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

Optimal. Leaf size=55 \[ \frac {2}{5} x^{5/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{9} B c x^{9/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {765} \begin {gather*} \frac {2}{5} x^{5/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{9} B c x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*x^(3/2))/3 + (2*(A*b + a*B)*x^(5/2))/5 + (2*(b*B + A*c)*x^(7/2))/7 + (2*B*c*x^(9/2))/9

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 (a+b x+c x^2\right ) \, dx &=\int \left (a A \sqrt {x}+(A b+a B) x^{3/2}+(b B+A c) x^{5/2}+B c x^{7/2}\right ) \, dx\\ &=\frac {2}{3} a A x^{3/2}+\frac {2}{5} (A b+a B) x^{5/2}+\frac {2}{7} (b B+A c) x^{7/2}+\frac {2}{9} B c x^{9/2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.85 \begin {gather*} \frac {2}{315} x^{3/2} (21 a (5 A+3 B x)+x (9 A (7 b+5 c x)+5 B x (9 b+7 c x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(21*a*(5*A + 3*B*x) + x*(9*A*(7*b + 5*c*x) + 5*B*x*(9*b + 7*c*x))))/315

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IntegrateAlgebraic [A]  time = 0.03, size = 59, normalized size = 1.07 \begin {gather*} \frac {2}{315} \left (105 a A x^{3/2}+63 a B x^{5/2}+63 A b x^{5/2}+45 A c x^{7/2}+45 b B x^{7/2}+35 B c x^{9/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(105*a*A*x^(3/2) + 63*A*b*x^(5/2) + 63*a*B*x^(5/2) + 45*b*B*x^(7/2) + 45*A*c*x^(7/2) + 35*B*c*x^(9/2)))/315

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fricas [A]  time = 0.41, size = 42, normalized size = 0.76 \begin {gather*} \frac {2}{315} \, {\left (35 \, B c x^{4} + 45 \, {\left (B b + A c\right )} x^{3} + 105 \, A a x + 63 \, {\left (B a + A b\right )} x^{2}\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+a)*x^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*c*x^4 + 45*(B*b + A*c)*x^3 + 105*A*a*x + 63*(B*a + A*b)*x^2)*sqrt(x)

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giac [A]  time = 0.22, size = 43, normalized size = 0.78 \begin {gather*} \frac {2}{9} \, B c x^{\frac {9}{2}} + \frac {2}{7} \, B b x^{\frac {7}{2}} + \frac {2}{7} \, A c x^{\frac {7}{2}} + \frac {2}{5} \, B a x^{\frac {5}{2}} + \frac {2}{5} \, A b x^{\frac {5}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*c*x^(9/2) + 2/7*B*b*x^(7/2) + 2/7*A*c*x^(7/2) + 2/5*B*a*x^(5/2) + 2/5*A*b*x^(5/2) + 2/3*A*a*x^(3/2)

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maple [A]  time = 0.05, size = 42, normalized size = 0.76 \begin {gather*} \frac {2 \left (35 B c \,x^{3}+45 A c \,x^{2}+45 B b \,x^{2}+63 A b x +63 B a x +105 A a \right ) x^{\frac {3}{2}}}{315} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*x^(3/2)*(35*B*c*x^3+45*A*c*x^2+45*B*b*x^2+63*A*b*x+63*B*a*x+105*A*a)

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maxima [A]  time = 0.50, size = 39, normalized size = 0.71 \begin {gather*} \frac {2}{9} \, B c x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B b + A c\right )} x^{\frac {7}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} + \frac {2}{5} \, {\left (B a + A b\right )} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*c*x^(9/2) + 2/7*(B*b + A*c)*x^(7/2) + 2/3*A*a*x^(3/2) + 2/5*(B*a + A*b)*x^(5/2)

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mupad [B]  time = 0.04, size = 41, normalized size = 0.75 \begin {gather*} x^{5/2}\,\left (\frac {2\,A\,b}{5}+\frac {2\,B\,a}{5}\right )+x^{7/2}\,\left (\frac {2\,A\,c}{7}+\frac {2\,B\,b}{7}\right )+\frac {2\,A\,a\,x^{3/2}}{3}+\frac {2\,B\,c\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(5/2)*((2*A*b)/5 + (2*B*a)/5) + x^(7/2)*((2*A*c)/7 + (2*B*b)/7) + (2*A*a*x^(3/2))/3 + (2*B*c*x^(9/2))/9

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sympy [A]  time = 2.79, size = 53, normalized size = 0.96 \begin {gather*} \frac {2 A a x^{\frac {3}{2}}}{3} + \frac {2 B c x^{\frac {9}{2}}}{9} + \frac {2 x^{\frac {7}{2}} \left (A c + B b\right )}{7} + \frac {2 x^{\frac {5}{2}} \left (A b + B a\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*x**(3/2)/3 + 2*B*c*x**(9/2)/9 + 2*x**(7/2)*(A*c + B*b)/7 + 2*x**(5/2)*(A*b + B*a)/5

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