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

Optimal. Leaf size=71 \[ -\frac {(3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}-\frac {x (A b-a B)}{2 a^2 \left (a+b x^2\right )}-\frac {A}{a^2 x} \]

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Rubi [A]  time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {456, 453, 205} \begin {gather*} -\frac {x (A b-a B)}{2 a^2 \left (a+b x^2\right )}-\frac {(3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}-\frac {A}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 456

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[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 70, normalized size = 0.99 \begin {gather*} \frac {(a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {x (a B-A b)}{2 a^2 \left (a+b x^2\right )}-\frac {A}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 210, normalized size = 2.96 \begin {gather*} \left [-\frac {4 \, A a^{2} b - 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (B a b - 3 \, A b^{2}\right )} x^{3} + {\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{4 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}, -\frac {2 \, A a^{2} b - {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (B a b - 3 \, A b^{2}\right )} x^{3} + {\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 62, normalized size = 0.87 \begin {gather*} \frac {{\left (B a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} + \frac {B a x^{2} - 3 \, A b x^{2} - 2 \, A a}{2 \, {\left (b x^{3} + a x\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 85, normalized size = 1.20 \begin {gather*} -\frac {A b x}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {3 A b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a^{2}}+\frac {B x}{2 \left (b \,x^{2}+a \right ) a}+\frac {B \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a}-\frac {A}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.15, size = 63, normalized size = 0.89 \begin {gather*} \frac {{\left (B a - 3 \, A b\right )} x^{2} - 2 \, A a}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}} + \frac {{\left (B a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 63, normalized size = 0.89 \begin {gather*} -\frac {\frac {A}{a}+\frac {x^2\,\left (3\,A\,b-B\,a\right )}{2\,a^2}}{b\,x^3+a\,x}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,A\,b-B\,a\right )}{2\,a^{5/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.49, size = 114, normalized size = 1.61 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{5} b}} \left (- 3 A b + B a\right ) \log {\left (- a^{3} \sqrt {- \frac {1}{a^{5} b}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{5} b}} \left (- 3 A b + B a\right ) \log {\left (a^{3} \sqrt {- \frac {1}{a^{5} b}} + x \right )}}{4} + \frac {- 2 A a + x^{2} \left (- 3 A b + B a\right )}{2 a^{3} x + 2 a^{2} b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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