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

Optimal. Leaf size=53 \[ \frac{2 a^2 (a+b x)^{7/2}}{7 b^3}+\frac{2 (a+b x)^{11/2}}{11 b^3}-\frac{4 a (a+b x)^{9/2}}{9 b^3} \]

[Out]

(2*a^2*(a + b*x)^(7/2))/(7*b^3) - (4*a*(a + b*x)^(9/2))/(9*b^3) + (2*(a + b*x)^(11/2))/(11*b^3)

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Rubi [A]  time = 0.0124758, antiderivative size = 53, 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} \[ \frac{2 a^2 (a+b x)^{7/2}}{7 b^3}+\frac{2 (a+b x)^{11/2}}{11 b^3}-\frac{4 a (a+b x)^{9/2}}{9 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^(5/2),x]

[Out]

(2*a^2*(a + b*x)^(7/2))/(7*b^3) - (4*a*(a + b*x)^(9/2))/(9*b^3) + (2*(a + b*x)^(11/2))/(11*b^3)

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^2 (a+b x)^{5/2} \, dx &=\int \left (\frac{a^2 (a+b x)^{5/2}}{b^2}-\frac{2 a (a+b x)^{7/2}}{b^2}+\frac{(a+b x)^{9/2}}{b^2}\right ) \, dx\\ &=\frac{2 a^2 (a+b x)^{7/2}}{7 b^3}-\frac{4 a (a+b x)^{9/2}}{9 b^3}+\frac{2 (a+b x)^{11/2}}{11 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0264435, size = 35, normalized size = 0.66 \[ \frac{2 (a+b x)^{7/2} \left (8 a^2-28 a b x+63 b^2 x^2\right )}{693 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x)^(5/2),x]

[Out]

(2*(a + b*x)^(7/2)*(8*a^2 - 28*a*b*x + 63*b^2*x^2))/(693*b^3)

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Maple [A]  time = 0.004, size = 32, normalized size = 0.6 \begin{align*}{\frac{126\,{b}^{2}{x}^{2}-56\,abx+16\,{a}^{2}}{693\,{b}^{3}} \left ( bx+a \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*(b*x+a)^(7/2)*(63*b^2*x^2-28*a*b*x+8*a^2)/b^3

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Maxima [A]  time = 1.02675, size = 55, normalized size = 1.04 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{\frac{11}{2}}}{11 \, b^{3}} - \frac{4 \,{\left (b x + a\right )}^{\frac{9}{2}} a}{9 \, b^{3}} + \frac{2 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2}}{7 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*(b*x + a)^(11/2)/b^3 - 4/9*(b*x + a)^(9/2)*a/b^3 + 2/7*(b*x + a)^(7/2)*a^2/b^3

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Fricas [A]  time = 1.50952, size = 146, normalized size = 2.75 \begin{align*} \frac{2 \,{\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt{b x + a}}{693 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*(63*b^5*x^5 + 161*a*b^4*x^4 + 113*a^2*b^3*x^3 + 3*a^3*b^2*x^2 - 4*a^4*b*x + 8*a^5)*sqrt(b*x + a)/b^3

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Sympy [A]  time = 4.74117, size = 124, normalized size = 2.34 \begin{align*} \begin{cases} \frac{16 a^{5} \sqrt{a + b x}}{693 b^{3}} - \frac{8 a^{4} x \sqrt{a + b x}}{693 b^{2}} + \frac{2 a^{3} x^{2} \sqrt{a + b x}}{231 b} + \frac{226 a^{2} x^{3} \sqrt{a + b x}}{693} + \frac{46 a b x^{4} \sqrt{a + b x}}{99} + \frac{2 b^{2} x^{5} \sqrt{a + b x}}{11} & \text{for}\: b \neq 0 \\\frac{a^{\frac{5}{2}} x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*a**5*sqrt(a + b*x)/(693*b**3) - 8*a**4*x*sqrt(a + b*x)/(693*b**2) + 2*a**3*x**2*sqrt(a + b*x)/(2
31*b) + 226*a**2*x**3*sqrt(a + b*x)/693 + 46*a*b*x**4*sqrt(a + b*x)/99 + 2*b**2*x**5*sqrt(a + b*x)/11, Ne(b, 0
)), (a**(5/2)*x**3/3, True))

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Giac [B]  time = 1.18155, size = 211, normalized size = 3.98 \begin{align*} \frac{2 \,{\left (\frac{33 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} a^{2}}{b^{2}} + \frac{22 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} a}{b^{2}} + \frac{315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}}{b^{2}}\right )}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(33*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*a^2/b^2 + 22*(35*(b*x + a)^(9/
2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)*a/b^2 + (315*(b*x + a)^(11/2)
- 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(7/2)*a^2 - 2772*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4)/b^2
)/b