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

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

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Rubi [A]  time = 0.01, antiderivative size = 53, 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 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} \end {gather*}

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*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.66 \begin {gather*} \frac {2 (a+b x)^{7/2} \left (8 a^2-28 a b x+63 b^2 x^2\right )}{693 b^3} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.02, size = 39, normalized size = 0.74 \begin {gather*} \frac {2 (a+b x)^{7/2} \left (99 a^2-154 a (a+b x)+63 (a+b x)^2\right )}{693 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(7/2)*(99*a^2 - 154*a*(a + b*x) + 63*(a + b*x)^2))/(693*b^3)

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fricas [A]  time = 1.17, size = 64, normalized size = 1.21 \begin {gather*} \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 {gather*}

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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giac [B]  time = 1.09, size = 233, normalized size = 4.40 \begin {gather*} \frac {2 \, {\left (\frac {231 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{3}}{b^{2}} + \frac {297 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2}}{b^{2}} + \frac {33 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{2}} + \frac {5 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{2}}\right )}}{3465 \, b} \end {gather*}

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*(231*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^3/b^2 + 297*(5*(b*x + a)^(7/2)
- 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a^2/b^2 + 33*(35*(b*x + a)^(9/2) - 18
0*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a/b^2 + 5*(63
*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a
)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)/b^2)/b

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maple [A]  time = 0.00, size = 32, normalized size = 0.60 \begin {gather*} \frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (63 b^{2} x^{2}-28 a b x +8 a^{2}\right )}{693 b^{3}} \end {gather*}

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 = 1.37, size = 41, normalized size = 0.77 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.04, size = 37, normalized size = 0.70 \begin {gather*} \frac {126\,{\left (a+b\,x\right )}^{11/2}-308\,a\,{\left (a+b\,x\right )}^{9/2}+198\,a^2\,{\left (a+b\,x\right )}^{7/2}}{693\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x)^(5/2),x)

[Out]

(126*(a + b*x)^(11/2) - 308*a*(a + b*x)^(9/2) + 198*a^2*(a + b*x)^(7/2))/(693*b^3)

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sympy [A]  time = 3.77, size = 124, normalized size = 2.34 \begin {gather*} \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 {gather*}

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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