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

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

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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} \begin {gather*} \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 {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.78 \begin {gather*} \frac {2}{693} x^{7/2} \left (99 a^2+154 a b x+63 b^2 x^2\right ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.01, size = 34, normalized size = 0.94 \begin {gather*} \frac {2}{693} \left (99 a^2 x^{7/2}+154 a b x^{9/2}+63 b^2 x^{11/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(a + b*x)^2,x]

[Out]

(2*(99*a^2*x^(7/2) + 154*a*b*x^(9/2) + 63*b^2*x^(11/2)))/693

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fricas [A]  time = 0.71, size = 29, normalized size = 0.81 \begin {gather*} \frac {2}{693} \, {\left (63 \, b^{2} x^{5} + 154 \, a b x^{4} + 99 \, a^{2} x^{3}\right )} \sqrt {x} \end {gather*}

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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giac [A]  time = 1.16, size = 24, normalized size = 0.67 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.00, size = 25, normalized size = 0.69 \begin {gather*} \frac {2 \left (63 b^{2} x^{2}+154 a b x +99 a^{2}\right ) x^{\frac {7}{2}}}{693} \end {gather*}

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 = 1.27, size = 24, normalized size = 0.67 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.10, size = 24, normalized size = 0.67 \begin {gather*} \frac {2\,x^{7/2}\,\left (99\,a^2+154\,a\,b\,x+63\,b^2\,x^2\right )}{693} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x)^2,x)

[Out]

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

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sympy [A]  time = 2.60, size = 34, normalized size = 0.94 \begin {gather*} \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 {gather*}

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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