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

Optimal. Leaf size=133 \[ -\frac {16 b^2 \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^{3/2}}+\frac {8 b \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{105 c^3 \sqrt {x}}-\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c} \]

[Out]

-16/315*b^2*(-3*A*c+2*B*b)*(c*x^2+b*x)^(3/2)/c^4/x^(3/2)+2/9*B*x^(3/2)*(c*x^2+b*x)^(3/2)/c+8/105*b*(-3*A*c+2*B
*b)*(c*x^2+b*x)^(3/2)/c^3/x^(1/2)-2/21*(-3*A*c+2*B*b)*(c*x^2+b*x)^(3/2)*x^(1/2)/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {794, 656, 648} \[ -\frac {16 b^2 \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^{3/2}}+\frac {8 b \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{105 c^3 \sqrt {x}}-\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2*(2*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(315*c^4*x^(3/2)) + (8*b*(2*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(1
05*c^3*Sqrt[x]) - (2*(2*b*B - 3*A*c)*Sqrt[x]*(b*x + c*x^2)^(3/2))/(21*c^2) + (2*B*x^(3/2)*(b*x + c*x^2)^(3/2))
/(9*c)

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]

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 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])

Rubi steps

\begin {align*} \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx &=\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}+\frac {\left (2 \left (\frac {3}{2} (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right )\right ) \int x^{3/2} \sqrt {b x+c x^2} \, dx}{9 c}\\ &=-\frac {2 (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}+\frac {(4 b (2 b B-3 A c)) \int \sqrt {x} \sqrt {b x+c x^2} \, dx}{21 c^2}\\ &=\frac {8 b (2 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt {x}}-\frac {2 (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {\left (8 b^2 (2 b B-3 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{105 c^3}\\ &=-\frac {16 b^2 (2 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{315 c^4 x^{3/2}}+\frac {8 b (2 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt {x}}-\frac {2 (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 72, normalized size = 0.54 \[ \frac {2 (x (b+c x))^{3/2} \left (24 b^2 c (A+B x)-6 b c^2 x (6 A+5 B x)+5 c^3 x^2 (9 A+7 B x)-16 b^3 B\right )}{315 c^4 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(-16*b^3*B + 24*b^2*c*(A + B*x) - 6*b*c^2*x*(6*A + 5*B*x) + 5*c^3*x^2*(9*A + 7*B*x)))/(
315*c^4*x^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.83, size = 102, normalized size = 0.77 \[ \frac {2 \, {\left (35 \, B c^{4} x^{4} - 16 \, B b^{4} + 24 \, A b^{3} c + 5 \, {\left (B b c^{3} + 9 \, A c^{4}\right )} x^{3} - 3 \, {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{2} + 4 \, {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{4} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*c^4*x^4 - 16*B*b^4 + 24*A*b^3*c + 5*(B*b*c^3 + 9*A*c^4)*x^3 - 3*(2*B*b^2*c^2 - 3*A*b*c^3)*x^2 + 4*
(2*B*b^3*c - 3*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 110, normalized size = 0.83 \[ \frac {2}{315} \, B {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} - \frac {2}{105} \, A {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2/315*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) - 2/105*A*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b
^2)/c^3)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 83, normalized size = 0.62 \[ \frac {2 \left (c x +b \right ) \left (35 B \,c^{3} x^{3}+45 A \,c^{3} x^{2}-30 B b \,c^{2} x^{2}-36 A b \,c^{2} x +24 B \,b^{2} c x +24 A \,b^{2} c -16 b^{3} B \right ) \sqrt {c \,x^{2}+b x}}{315 c^{4} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x)

[Out]

2/315*(c*x+b)*(35*B*c^3*x^3+45*A*c^3*x^2-30*B*b*c^2*x^2-36*A*b*c^2*x+24*B*b^2*c*x+24*A*b^2*c-16*B*b^3)*(c*x^2+
b*x)^(1/2)/c^4/x^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 98, normalized size = 0.74 \[ \frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b} A}{105 \, c^{3}} + \frac {2 \, {\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x + b} B}{315 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x + b)*A/c^3 + 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)*B/c^4

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x + c*x^2)^(1/2)*(A + B*x),x)

[Out]

int(x^(3/2)*(b*x + c*x^2)^(1/2)*(A + B*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \sqrt {x \left (b + c x\right )} \left (A + B x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(B*x+A)*(c*x**2+b*x)**(1/2),x)

[Out]

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

________________________________________________________________________________________