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

Optimal. Leaf size=32 \[ \frac {2 a}{3 b^2 (a+b x)^{3/2}}-\frac {2}{b^2 \sqrt {a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {43} \[ \frac {2 a}{3 b^2 (a+b x)^{3/2}}-\frac {2}{b^2 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.75 \[ -\frac {2 (2 a+3 b x)}{3 b^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 41, normalized size = 1.28 \[ -\frac {2 \, {\left (3 \, b x + 2 \, a\right )} \sqrt {b x + a}}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.06, size = 20, normalized size = 0.62 \[ -\frac {2 \, {\left (3 \, b x + 2 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 21, normalized size = 0.66 \[ -\frac {2 \left (3 b x +2 a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 26, normalized size = 0.81 \[ -\frac {2}{\sqrt {b x + a} b^{2}} + \frac {2 \, a}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 20, normalized size = 0.62 \[ -\frac {4\,a+6\,b\,x}{3\,b^2\,{\left (a+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x)^(5/2),x)

[Out]

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

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sympy [A]  time = 1.13, size = 80, normalized size = 2.50 \[ \begin {cases} - \frac {4 a}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} - \frac {6 b x}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*a/(3*a*b**2*sqrt(a + b*x) + 3*b**3*x*sqrt(a + b*x)) - 6*b*x/(3*a*b**2*sqrt(a + b*x) + 3*b**3*x*s
qrt(a + b*x)), Ne(b, 0)), (x**2/(2*a**(5/2)), True))

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