\(\int \frac {x}{(a x+b x^2)^{5/2}} \, dx\) [65]

Optimal result

Integrand size = 15, antiderivative size = 48 \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac {8 (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}} \]

[Out]

2/3*x/a/(b*x^2+a*x)^(3/2)-8/3*(2*b*x+a)/a^3/(b*x^2+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {652, 627} \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac {8 (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}} \]

[In]

Int[x/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x)/(3*a*(a*x + b*x^2)^(3/2)) - (8*(a + 2*b*x))/(3*a^3*Sqrt[a*x + b*x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}+\frac {4 \int \frac {1}{\left (a x+b x^2\right )^{3/2}} \, dx}{3 a} \\ & = \frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac {8 (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 x \left (3 a^2+12 a b x+8 b^2 x^2\right )}{3 a^3 (x (a+b x))^{3/2}} \]

[In]

Integrate[x/(a*x + b*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\frac {2 \left (\frac {8}{3} b^{2} x^{2}+4 a b x +a^{2}\right )}{\sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{3}}\)	\(39\)
gosper	\(-\frac {2 x^{2} \left (b x +a \right ) \left (8 b^{2} x^{2}+12 a b x +3 a^{2}\right )}{3 a^{3} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\)	\(44\)
trager	\(-\frac {2 \left (8 b^{2} x^{2}+12 a b x +3 a^{2}\right ) \sqrt {b \,x^{2}+a x}}{3 a^{3} x \left (b x +a \right )^{2}}\)	\(46\)
risch	\(-\frac {2 \left (b x +a \right )}{a^{3} \sqrt {x \left (b x +a \right )}}-\frac {2 b \left (5 b x +6 a \right ) x}{3 \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{3}}\)	\(52\)
default	\(-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\)	\(70\)

[In]

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

[Out]

-2*(8/3*b^2*x^2+4*a*b*x+a^2)/(x*(b*x+a))^(1/2)/(b*x+a)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x/(x*(a + b*x))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} - \frac {16 \, b x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {8}{3 \, \sqrt {b x^{2} + a x} a^{2}} \]

[In]

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

[Out]

2/3*x/((b*x^2 + a*x)^(3/2)*a) - 16/3*b*x/(sqrt(b*x^2 + a*x)*a^3) - 8/3/(sqrt(b*x^2 + a*x)*a^2)

Giac [F]

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

[In]

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

[Out]

integrate(x/(b*x^2 + a*x)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^2+12\,a\,b\,x+8\,b^2\,x^2\right )}{3\,a^3\,x\,{\left (a+b\,x\right )}^2} \]

[In]

int(x/(a*x + b*x^2)^(5/2),x)

[Out]

-(2*(a*x + b*x^2)^(1/2)*(3*a^2 + 8*b^2*x^2 + 12*a*b*x))/(3*a^3*x*(a + b*x)^2)