693 \,{\left (b x + a\right )}^{\frac{13}{2}} B - 819 \,{\left (3 \, B a - A b\right )}{\left (b x + a\right )}^{\frac{11}{2}} + 1001 \,{\left (3 \, B a^{2} - 2 \, A a b\right )}{\left (b x + a\right )}^{\frac{9}{2}} - 1287 \,{\left (B a^{3} - A a^{2} b\right )}{\left (b x + a\right )}^{\frac{7}{2}}\right )}}{9009 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="maxima")

[Out]

2/9009*(693*(b*x + a)^(13/2)*B - 819*(3*B*a - A*b)*(b*x + a)^(11/2) + 1001*(3*B*a^2 - 2*A*a*b)*(b*x + a)^(9/2)
- 1287*(B*a^3 - A*a^2*b)*(b*x + a)^(7/2))/b^4

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Fricas [A]  time = 2.37343, size = 332, normalized size = 3.49 \begin{align*} \frac{2 \,{\left (693 \, B b^{6} x^{6} - 48 \, B a^{6} + 104 \, A a^{5} b + 63 \,{\left (27 \, B a b^{5} + 13 \, A b^{6}\right )} x^{5} + 7 \,{\left (159 \, B a^{2} b^{4} + 299 \, A a b^{5}\right )} x^{4} +{\left (15 \, B a^{3} b^{3} + 1469 \, A a^{2} b^{4}\right )} x^{3} - 3 \,{\left (6 \, B a^{4} b^{2} - 13 \, A a^{3} b^{3}\right )} x^{2} + 4 \,{\left (6 \, B a^{5} b - 13 \, A a^{4} b^{2}\right )} x\right )} \sqrt{b x + a}}{9009 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="fricas")

[Out]

2/9009*(693*B*b^6*x^6 - 48*B*a^6 + 104*A*a^5*b + 63*(27*B*a*b^5 + 13*A*b^6)*x^5 + 7*(159*B*a^2*b^4 + 299*A*a*b
^5)*x^4 + (15*B*a^3*b^3 + 1469*A*a^2*b^4)*x^3 - 3*(6*B*a^4*b^2 - 13*A*a^3*b^3)*x^2 + 4*(6*B*a^5*b - 13*A*a^4*b
^2)*x)*sqrt(b*x + a)/b^4

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Sympy [A]  time = 4.48589, size = 292, normalized size = 3.07 \begin{align*} \begin{cases} \frac{16 A a^{5} \sqrt{a + b x}}{693 b^{3}} - \frac{8 A a^{4} x \sqrt{a + b x}}{693 b^{2}} + \frac{2 A a^{3} x^{2} \sqrt{a + b x}}{231 b} + \frac{226 A a^{2} x^{3} \sqrt{a + b x}}{693} + \frac{46 A a b x^{4} \sqrt{a + b x}}{99} + \frac{2 A b^{2} x^{5} \sqrt{a + b x}}{11} - \frac{32 B a^{6} \sqrt{a + b x}}{3003 b^{4}} + \frac{16 B a^{5} x \sqrt{a + b x}}{3003 b^{3}} - \frac{4 B a^{4} x^{2} \sqrt{a + b x}}{1001 b^{2}} + \frac{10 B a^{3} x^{3} \sqrt{a + b x}}{3003 b} + \frac{106 B a^{2} x^{4} \sqrt{a + b x}}{429} + \frac{54 B a b x^{5} \sqrt{a + b x}}{143} + \frac{2 B b^{2} x^{6} \sqrt{a + b x}}{13} & \text{for}\: b \neq 0 \\a^{\frac{5}{2}} \left (\frac{A x^{3}}{3} + \frac{B x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**(5/2)*(B*x+A),x)

[Out]

Piecewise((16*A*a**5*sqrt(a + b*x)/(693*b**3) - 8*A*a**4*x*sqrt(a + b*x)/(693*b**2) + 2*A*a**3*x**2*sqrt(a + b
*x)/(231*b) + 226*A*a**2*x**3*sqrt(a + b*x)/693 + 46*A*a*b*x**4*sqrt(a + b*x)/99 + 2*A*b**2*x**5*sqrt(a + b*x)
/11 - 32*B*a**6*sqrt(a + b*x)/(3003*b**4) + 16*B*a**5*x*sqrt(a + b*x)/(3003*b**3) - 4*B*a**4*x**2*sqrt(a + b*x
)/(1001*b**2) + 10*B*a**3*x**3*sqrt(a + b*x)/(3003*b) + 106*B*a**2*x**4*sqrt(a + b*x)/429 + 54*B*a*b*x**5*sqrt
(a + b*x)/143 + 2*B*b**2*x**6*sqrt(a + b*x)/13, Ne(b, 0)), (a**(5/2)*(A*x**3/3 + B*x**4/4), True))

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Giac [B]  time = 1.18191, size = 473, normalized size = 4.98 \begin{align*} \frac{2 \,{\left (\frac{429 \,{\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^{2}}{b^{2}} + \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 )} B a^{2}}{b^{3}} + \frac{286 \,{\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^{2}} + \frac{26 \,{\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^{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 )} A}{b^{2}} + \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^{3}}\right )}}{45045 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)*(B*x+A),x, algorithm="giac")

[Out]

2/45045*(429*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*A*a^2/b^2 + 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)*B*a^2/b^3 + 286*(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^2 + 26*(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^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)*A/b^2 + 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^3)/b