\(\int \frac {x}{(a+b x)^{5/2}} \, dx\) [354]

Optimal result

Integrand size = 11, antiderivative size = 32 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=\frac {2 a}{3 b^2 (a+b x)^{3/2}}-\frac {2}{b^2 \sqrt {a+b x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=\frac {2 a}{3 b^2 (a+b x)^{3/2}}-\frac {2}{b^2 \sqrt {a+b x}} \]

[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 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=-\frac {2 (2 a+3 b x)}{3 b^2 (a+b x)^{3/2}} \]

[In]

Integrate[x/(a + b*x)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66

method	result	size
gosper	\(-\frac {2 \left (3 b x +2 a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(21\)
trager	\(-\frac {2 \left (3 b x +2 a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(21\)
pseudoelliptic	\(\frac {-6 b x -4 a}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(21\)
derivativedivides	\(\frac {-\frac {2}{\sqrt {b x +a}}+\frac {2 a}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{2}}\)	\(26\)
default	\(\frac {-\frac {2}{\sqrt {b x +a}}+\frac {2 a}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{2}}\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b x + 2 \, a\right )} \sqrt {b x + a}}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (29) = 58\).

Time = 0.42 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.50 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=\begin {cases} - \frac {4 a}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} - \frac {6 b x}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=-\frac {2}{\sqrt {b x + a} b^{2}} + \frac {2 \, a}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b x + 2 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {x}{(a+b x)^{5/2}} \, dx=-\frac {4\,a+6\,b\,x}{3\,b^2\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int(x/(a + b*x)^(5/2),x)

[Out]

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