\(\int \frac {x^2}{(a x+b x^2)^{5/2}} \, dx\) [64]

Optimal result

Integrand size = 17, antiderivative size = 51 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 x}{3 b \left (a x+b x^2\right )^{3/2}}+\frac {2 (a+2 b x)}{3 a^2 b \sqrt {a x+b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {666, 627} \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 (a+2 b x)}{3 a^2 b \sqrt {a x+b x^2}}-\frac {2 x}{3 b \left (a x+b x^2\right )^{3/2}} \]

[In]

Int[x^2/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*x)/(3*b*(a*x + b*x^2)^(3/2)) + (2*(a + 2*b*x))/(3*a^2*b*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 666

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x^2 (3 a+2 b x)}{3 a^2 (x (a+b x))^{3/2}} \]

[In]

Integrate[x^2/(a*x + b*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(\frac {2 x \left (\frac {2 b x}{3}+a \right )}{\sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{2}}\)	\(29\)
trager	\(\frac {2 \left (2 b x +3 a \right ) \sqrt {b \,x^{2}+a x}}{3 a^{2} \left (b x +a \right )^{2}}\)	\(32\)
gosper	\(\frac {2 x^{3} \left (b x +a \right ) \left (2 b x +3 a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\)	\(33\)
default	\(-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, \sqrt {b x^{2} + a x} {\left (2 \, b x + 3 \, a\right )}}{3 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {4 \, x}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {2 \, x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {2}{3 \, \sqrt {b x^{2} + a x} a b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + 2 \, a\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )}^{3} \sqrt {b}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {x^2}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a+2\,b\,x\right )}{3\,a^2\,{\left (a+b\,x\right )}^2} \]

[In]

int(x^2/(a*x + b*x^2)^(5/2),x)

[Out]

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