3.322 \(\int \sqrt{x} (a+b x) (A+B x) \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{5} x^{5/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{7} b B x^{7/2} \]

[Out]

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

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Rubi [A]  time = 0.0135763, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {76} \[ \frac{2}{5} x^{5/2} (a B+A b)+\frac{2}{3} a A x^{3/2}+\frac{2}{7} b B x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x)*(A + B*x),x]

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0109327, size = 33, normalized size = 0.85 \[ \frac{2}{105} x^{3/2} (7 a (5 A+3 B x)+3 b x (7 A+5 B x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(7*a*(5*A + 3*B*x) + 3*b*x*(7*A + 5*B*x)))/105

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Maple [A]  time = 0.003, size = 28, normalized size = 0.7 \begin{align*}{\frac{30\,bB{x}^{2}+42\,Abx+42\,Bax+70\,Aa}{105}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*x^(3/2)*(15*B*b*x^2+21*A*b*x+21*B*a*x+35*A*a)

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Maxima [A]  time = 1.09683, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{7} \, B b 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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.54123, size = 81, normalized size = 2.08 \begin{align*} \frac{2}{105} \,{\left (15 \, B b x^{3} + 35 \, A a x + 21 \,{\left (B a + A b\right )} x^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*b*x^3 + 35*A*a*x + 21*(B*a + A*b)*x^2)*sqrt(x)

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Sympy [A]  time = 2.57499, size = 37, normalized size = 0.95 \begin{align*} \frac{2 A a x^{\frac{3}{2}}}{3} + \frac{2 B b x^{\frac{7}{2}}}{7} + \frac{2 x^{\frac{5}{2}} \left (A b + B a\right )}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1671, size = 39, normalized size = 1. \begin{align*} \frac{2}{7} \, B b 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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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