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

Optimal. Leaf size=97 \[ -\frac{x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac{x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b}}-\frac{A}{a^3 x} \]

[Out]

-(A/(a^3*x)) - ((A*b - a*B)*x)/(4*a^2*(a + b*x^2)^2) - ((7*A*b - 3*a*B)*x)/(8*a^3*(a + b*x^2)) - (3*(5*A*b - a
*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b])

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Rubi [A]  time = 0.100256, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {456, 453, 205} \[ -\frac{x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac{x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b}}-\frac{A}{a^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^3*x)) - ((A*b - a*B)*x)/(4*a^2*(a + b*x^2)^2) - ((7*A*b - 3*a*B)*x)/(8*a^3*(a + b*x^2)) - (3*(5*A*b - a
*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b])

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])

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 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]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx &=-\frac{(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac{1}{4} \int \frac{-\frac{4 A}{a}+\frac{3 (A b-a B) x^2}{a^2}}{x^2 \left (a+b x^2\right )^2} \, dx\\ &=-\frac{(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac{(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac{1}{8} \int \frac{\frac{8 A}{a^2}-\frac{(7 A b-3 a B) x^2}{a^3}}{x^2 \left (a+b x^2\right )} \, dx\\ &=-\frac{A}{a^3 x}-\frac{(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac{(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac{(3 (5 A b-a B)) \int \frac{1}{a+b x^2} \, dx}{8 a^3}\\ &=-\frac{A}{a^3 x}-\frac{(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac{(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.057687, size = 96, normalized size = 0.99 \[ \frac{x (3 a B-7 A b)}{8 a^3 \left (a+b x^2\right )}+\frac{x (a B-A b)}{4 a^2 \left (a+b x^2\right )^2}+\frac{3 (a B-5 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b}}-\frac{A}{a^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^3*x)) + ((-(A*b) + a*B)*x)/(4*a^2*(a + b*x^2)^2) + ((-7*A*b + 3*a*B)*x)/(8*a^3*(a + b*x^2)) + (3*(-5*A*
b + a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b])

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Maple [A]  time = 0.012, size = 125, normalized size = 1.3 \begin{align*} -{\frac{A}{{a}^{3}x}}-{\frac{7\,A{x}^{3}{b}^{2}}{8\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,bB{x}^{3}}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,Abx}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,Bx}{8\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,Ab}{8\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,B}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A/a^3/x-7/8/a^3/(b*x^2+a)^2*A*x^3*b^2+3/8/a^2/(b*x^2+a)^2*B*x^3*b-9/8/a^2/(b*x^2+a)^2*b*A*x+5/8/a/(b*x^2+a)^2
*B*x-15/8/a^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A*b+3/8/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29727, size = 686, normalized size = 7.07 \begin{align*} \left [-\frac{16 \, A a^{3} b - 6 \,{\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 10 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \,{\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \,{\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} +{\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x^{2} + 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{16 \,{\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac{8 \, A a^{3} b - 3 \,{\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 5 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \,{\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \,{\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} +{\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{8 \,{\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*A*a^3*b - 6*(B*a^2*b^2 - 5*A*a*b^3)*x^4 - 10*(B*a^3*b - 5*A*a^2*b^2)*x^2 - 3*((B*a*b^2 - 5*A*b^3)*x
^5 + 2*(B*a^2*b - 5*A*a*b^2)*x^3 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 +
a)))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x), -1/8*(8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^4 - 5*(B*a^3*b -
5*A*a^2*b^2)*x^2 - 3*((B*a*b^2 - 5*A*b^3)*x^5 + 2*(B*a^2*b - 5*A*a*b^2)*x^3 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(a*b)
*arctan(sqrt(a*b)*x/a))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x)]

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Sympy [B]  time = 0.87941, size = 194, normalized size = 2. \begin{align*} - \frac{3 \sqrt{- \frac{1}{a^{7} b}} \left (- 5 A b + B a\right ) \log{\left (- \frac{3 a^{4} \sqrt{- \frac{1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{a^{7} b}} \left (- 5 A b + B a\right ) \log{\left (\frac{3 a^{4} \sqrt{- \frac{1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac{- 8 A a^{2} + x^{4} \left (- 15 A b^{2} + 3 B a b\right ) + x^{2} \left (- 25 A a b + 5 B a^{2}\right )}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log(-3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + 3
*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log(3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + (-8*
A*a**2 + x**4*(-15*A*b**2 + 3*B*a*b) + x**2*(-25*A*a*b + 5*B*a**2))/(8*a**5*x + 16*a**4*b*x**3 + 8*a**3*b**2*x
**5)

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Giac [A]  time = 1.18626, size = 111, normalized size = 1.14 \begin{align*} \frac{3 \,{\left (B a - 5 \, A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{3}} - \frac{A}{a^{3} x} + \frac{3 \, B a b x^{3} - 7 \, A b^{2} x^{3} + 5 \, B a^{2} x - 9 \, A a b x}{8 \,{\left (b x^{2} + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(B*a - 5*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - A/(a^3*x) + 1/8*(3*B*a*b*x^3 - 7*A*b^2*x^3 + 5*B*a^2
*x - 9*A*a*b*x)/((b*x^2 + a)^2*a^3)