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

Optimal. Leaf size=39 \[ \frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{5} A b x^{5/2}+\frac {2}{9} B c x^{9/2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {765} \[ \frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{5} A b x^{5/2}+\frac {2}{9} B c x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*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 (b x+c x^2\right ) \, dx &=\int \left (A b x^{3/2}+(b B+A c) x^{5/2}+B c x^{7/2}\right ) \, dx\\ &=\frac {2}{5} 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 = 0.01, size = 33, normalized size = 0.85 \[ \frac {2}{315} x^{5/2} (9 A (7 b+5 c x)+5 B x (9 b+7 c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.99, size = 32, normalized size = 0.82 \[ \frac {2}{315} \, {\left (35 \, B c x^{4} + 63 \, A b x^{2} + 45 \, {\left (B b + A c\right )} x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 29, normalized size = 0.74 \[ \frac {2}{9} \, B c x^{\frac {9}{2}} + \frac {2}{7} \, B b x^{\frac {7}{2}} + \frac {2}{7} \, A c x^{\frac {7}{2}} + \frac {2}{5} \, A b x^{\frac {5}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)*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*A*b*x^(5/2)

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maple [A]  time = 0.04, size = 28, normalized size = 0.72 \[ \frac {2 \left (35 B c \,x^{2}+45 A c x +45 B b x +63 A b \right ) x^{\frac {5}{2}}}{315} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*x^(5/2)*(35*B*c*x^2+45*A*c*x+45*B*b*x+63*A*b)

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maxima [A]  time = 0.58, size = 27, normalized size = 0.69 \[ \frac {2}{9} \, B c x^{\frac {9}{2}} + \frac {2}{5} \, A b x^{\frac {5}{2}} + \frac {2}{7} \, {\left (B b + A c\right )} x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 27, normalized size = 0.69 \[ \frac {2\,x^{5/2}\,\left (63\,A\,b+45\,A\,c\,x+45\,B\,b\,x+35\,B\,c\,x^2\right )}{315} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^(5/2)*(63*A*b + 45*A*c*x + 45*B*b*x + 35*B*c*x^2))/315

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sympy [A]  time = 2.33, size = 37, normalized size = 0.95 \[ \frac {2 A b x^{\frac {5}{2}}}{5} + \frac {2 B c x^{\frac {9}{2}}}{9} + \frac {2 x^{\frac {7}{2}} \left (A c + B b\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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