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

Optimal. Leaf size=63 \[ \frac {2}{7} A b^2 x^{7/2}+\frac {2}{11} c x^{11/2} (A c+2 b B)+\frac {2}{9} b x^{9/2} (2 A c+b B)+\frac {2}{13} B c^2 x^{13/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {765} \begin {gather*} \frac {2}{7} A b^2 x^{7/2}+\frac {2}{11} c x^{11/2} (A c+2 b B)+\frac {2}{9} b x^{9/2} (2 A c+b B)+\frac {2}{13} B c^2 x^{13/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b^2*x^(7/2))/7 + (2*b*(b*B + 2*A*c)*x^(9/2))/9 + (2*c*(2*b*B + A*c)*x^(11/2))/11 + (2*B*c^2*x^(13/2))/13

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.87 \begin {gather*} \frac {2 x^{7/2} \left (13 A \left (99 b^2+154 b c x+63 c^2 x^2\right )+7 B x \left (143 b^2+234 b c x+99 c^2 x^2\right )\right )}{9009} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(13*A*(99*b^2 + 154*b*c*x + 63*c^2*x^2) + 7*B*x*(143*b^2 + 234*b*c*x + 99*c^2*x^2)))/9009

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (1287 A b^2 x^{7/2}+2002 A b c x^{9/2}+819 A c^2 x^{11/2}+1001 b^2 B x^{9/2}+1638 b B c x^{11/2}+693 B c^2 x^{13/2}\right )}{9009} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*(1287*A*b^2*x^(7/2) + 1001*b^2*B*x^(9/2) + 2002*A*b*c*x^(9/2) + 1638*b*B*c*x^(11/2) + 819*A*c^2*x^(11/2) +
693*B*c^2*x^(13/2)))/9009

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fricas [A]  time = 0.39, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{9009} \, {\left (693 \, B c^{2} x^{6} + 1287 \, A b^{2} x^{3} + 819 \, {\left (2 \, B b c + A c^{2}\right )} x^{5} + 1001 \, {\left (B b^{2} + 2 \, A b c\right )} x^{4}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9009*(693*B*c^2*x^6 + 1287*A*b^2*x^3 + 819*(2*B*b*c + A*c^2)*x^5 + 1001*(B*b^2 + 2*A*b*c)*x^4)*sqrt(x)

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giac [A]  time = 0.15, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{13} \, B c^{2} x^{\frac {13}{2}} + \frac {4}{11} \, B b c x^{\frac {11}{2}} + \frac {2}{11} \, A c^{2} x^{\frac {11}{2}} + \frac {2}{9} \, B b^{2} x^{\frac {9}{2}} + \frac {4}{9} \, A b c x^{\frac {9}{2}} + \frac {2}{7} \, A b^{2} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (693 B \,c^{2} x^{3}+819 A \,c^{2} x^{2}+1638 B b c \,x^{2}+2002 A b c x +1001 B \,b^{2} x +1287 A \,b^{2}\right ) x^{\frac {7}{2}}}{9009} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9009*x^(7/2)*(693*B*c^2*x^3+819*A*c^2*x^2+1638*B*b*c*x^2+2002*A*b*c*x+1001*B*b^2*x+1287*A*b^2)

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maxima [A]  time = 0.53, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{13} \, B c^{2} x^{\frac {13}{2}} + \frac {2}{7} \, A b^{2} x^{\frac {7}{2}} + \frac {2}{11} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {9}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*c^2*x^(13/2) + 2/7*A*b^2*x^(7/2) + 2/11*(2*B*b*c + A*c^2)*x^(11/2) + 2/9*(B*b^2 + 2*A*b*c)*x^(9/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{9/2}\,\left (\frac {2\,B\,b^2}{9}+\frac {4\,A\,c\,b}{9}\right )+x^{11/2}\,\left (\frac {2\,A\,c^2}{11}+\frac {4\,B\,b\,c}{11}\right )+\frac {2\,A\,b^2\,x^{7/2}}{7}+\frac {2\,B\,c^2\,x^{13/2}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.24, size = 66, normalized size = 1.05 \begin {gather*} \frac {2 A b^{2} x^{\frac {7}{2}}}{7} + \frac {2 B c^{2} x^{\frac {13}{2}}}{13} + \frac {2 x^{\frac {11}{2}} \left (A c^{2} + 2 B b c\right )}{11} + \frac {2 x^{\frac {9}{2}} \left (2 A b c + B b^{2}\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*b**2*x**(7/2)/7 + 2*B*c**2*x**(13/2)/13 + 2*x**(11/2)*(A*c**2 + 2*B*b*c)/11 + 2*x**(9/2)*(2*A*b*c + B*b**2
)/9

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