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

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.85 \begin {gather*} \frac {2}{105} x^{3/2} (7 a (5 A+3 B x)+3 b x (7 A+5 B x)) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.02, size = 41, normalized size = 1.05 \begin {gather*} \frac {2}{105} \left (35 a A x^{3/2}+21 a B x^{5/2}+21 A b x^{5/2}+15 b B x^{7/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(35*a*A*x^(3/2) + 21*A*b*x^(5/2) + 21*a*B*x^(5/2) + 15*b*B*x^(7/2)))/105

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fricas [A]  time = 0.99, size = 30, normalized size = 0.77 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.23, size = 29, normalized size = 0.74 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.00, size = 28, normalized size = 0.72 \begin {gather*} \frac {2 \left (15 B b \,x^{2}+21 A b x +21 B a x +35 A a \right ) x^{\frac {3}{2}}}{105} \end {gather*}

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 = 0.89, size = 27, normalized size = 0.69 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.34, size = 27, normalized size = 0.69 \begin {gather*} \frac {2\,x^{3/2}\,\left (35\,A\,a+21\,A\,b\,x+21\,B\,a\,x+15\,B\,b\,x^2\right )}{105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(A + B*x)*(a + b*x),x)

[Out]

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

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sympy [A]  time = 2.88, size = 37, normalized size = 0.95 \begin {gather*} \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 {gather*}

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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