3.438 \(\int \sqrt {x} (a+b x)^2 \, dx\)

Optimal. Leaf size=36 \[ \frac {2}{3} a^2 x^{3/2}+\frac {4}{5} a b x^{5/2}+\frac {2}{7} b^2 x^{7/2} \]

[Out]

2/3*a^2*x^(3/2)+4/5*a*b*x^(5/2)+2/7*b^2*x^(7/2)

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \[ \frac {2}{3} a^2 x^{3/2}+\frac {4}{5} a b x^{5/2}+\frac {2}{7} b^2 x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x)^2,x]

[Out]

(2*a^2*x^(3/2))/3 + (4*a*b*x^(5/2))/5 + (2*b^2*x^(7/2))/7

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \sqrt {x} (a+b x)^2 \, dx &=\int \left (a^2 \sqrt {x}+2 a b x^{3/2}+b^2 x^{5/2}\right ) \, dx\\ &=\frac {2}{3} a^2 x^{3/2}+\frac {4}{5} a b x^{5/2}+\frac {2}{7} b^2 x^{7/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.78 \[ \frac {2}{105} x^{3/2} \left (35 a^2+42 a b x+15 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x)^2,x]

[Out]

(2*x^(3/2)*(35*a^2 + 42*a*b*x + 15*b^2*x^2))/105

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fricas [A]  time = 0.43, size = 27, normalized size = 0.75 \[ \frac {2}{105} \, {\left (15 \, b^{2} x^{3} + 42 \, a b x^{2} + 35 \, a^{2} x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^2*x^3 + 42*a*b*x^2 + 35*a^2*x)*sqrt(x)

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giac [A]  time = 0.90, size = 24, normalized size = 0.67 \[ \frac {2}{7} \, b^{2} x^{\frac {7}{2}} + \frac {4}{5} \, a b x^{\frac {5}{2}} + \frac {2}{3} \, a^{2} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^2*x^(7/2) + 4/5*a*b*x^(5/2) + 2/3*a^2*x^(3/2)

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maple [A]  time = 0.00, size = 25, normalized size = 0.69 \[ \frac {2 \left (15 b^{2} x^{2}+42 a b x +35 a^{2}\right ) x^{\frac {3}{2}}}{105} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*x^(3/2)*(15*b^2*x^2+42*a*b*x+35*a^2)

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maxima [A]  time = 1.29, size = 24, normalized size = 0.67 \[ \frac {2}{7} \, b^{2} x^{\frac {7}{2}} + \frac {4}{5} \, a b x^{\frac {5}{2}} + \frac {2}{3} \, a^{2} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^2*x^(7/2) + 4/5*a*b*x^(5/2) + 2/3*a^2*x^(3/2)

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mupad [B]  time = 0.04, size = 24, normalized size = 0.67 \[ \frac {2\,x^{3/2}\,\left (35\,a^2+42\,a\,b\,x+15\,b^2\,x^2\right )}{105} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a + b*x)^2,x)

[Out]

(2*x^(3/2)*(35*a^2 + 15*b^2*x^2 + 42*a*b*x))/105

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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