3.4.53 \(\int \frac {x^2}{(a+b x)^{5/2}} \, dx\) [353]

Optimal. Leaf size=49 \[ -\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3} \]

[Out]

-2/3*a^2/b^3/(b*x+a)^(3/2)+4*a/b^3/(b*x+a)^(1/2)+2*(b*x+a)^(1/2)/b^3

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Rubi [A]
time = 0.01, antiderivative size = 49, 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 = {45} \begin {gather*} -\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x)^(5/2),x]

[Out]

(-2*a^2)/(3*b^3*(a + b*x)^(3/2)) + (4*a)/(b^3*Sqrt[a + b*x]) + (2*Sqrt[a + b*x])/b^3

Rule 45

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 \frac {x^2}{(a+b x)^{5/2}} \, dx &=\int \left (\frac {a^2}{b^2 (a+b x)^{5/2}}-\frac {2 a}{b^2 (a+b x)^{3/2}}+\frac {1}{b^2 \sqrt {a+b x}}\right ) \, dx\\ &=-\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.71 \begin {gather*} \frac {2 \left (8 a^2+12 a b x+3 b^2 x^2\right )}{3 b^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x)^(5/2),x]

[Out]

(2*(8*a^2 + 12*a*b*x + 3*b^2*x^2))/(3*b^3*(a + b*x)^(3/2))

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Maple [A]
time = 0.10, size = 36, normalized size = 0.73

method	result	size
gosper	\(\frac {2 x^{2} b^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(32\)
trager	\(\frac {2 x^{2} b^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {2 \sqrt {b x +a}}{b^{3}}+\frac {2 a \left (6 b x +5 a \right )}{3 b^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(35\)
derivativedivides	\(\frac {2 \sqrt {b x +a}+\frac {4 a}{\sqrt {b x +a}}-\frac {2 a^{2}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{3}}\)	\(36\)
default	\(\frac {2 \sqrt {b x +a}+\frac {4 a}{\sqrt {b x +a}}-\frac {2 a^{2}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{3}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^3*((b*x+a)^(1/2)+2*a/(b*x+a)^(1/2)-1/3*a^2/(b*x+a)^(3/2))

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Maxima [A]
time = 0.28, size = 41, normalized size = 0.84 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b^{3}} + \frac {4 \, a}{\sqrt {b x + a} b^{3}} - \frac {2 \, a^{2}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} 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*sqrt(b*x + a)/b^3 + 4*a/(sqrt(b*x + a)*b^3) - 2/3*a^2/((b*x + a)^(3/2)*b^3)

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Fricas [A]
time = 0.42, size = 52, normalized size = 1.06 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \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]

2/3*(3*b^2*x^2 + 12*a*b*x + 8*a^2)*sqrt(b*x + a)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (46) = 92\).
time = 0.40, size = 121, normalized size = 2.47 \begin {gather*} \begin {cases} \frac {16 a^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {24 a b x}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {6 b^{2} x^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 a^{\frac {5}{2}}} & \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**2/(3*a*b**3*sqrt(a + b*x) + 3*b**4*x*sqrt(a + b*x)) + 24*a*b*x/(3*a*b**3*sqrt(a + b*x) + 3*b*
*4*x*sqrt(a + b*x)) + 6*b**2*x**2/(3*a*b**3*sqrt(a + b*x) + 3*b**4*x*sqrt(a + b*x)), Ne(b, 0)), (x**3/(3*a**(5
/2)), True))

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Giac [A]
time = 0.63, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b^{3}} + \frac {2 \, {\left (6 \, {\left (b x + a\right )} a - a^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} 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="giac")

[Out]

2*sqrt(b*x + a)/b^3 + 2/3*(6*(b*x + a)*a - a^2)/((b*x + a)^(3/2)*b^3)

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Mupad [B]
time = 0.08, size = 35, normalized size = 0.71 \begin {gather*} \frac {6\,{\left (a+b\,x\right )}^2+12\,a\,\left (a+b\,x\right )-2\,a^2}{3\,b^3\,{\left (a+b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x)^(5/2),x)

[Out]

(6*(a + b*x)^2 + 12*a*(a + b*x) - 2*a^2)/(3*b^3*(a + b*x)^(3/2))

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