\(\int \frac {x^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [192]

Optimal result

Integrand size = 24, antiderivative size = 99 \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 a}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {a^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[x^2/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.73 \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\frac {b \left (a x \sqrt {(a+b x)^2} \left (-2 a^2-a b x+b^2 x^2\right )+\sqrt {a^2} x \left (2 a^3+3 a^2 b x+b^3 x^3\right )\right )}{a^2 (a+b x) \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}+2 \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-2 \log \left (b^3 \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{2 b^3} \]

[In]

Integrate[x^2/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((b*(a*x*Sqrt[(a + b*x)^2]*(-2*a^2 - a*b*x + b^2*x^2) + Sqrt[a^2]*x*(2*a^3 + 3*a^2*b*x + b^3*x^3)))/(a^2*(a +
b*x)*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) + 2*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] - 2*Log[b^3*(Sq
rt[a^2] + b*x - Sqrt[(a + b*x)^2])])/(2*b^3)

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {3 a^{2}}{2 b^{3}}+\frac {2 a x}{b^{2}}\right )}{\left (b x +a \right )^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) b^{3}}\)	\(61\)
default	\(\frac {\left (2 b^{2} \ln \left (b x +a \right ) x^{2}+4 \ln \left (b x +a \right ) x a b +2 a^{2} \ln \left (b x +a \right )+4 a b x +3 a^{2}\right ) \left (b x +a \right )}{2 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(67\)

[In]

int(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {4 \, a b x + 3 \, a^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

Sympy [F]

\[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**2/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**2/((a + b*x)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.46 \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

log(x + a/b)/b^3 + 2*a*x/(b^4*(x + a/b)^2) + 3/2*a^2/(b^5*(x + a/b)^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.54 \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {4 \, a x + \frac {3 \, a^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

log(abs(b*x + a))/(b^3*sgn(b*x + a)) + 1/2*(4*a*x + 3*a^2/b)/((b*x + a)^2*b^2*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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