\(\int \frac {x}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [203]

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {a}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {654, 621} \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 654

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(212\) vs. \(2(63)=126\).

Time = 0.67 (sec) , antiderivative size = 212, normalized size of antiderivative = 3.37 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x^2 \left (3 \sqrt {a^2} b^6 x^6+3 a^3 b^3 x^3 \sqrt {(a+b x)^2}-3 a^2 b^4 x^4 \sqrt {(a+b x)^2}+3 a b^5 x^5 \sqrt {(a+b x)^2}+a^4 b^2 x^2 \left (\sqrt {a^2}-3 \sqrt {(a+b x)^2}\right )+6 a^6 \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )+2 a^5 b x \left (2 \sqrt {a^2}+\sqrt {(a+b x)^2}\right )\right )}{12 a^7 (a+b x)^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

-1/12*(x^2*(3*Sqrt[a^2]*b^6*x^6 + 3*a^3*b^3*x^3*Sqrt[(a + b*x)^2] - 3*a^2*b^4*x^4*Sqrt[(a + b*x)^2] + 3*a*b^5*
x^5*Sqrt[(a + b*x)^2] + a^4*b^2*x^2*(Sqrt[a^2] - 3*Sqrt[(a + b*x)^2]) + 6*a^6*(Sqrt[a^2] - Sqrt[(a + b*x)^2])
+ 2*a^5*b*x*(2*Sqrt[a^2] + Sqrt[(a + b*x)^2])))/(a^7*(a + b*x)^3*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.41

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (4 b x +a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(26\)
default	\(-\frac {\left (b x +a \right ) \left (4 b x +a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(26\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {x}{3 b}-\frac {a}{12 b^{2}}\right )}{\left (b x +a \right )^{5}}\)	\(31\)

[In]

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

[Out]

-1/12*(b*x+a)*(4*b*x+a)/b^2/((b*x+a)^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b x + a}{12 \, {\left (b^{6} x^{4} + 4 \, a b^{5} x^{3} + 6 \, a^{2} b^{4} x^{2} + 4 \, a^{3} b^{3} x + a^{4} b^{2}\right )}} \]

[In]

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

[Out]

-1/12*(4*b*x + a)/(b^6*x^4 + 4*a*b^5*x^3 + 6*a^2*b^4*x^2 + 4*a^3*b^3*x + a^4*b^2)

Sympy [F]

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

[In]

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

[Out]

Integral(x/((a + b*x)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.62 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

integrate(x/(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^2) + 1/4*a/(b^6*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.41 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b x + a}{12 \, {\left (b x + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

-1/12*(4*b*x + a)/((b*x + a)^4*b^2*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (a+4\,b\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{12\,b^2\,{\left (a+b\,x\right )}^5} \]

[In]

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

[Out]

-((a + 4*b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(12*b^2*(a + b*x)^5)