3.195 \(\int x^{5/2} (A+B x) \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=170 \[ \frac{32 b^3 \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{3465 c^5 x^{3/2}}-\frac{16 b^2 \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{1155 c^4 \sqrt{x}}+\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{231 c^3}-\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c} \]

[Out]

(32*b^3*(8*b*B - 11*A*c)*(b*x + c*x^2)^(3/2))/(3465*c^5*x^(3/2)) - (16*b^2*(8*b*B - 11*A*c)*(b*x + c*x^2)^(3/2
))/(1155*c^4*Sqrt[x]) + (4*b*(8*b*B - 11*A*c)*Sqrt[x]*(b*x + c*x^2)^(3/2))/(231*c^3) - (2*(8*b*B - 11*A*c)*x^(
3/2)*(b*x + c*x^2)^(3/2))/(99*c^2) + (2*B*x^(5/2)*(b*x + c*x^2)^(3/2))/(11*c)

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Rubi [A]  time = 0.158624, 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} \[ \frac{32 b^3 \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{3465 c^5 x^{3/2}}-\frac{16 b^2 \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{1155 c^4 \sqrt{x}}+\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{231 c^3}-\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2} (8 b B-11 A c)}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(32*b^3*(8*b*B - 11*A*c)*(b*x + c*x^2)^(3/2))/(3465*c^5*x^(3/2)) - (16*b^2*(8*b*B - 11*A*c)*(b*x + c*x^2)^(3/2
))/(1155*c^4*Sqrt[x]) + (4*b*(8*b*B - 11*A*c)*Sqrt[x]*(b*x + c*x^2)^(3/2))/(231*c^3) - (2*(8*b*B - 11*A*c)*x^(
3/2)*(b*x + c*x^2)^(3/2))/(99*c^2) + (2*B*x^(5/2)*(b*x + c*x^2)^(3/2))/(11*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^{5/2} (A+B x) \sqrt{b x+c x^2} \, dx &=\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac{\left (2 \left (\frac{5}{2} (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right )\right ) \int x^{5/2} \sqrt{b x+c x^2} \, dx}{11 c}\\ &=-\frac{2 (8 b B-11 A c) x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac{(2 b (8 b B-11 A c)) \int x^{3/2} \sqrt{b x+c x^2} \, dx}{33 c^2}\\ &=\frac{4 b (8 b B-11 A c) \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{2 (8 b B-11 A c) x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac{\left (8 b^2 (8 b B-11 A c)\right ) \int \sqrt{x} \sqrt{b x+c x^2} \, dx}{231 c^3}\\ &=-\frac{16 b^2 (8 b B-11 A c) \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{4 b (8 b B-11 A c) \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{2 (8 b B-11 A c) x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac{\left (16 b^3 (8 b B-11 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{\sqrt{x}} \, dx}{1155 c^4}\\ &=\frac{32 b^3 (8 b B-11 A c) \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac{16 b^2 (8 b B-11 A c) \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{4 b (8 b B-11 A c) \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{2 (8 b B-11 A c) x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}\\ \end{align*}

Mathematica [A]  time = 0.0748335, size = 94, normalized size = 0.55 \[ \frac{2 (x (b+c x))^{3/2} \left (24 b^2 c^2 x (11 A+10 B x)-16 b^3 c (11 A+12 B x)-10 b c^3 x^2 (33 A+28 B x)+35 c^4 x^3 (11 A+9 B x)+128 b^4 B\right )}{3465 c^5 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(128*b^4*B + 35*c^4*x^3*(11*A + 9*B*x) + 24*b^2*c^2*x*(11*A + 10*B*x) - 16*b^3*c*(11*A
+ 12*B*x) - 10*b*c^3*x^2*(33*A + 28*B*x)))/(3465*c^5*x^(3/2))

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Maple [A]  time = 0.005, size = 107, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -315\,B{x}^{4}{c}^{4}-385\,A{c}^{4}{x}^{3}+280\,Bb{c}^{3}{x}^{3}+330\,Ab{c}^{3}{x}^{2}-240\,B{b}^{2}{c}^{2}{x}^{2}-264\,A{b}^{2}{c}^{2}x+192\,B{b}^{3}cx+176\,A{b}^{3}c-128\,{b}^{4}B \right ) }{3465\,{c}^{5}}\sqrt{c{x}^{2}+bx}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x)

[Out]

-2/3465*(c*x+b)*(-315*B*c^4*x^4-385*A*c^4*x^3+280*B*b*c^3*x^3+330*A*b*c^3*x^2-240*B*b^2*c^2*x^2-264*A*b^2*c^2*
x+192*B*b^3*c*x+176*A*b^3*c-128*B*b^4)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)

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Maxima [A]  time = 1.16849, size = 162, normalized size = 0.95 \begin{align*} \frac{2 \,{\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt{c x + b} A}{315 \, c^{4}} + \frac{2 \,{\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 )} \sqrt{c x + b} B}{3465 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*sqrt(c*x + b)*A/c^4 + 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)*sqrt(c*x + b)*B/c^5

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Fricas [A]  time = 1.6402, size = 293, normalized size = 1.72 \begin{align*} \frac{2 \,{\left (315 \, B c^{5} x^{5} + 128 \, B b^{5} - 176 \, A b^{4} c + 35 \,{\left (B b c^{4} + 11 \, A c^{5}\right )} x^{4} - 5 \,{\left (8 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{3} + 6 \,{\left (8 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{2} - 8 \,{\left (8 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{3465 \, c^{5} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*B*c^5*x^5 + 128*B*b^5 - 176*A*b^4*c + 35*(B*b*c^4 + 11*A*c^5)*x^4 - 5*(8*B*b^2*c^3 - 11*A*b*c^4)*x
^3 + 6*(8*B*b^3*c^2 - 11*A*b^2*c^3)*x^2 - 8*(8*B*b^4*c - 11*A*b^3*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**(5/2)*(B*x+A)*(c*x**2+b*x)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.15083, size = 181, normalized size = 1.06 \begin{align*} -\frac{2}{3465} \, 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}{315} \, A{\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^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-2/3465*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) + 2/315*A*(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)