\(\int \frac {x^2 (A+B x^2)}{(a+b x^2)^2} \, dx\) [78]

Optimal result

Integrand size = 20, antiderivative size = 67 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x}{b^2}-\frac {(A b-a B) x}{2 b^2 \left (a+b x^2\right )}+\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{5/2}} \]

[Out]

B*x/b^2-1/2*(A*b-B*a)*x/b^2/(b*x^2+a)+1/2*(A*b-3*B*a)*arctan(x*b^(1/2)/a^(1/2))/b^(5/2)/a^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {466, 396, 211} \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{5/2}}-\frac {x (A b-a B)}{2 b^2 \left (a+b x^2\right )}+\frac {B x}{b^2} \]

[In]

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

[Out]

(B*x)/b^2 - ((A*b - a*B)*x)/(2*b^2*(a + b*x^2)) + ((A*b - 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(5/
2))

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 396

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x}{b^2}-\frac {(A b-a B) x}{2 b^2 \left (a+b x^2\right )}-\frac {(-A b+3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{5/2}} \]

[In]

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

[Out]

(B*x)/b^2 - ((A*b - a*B)*x)/(2*b^2*(a + b*x^2)) - ((-(A*b) + 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^
(5/2))

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {B x}{b^{2}}+\frac {\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}}{b^{2}}\)	\(57\)
risch	\(\frac {B x}{b^{2}}+\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) x}{b^{2} \left (b \,x^{2}+a \right )}-\frac {\ln \left (b x +\sqrt {-a b}\right ) A}{4 b \sqrt {-a b}}+\frac {3 \ln \left (b x +\sqrt {-a b}\right ) B a}{4 b^{2} \sqrt {-a b}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) A}{4 b \sqrt {-a b}}-\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) B a}{4 b^{2} \sqrt {-a b}}\)	\(127\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 3.10 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, B a b^{2} x^{3} + {\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} x}{4 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}, \frac {2 \, B a b^{2} x^{3} - {\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, B a^{2} b - A a b^{2}\right )} x}{2 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x}{b^{2}} + \frac {x \left (- A b + B a\right )}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {\sqrt {- \frac {1}{a b^{5}}} \left (- A b + 3 B a\right ) \log {\left (- a b^{2} \sqrt {- \frac {1}{a b^{5}}} + x \right )}}{4} - \frac {\sqrt {- \frac {1}{a b^{5}}} \left (- A b + 3 B a\right ) \log {\left (a b^{2} \sqrt {- \frac {1}{a b^{5}}} + x \right )}}{4} \]

[In]

integrate(x**2*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

B*x/b**2 + x*(-A*b + B*a)/(2*a*b**2 + 2*b**3*x**2) + sqrt(-1/(a*b**5))*(-A*b + 3*B*a)*log(-a*b**2*sqrt(-1/(a*b
**5)) + x)/4 - sqrt(-1/(a*b**5))*(-A*b + 3*B*a)*log(a*b**2*sqrt(-1/(a*b**5)) + x)/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (B a - A b\right )} x}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {B x}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{2}} \]

[In]

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

[Out]

1/2*(B*a - A*b)*x/(b^3*x^2 + a*b^2) + B*x/b^2 - 1/2*(3*B*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{2}} + \frac {B a x - A b x}{2 \, {\left (b x^{2} + a\right )} b^{2}} \]

[In]

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

[Out]

B*x/b^2 - 1/2*(3*B*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) + 1/2*(B*a*x - A*b*x)/((b*x^2 + a)*b^2)

Mupad [B] (verification not implemented)

Time = 5.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B\,x}{b^2}-\frac {x\,\left (\frac {A\,b}{2}-\frac {B\,a}{2}\right )}{b^3\,x^2+a\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-3\,B\,a\right )}{2\,\sqrt {a}\,b^{5/2}} \]

[In]

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

[Out]

(B*x)/b^2 - (x*((A*b)/2 - (B*a)/2))/(a*b^2 + b^3*x^2) + (atan((b^(1/2)*x)/a^(1/2))*(A*b - 3*B*a))/(2*a^(1/2)*b
^(5/2))