3.436 \(\int x^{5/2} (a+b x)^2 \, dx\)

Optimal. Leaf size=36 \[ \frac{2}{7} a^2 x^{7/2}+\frac{4}{9} a b x^{9/2}+\frac{2}{11} b^2 x^{11/2} \]

[Out]

(2*a^2*x^(7/2))/7 + (4*a*b*x^(9/2))/9 + (2*b^2*x^(11/2))/11

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Rubi [A]  time = 0.0071863, antiderivative size = 36, normalized size of antiderivative = 1., 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} \[ \frac{2}{7} a^2 x^{7/2}+\frac{4}{9} a b x^{9/2}+\frac{2}{11} b^2 x^{11/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x)^2,x]

[Out]

(2*a^2*x^(7/2))/7 + (4*a*b*x^(9/2))/9 + (2*b^2*x^(11/2))/11

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 x^{5/2} (a+b x)^2 \, dx &=\int \left (a^2 x^{5/2}+2 a b x^{7/2}+b^2 x^{9/2}\right ) \, dx\\ &=\frac{2}{7} a^2 x^{7/2}+\frac{4}{9} a b x^{9/2}+\frac{2}{11} b^2 x^{11/2}\\ \end{align*}

Mathematica [A]  time = 0.0077778, size = 28, normalized size = 0.78 \[ \frac{2}{693} x^{7/2} \left (99 a^2+154 a b x+63 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x)^2,x]

[Out]

(2*x^(7/2)*(99*a^2 + 154*a*b*x + 63*b^2*x^2))/693

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*}{\frac{126\,{b}^{2}{x}^{2}+308\,abx+198\,{a}^{2}}{693}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x+a)^2,x)

[Out]

2/693*x^(7/2)*(63*b^2*x^2+154*a*b*x+99*a^2)

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Maxima [A]  time = 0.991102, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{11} \, b^{2} x^{\frac{11}{2}} + \frac{4}{9} \, a b x^{\frac{9}{2}} + \frac{2}{7} \, a^{2} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x+a)^2,x, algorithm="maxima")

[Out]

2/11*b^2*x^(11/2) + 4/9*a*b*x^(9/2) + 2/7*a^2*x^(7/2)

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Fricas [A]  time = 1.44803, size = 74, normalized size = 2.06 \begin{align*} \frac{2}{693} \,{\left (63 \, b^{2} x^{5} + 154 \, a b x^{4} + 99 \, a^{2} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x+a)^2,x, algorithm="fricas")

[Out]

2/693*(63*b^2*x^5 + 154*a*b*x^4 + 99*a^2*x^3)*sqrt(x)

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Sympy [A]  time = 3.74924, size = 34, normalized size = 0.94 \begin{align*} \frac{2 a^{2} x^{\frac{7}{2}}}{7} + \frac{4 a b x^{\frac{9}{2}}}{9} + \frac{2 b^{2} x^{\frac{11}{2}}}{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(b*x+a)**2,x)

[Out]

2*a**2*x**(7/2)/7 + 4*a*b*x**(9/2)/9 + 2*b**2*x**(11/2)/11

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Giac [A]  time = 1.2254, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{11} \, b^{2} x^{\frac{11}{2}} + \frac{4}{9} \, a b x^{\frac{9}{2}} + \frac{2}{7} \, a^{2} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x+a)^2,x, algorithm="giac")

[Out]

2/11*b^2*x^(11/2) + 4/9*a*b*x^(9/2) + 2/7*a^2*x^(7/2)