\(\int \frac {a+b x}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [2038]

Optimal result

Integrand size = 26, antiderivative size = 27 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {643} \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 b \left ((a+b x)^2\right )^{3/2}} \]

[In]

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

[Out]

-1/3*1/(b*((a + b*x)^2)^(3/2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b x +a \right )}{3 b \left (b x +a \right )^{3}}\)	\(19\)
gosper	\(-\frac {\left (b x +a \right )^{2}}{3 b \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(22\)
default	\(-\frac {\left (b x +a \right )^{2}}{3 b \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(22\)
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{3 \left (b x +a \right )^{4} b}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (26) = 52\).

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.59 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {1}{3 a^{2} b \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} + 6 a b^{2} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} + 3 b^{3} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}} & \text {for}\: b \neq 0 \\\frac {a x}{\left (a^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-1/(3*a**2*b*sqrt(a**2 + 2*a*b*x + b**2*x**2) + 6*a*b**2*x*sqrt(a**2 + 2*a*b*x + b**2*x**2) + 3*b**
3*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2)), Ne(b, 0)), (a*x/(a**2)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, {\left (b x + a\right )}^{3} b \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

-1/3/((b*x + a)^3*b*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 10.79 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,b\,{\left (a+b\,x\right )}^4} \]

[In]

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

[Out]

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