\(\int \frac {x^2 (a c+b c x^2)}{(a+b x^2)^3} \, dx\) [135]

Optimal result

Integrand size = 23, antiderivative size = 47 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {c x}{2 b \left (a+b x^2\right )}+\frac {c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {21, 294, 211} \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}}-\frac {c x}{2 b \left (a+b x^2\right )} \]

[In]

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

[Out]

-1/2*(c*x)/(b*(a + b*x^2)) + (c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=c \left (-\frac {x}{2 b \left (a+b x^2\right )}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}}\right ) \]

[In]

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

[Out]

c*(-1/2*x/(b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*Sqrt[a]*b^(3/2)))

Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81

method	result	size
default	\(c \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )}+\frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\right )\)	\(38\)
risch	\(-\frac {c x}{2 b \left (b \,x^{2}+a \right )}-\frac {c \ln \left (b x +\sqrt {-a b}\right )}{4 \sqrt {-a b}\, b}+\frac {c \ln \left (-b x +\sqrt {-a b}\right )}{4 \sqrt {-a b}\, b}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.72 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {2 \, a b c x + {\left (b c x^{2} + a c\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{4 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, -\frac {a b c x - {\left (b c x^{2} + a c\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/4*(2*a*b*c*x + (b*c*x^2 + a*c)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a*b^3*x^2 + a^2*
b^2), -1/2*(a*b*c*x - (b*c*x^2 + a*c)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a*b^3*x^2 + a^2*b^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (39) = 78\).

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.70 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=c \left (- \frac {x}{2 a b + 2 b^{2} x^{2}} - \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (- a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{4}\right ) \]

[In]

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

[Out]

c*(-x/(2*a*b + 2*b**2*x**2) - sqrt(-1/(a*b**3))*log(-a*b*sqrt(-1/(a*b**3)) + x)/4 + sqrt(-1/(a*b**3))*log(a*b*
sqrt(-1/(a*b**3)) + x)/4)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {c x}{2 \, {\left (b^{2} x^{2} + a b\right )}} + \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b} \]

[In]

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

[Out]

-1/2*c*x/(b^2*x^2 + a*b) + 1/2*c*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b} - \frac {c x}{2 \, {\left (b x^{2} + a\right )} b} \]

[In]

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

[Out]

1/2*c*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b) - 1/2*c*x/((b*x^2 + a)*b)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,b^{3/2}}-\frac {c\,x}{2\,b\,\left (b\,x^2+a\right )} \]

[In]

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

[Out]

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