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

Optimal. Leaf size=170 \[ \frac{32 b^3 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^{5/2}}-\frac{16 b^2 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^{3/2}}+\frac{4 b \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{429 c^3 \sqrt{x}}-\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c} \]

[Out]

(32*b^3*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(15015*c^5*x^(5/2)) - (16*b^2*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/
2))/(3003*c^4*x^(3/2)) + (4*b*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(429*c^3*Sqrt[x]) - (2*(8*b*B - 13*A*c)*Sq
rt[x]*(b*x + c*x^2)^(5/2))/(143*c^2) + (2*B*x^(3/2)*(b*x + c*x^2)^(5/2))/(13*c)

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Rubi [A]  time = 0.16146, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {794, 656, 648} \[ \frac{32 b^3 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^{5/2}}-\frac{16 b^2 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^{3/2}}+\frac{4 b \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{429 c^3 \sqrt{x}}-\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(15015*c^5*x^(5/2)) - (16*b^2*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/
2))/(3003*c^4*x^(3/2)) + (4*b*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(429*c^3*Sqrt[x]) - (2*(8*b*B - 13*A*c)*Sq
rt[x]*(b*x + c*x^2)^(5/2))/(143*c^2) + (2*B*x^(3/2)*(b*x + c*x^2)^(5/2))/(13*c)

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

\begin{align*} \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac{\left (2 \left (\frac{3}{2} (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right )\right ) \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx}{13 c}\\ &=-\frac{2 (8 b B-13 A c) \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac{(6 b (8 b B-13 A c)) \int \sqrt{x} \left (b x+c x^2\right )^{3/2} \, dx}{143 c^2}\\ &=\frac{4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{2 (8 b B-13 A c) \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac{\left (8 b^2 (8 b B-13 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{\sqrt{x}} \, dx}{429 c^3}\\ &=-\frac{16 b^2 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{2 (8 b B-13 A c) \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac{\left (16 b^3 (8 b B-13 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{3003 c^4}\\ &=\frac{32 b^3 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac{16 b^2 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{2 (8 b B-13 A c) \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}\\ \end{align*}

Mathematica [A]  time = 0.0787966, size = 94, normalized size = 0.55 \[ \frac{2 (x (b+c x))^{5/2} \left (40 b^2 c^2 x (13 A+14 B x)-16 b^3 c (13 A+20 B x)-70 b c^3 x^2 (13 A+12 B x)+105 c^4 x^3 (13 A+11 B x)+128 b^4 B\right )}{15015 c^5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 107, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -1155\,B{x}^{4}{c}^{4}-1365\,A{c}^{4}{x}^{3}+840\,Bb{c}^{3}{x}^{3}+910\,Ab{c}^{3}{x}^{2}-560\,B{b}^{2}{c}^{2}{x}^{2}-520\,A{b}^{2}{c}^{2}x+320\,B{b}^{3}cx+208\,A{b}^{3}c-128\,{b}^{4}B \right ) }{15015\,{c}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22939, size = 370, normalized size = 2.18 \begin{align*} \frac{2 \,{\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \,{\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt{c x + b} A}{3465 \, c^{4} x^{4}} + \frac{2 \,{\left (5 \,{\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 13 \,{\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4}\right )} \sqrt{c x + b} B}{45045 \, c^{5} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 11*(35*b*c
^4*x^5 + 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)*A/(c^4*x^4) + 2/45045*(5*(
693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^4 + 80*b^3*c^3*x^3 - 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*b^6)*x^5 + 1
3*(315*b*c^5*x^6 + 35*b^2*c^4*x^5 - 40*b^3*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^4)*sqrt(c*x
+ b)*B/(c^5*x^5)

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Fricas [A]  time = 1.49967, size = 350, normalized size = 2.06 \begin{align*} \frac{2 \,{\left (1155 \, B c^{6} x^{6} + 128 \, B b^{6} - 208 \, A b^{5} c + 105 \,{\left (14 \, B b c^{5} + 13 \, A c^{6}\right )} x^{5} + 35 \,{\left (B b^{2} c^{4} + 52 \, A b c^{5}\right )} x^{4} - 5 \,{\left (8 \, B b^{3} c^{3} - 13 \, A b^{2} c^{4}\right )} x^{3} + 6 \,{\left (8 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{2} - 8 \,{\left (8 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{15015 \, c^{5} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.23521, size = 398, normalized size = 2.34 \begin{align*} \frac{2}{9009} \, B c{\left (\frac{256 \, b^{\frac{13}{2}}}{c^{6}} + \frac{693 \,{\left (c x + b\right )}^{\frac{13}{2}} - 4095 \,{\left (c x + b\right )}^{\frac{11}{2}} b + 10010 \,{\left (c x + b\right )}^{\frac{9}{2}} b^{2} - 12870 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{3} + 9009 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{4} - 3003 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{5}}{c^{6}}\right )} - \frac{2}{3465} \, B b{\left (\frac{128 \, b^{\frac{11}{2}}}{c^{5}} - \frac{315 \,{\left (c x + b\right )}^{\frac{11}{2}} - 1540 \,{\left (c x + b\right )}^{\frac{9}{2}} b + 2970 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{2} - 2772 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{3} + 1155 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{4}}{c^{5}}\right )} - \frac{2}{3465} \, A c{\left (\frac{128 \, b^{\frac{11}{2}}}{c^{5}} - \frac{315 \,{\left (c x + b\right )}^{\frac{11}{2}} - 1540 \,{\left (c x + b\right )}^{\frac{9}{2}} b + 2970 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{2} - 2772 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{3} + 1155 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{4}}{c^{5}}\right )} + \frac{2}{315} \, A b{\left (\frac{16 \, b^{\frac{9}{2}}}{c^{4}} + \frac{35 \,{\left (c x + b\right )}^{\frac{9}{2}} - 135 \,{\left (c x + b\right )}^{\frac{7}{2}} b + 189 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{3}}{c^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9009*B*c*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^2 - 1
2870*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) - 2/3465*B*b*(128*b^(11/2
)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 +
1155*(c*x + b)^(3/2)*b^4)/c^5) - 2/3465*A*c*(128*b^(11/2)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*
b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5) + 2/315*A*b*(16*b^(9/
2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4)