\(\int \frac {x^2 (A+B x^2)}{(a+b x^2)^3} \, dx\) [102]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {(A b-a B) x}{4 b^2 \left (a+b x^2\right )^2}+\frac {(A b-5 a B) x}{8 a b^2 \left (a+b x^2\right )}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}} \]

[Out]

-1/4*(A*b-B*a)*x/b^2/(b*x^2+a)^2+1/8*(A*b-5*B*a)*x/a/b^2/(b*x^2+a)+1/8*(A*b+3*B*a)*arctan(x*b^(1/2)/a^(1/2))/a
^(3/2)/b^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, 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, 393, 211} \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(3 a B+A b) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}+\frac {x (A b-5 a B)}{8 a b^2 \left (a+b x^2\right )}-\frac {x (A b-a B)}{4 b^2 \left (a+b x^2\right )^2} \]

[In]

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

[Out]

-1/4*((A*b - a*B)*x)/(b^2*(a + b*x^2)^2) + ((A*b - 5*a*B)*x)/(8*a*b^2*(a + b*x^2)) + ((A*b + 3*a*B)*ArcTan[(Sq
rt[b]*x)/Sqrt[a]])/(8*a^(3/2)*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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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}{4 b^2 \left (a+b x^2\right )^2}-\frac {\int \frac {-A b+a B-4 b B x^2}{\left (a+b x^2\right )^2} \, dx}{4 b^2} \\ & = -\frac {(A b-a B) x}{4 b^2 \left (a+b x^2\right )^2}+\frac {(A b-5 a B) x}{8 a b^2 \left (a+b x^2\right )}+\frac {(A b+3 a B) \int \frac {1}{a+b x^2} \, dx}{8 a b^2} \\ & = -\frac {(A b-a B) x}{4 b^2 \left (a+b x^2\right )^2}+\frac {(A b-5 a B) x}{8 a b^2 \left (a+b x^2\right )}+\frac {(A b+3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {\sqrt {b} x \left (-3 a^2 B+A b^2 x^2-a b \left (A+5 B x^2\right )\right )}{a \left (a+b x^2\right )^2}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}}{8 b^{5/2}} \]

[In]

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

[Out]

((Sqrt[b]*x*(-3*a^2*B + A*b^2*x^2 - a*b*(A + 5*B*x^2)))/(a*(a + b*x^2)^2) + ((A*b + 3*a*B)*ArcTan[(Sqrt[b]*x)/
Sqrt[a]])/a^(3/2))/(8*b^(5/2))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {\frac {\left (A b -5 B a \right ) x^{3}}{8 a b}-\frac {\left (A b +3 B a \right ) x}{8 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (A b +3 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 b^{2} a \sqrt {a b}}\)	\(76\)
risch	\(\frac {\frac {\left (A b -5 B a \right ) x^{3}}{8 a b}-\frac {\left (A b +3 B a \right ) x}{8 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, b a}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b^{2}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, b a}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b^{2}}\)	\(146\)

[In]

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

[Out]

(1/8*(A*b-5*B*a)/a/b*x^3-1/8*(A*b+3*B*a)/b^2*x)/(b*x^2+a)^2+1/8*(A*b+3*B*a)/b^2/a/(a*b)^(1/2)*arctan(b*x/(a*b)
^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.38 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} + {\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac {{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} - {\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/16*(2*(5*B*a^2*b^2 - A*a*b^3)*x^3 + ((3*B*a*b^2 + A*b^3)*x^4 + 3*B*a^3 + A*a^2*b + 2*(3*B*a^2*b + A*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^3*b + A*a^2*b^2)*x)/(a^2*b^5*x^4 + 2
*a^3*b^4*x^2 + a^4*b^3), -1/8*((5*B*a^2*b^2 - A*a*b^3)*x^3 - ((3*B*a*b^2 + A*b^3)*x^4 + 3*B*a^3 + A*a^2*b + 2*
(3*B*a^2*b + A*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (3*B*a^3*b + A*a^2*b^2)*x)/(a^2*b^5*x^4 + 2*a^3*b
^4*x^2 + a^4*b^3)]

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.74 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (A b + 3 B a\right ) \log {\left (- a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (A b + 3 B a\right ) \log {\left (a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {x^{3} \left (A b^{2} - 5 B a b\right ) + x \left (- A a b - 3 B a^{2}\right )}{8 a^{3} b^{2} + 16 a^{2} b^{3} x^{2} + 8 a b^{4} x^{4}} \]

[In]

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

[Out]

-sqrt(-1/(a**3*b**5))*(A*b + 3*B*a)*log(-a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + sqrt(-1/(a**3*b**5))*(A*b +
3*B*a)*log(a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + (x**3*(A*b**2 - 5*B*a*b) + x*(-A*a*b - 3*B*a**2))/(8*a**3*
b**2 + 16*a**2*b**3*x**2 + 8*a*b**4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (5 \, B a b - A b^{2}\right )} x^{3} + {\left (3 \, B a^{2} + A a b\right )} x}{8 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} \]

[In]

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

[Out]

-1/8*((5*B*a*b - A*b^2)*x^3 + (3*B*a^2 + A*a*b)*x)/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2) + 1/8*(3*B*a + A*b)*a
rctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} - \frac {5 \, B a b x^{3} - A b^{2} x^{3} + 3 \, B a^{2} x + A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} \]

[In]

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

[Out]

1/8*(3*B*a + A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2) - 1/8*(5*B*a*b*x^3 - A*b^2*x^3 + 3*B*a^2*x + A*a*b*x
)/((b*x^2 + a)^2*a*b^2)

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b+3\,B\,a\right )}{8\,a^{3/2}\,b^{5/2}}-\frac {\frac {x\,\left (A\,b+3\,B\,a\right )}{8\,b^2}-\frac {x^3\,\left (A\,b-5\,B\,a\right )}{8\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4} \]

[In]

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

[Out]

(atan((b^(1/2)*x)/a^(1/2))*(A*b + 3*B*a))/(8*a^(3/2)*b^(5/2)) - ((x*(A*b + 3*B*a))/(8*b^2) - (x^3*(A*b - 5*B*a
))/(8*a*b))/(a^2 + b^2*x^4 + 2*a*b*x^2)