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

Optimal. Leaf size=133 \[ -\frac {16 b^2 \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^{5/2}}+\frac {8 b \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^{3/2}}-\frac {2 \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{99 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

[Out]

-16/3465*b^2*(-11*A*c+6*B*b)*(c*x^2+b*x)^(5/2)/c^4/x^(5/2)+8/693*b*(-11*A*c+6*B*b)*(c*x^2+b*x)^(5/2)/c^3/x^(3/
2)-2/99*(-11*A*c+6*B*b)*(c*x^2+b*x)^(5/2)/c^2/x^(1/2)+2/11*B*(c*x^2+b*x)^(5/2)*x^(1/2)/c

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Rubi [A]  time = 0.11, 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} \[ -\frac {16 b^2 \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^{5/2}}-\frac {2 \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{99 c^2 \sqrt {x}}+\frac {8 b \left (b x+c x^2\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^{3/2}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2*(6*b*B - 11*A*c)*(b*x + c*x^2)^(5/2))/(3465*c^4*x^(5/2)) + (8*b*(6*b*B - 11*A*c)*(b*x + c*x^2)^(5/2))
/(693*c^3*x^(3/2)) - (2*(6*b*B - 11*A*c)*(b*x + c*x^2)^(5/2))/(99*c^2*Sqrt[x]) + (2*B*Sqrt[x]*(b*x + c*x^2)^(5
/2))/(11*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 \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}+\frac {\left (2 \left (\frac {1}{2} (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right )\right ) \int \sqrt {x} \left (b x+c x^2\right )^{3/2} \, dx}{11 c}\\ &=-\frac {2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{99 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}+\frac {(4 b (6 b B-11 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx}{99 c^2}\\ &=\frac {8 b (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{693 c^3 x^{3/2}}-\frac {2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{99 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {\left (8 b^2 (6 b B-11 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{693 c^3}\\ &=-\frac {16 b^2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{3465 c^4 x^{5/2}}+\frac {8 b (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{693 c^3 x^{3/2}}-\frac {2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{99 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 75, normalized size = 0.56 \[ \frac {2 (x (b+c x))^{5/2} \left (8 b^2 c (11 A+15 B x)-10 b c^2 x (22 A+21 B x)+35 c^3 x^2 (11 A+9 B x)-48 b^3 B\right )}{3465 c^4 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(-48*b^3*B + 35*c^3*x^2*(11*A + 9*B*x) + 8*b^2*c*(11*A + 15*B*x) - 10*b*c^2*x*(22*A + 2
1*B*x)))/(3465*c^4*x^(5/2))

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fricas [A]  time = 1.13, size = 127, normalized size = 0.95 \[ \frac {2 \, {\left (315 \, B c^{5} x^{5} - 48 \, B b^{5} + 88 \, A b^{4} c + 35 \, {\left (12 \, B b c^{4} + 11 \, A c^{5}\right )} x^{4} + 5 \, {\left (3 \, B b^{2} c^{3} + 110 \, A b c^{4}\right )} x^{3} - 3 \, {\left (6 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3465 \, c^{4} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*B*c^5*x^5 - 48*B*b^5 + 88*A*b^4*c + 35*(12*B*b*c^4 + 11*A*c^5)*x^4 + 5*(3*B*b^2*c^3 + 110*A*b*c^4)
*x^3 - 3*(6*B*b^3*c^2 - 11*A*b^2*c^3)*x^2 + 4*(6*B*b^4*c - 11*A*b^3*c^2)*x)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))

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giac [B]  time = 0.22, size = 247, normalized size = 1.86 \[ -\frac {2}{3465} \, B c {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} + \frac {2}{315} \, B b {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} + \frac {2}{315} \, A c {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} - \frac {2}{105} \, A b {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*B*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 - 27
72*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5) + 2/315*B*b*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 13
5*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4) + 2/315*A*c*(16*b^(9/2)/c^4 + (3
5*(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*
b*(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)

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maple [A]  time = 0.05, size = 83, normalized size = 0.62 \[ \frac {2 \left (c x +b \right ) \left (315 B \,c^{3} x^{3}+385 A \,c^{3} x^{2}-210 B b \,c^{2} x^{2}-220 A b \,c^{2} x +120 B \,b^{2} c x +88 A \,b^{2} c -48 b^{3} B \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3465 c^{4} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(c*x+b)*(315*B*c^3*x^3+385*A*c^3*x^2-210*B*b*c^2*x^2-220*A*b*c^2*x+120*B*b^2*c*x+88*A*b^2*c-48*B*b^3)*(
c*x^2+b*x)^(3/2)/c^4/x^(3/2)

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maxima [B]  time = 0.66, size = 229, normalized size = 1.72 \[ \frac {2 \, {\left ({\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} x^{3} + 3 \, {\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2}\right )} \sqrt {c x + b} A}{315 \, c^{3} x^{3}} + \frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt {c x + b} B}{3465 \, c^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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