3.66 \(\int \frac{A+B x^2}{x^4 (a+b x^2)} \, dx\)

Optimal. Leaf size=59 \[ \frac{A b-a B}{a^2 x}+\frac{\sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2}}-\frac{A}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.0393545, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {453, 325, 205} \[ \frac{A b-a B}{a^2 x}+\frac{\sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2}}-\frac{A}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^4*(a + b*x^2)),x]

[Out]

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

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 325

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

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

Mathematica [A]  time = 0.0523703, size = 60, normalized size = 1.02 \[ \frac{A b-a B}{a^2 x}-\frac{\sqrt{b} (a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2}}-\frac{A}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^4*(a + b*x^2)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 72, normalized size = 1.2 \begin{align*} -{\frac{A}{3\,a{x}^{3}}}+{\frac{Ab}{{a}^{2}x}}-{\frac{B}{ax}}+{\frac{{b}^{2}A}{{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{bB}{a}\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^4/(b*x^2+a),x)

[Out]

-1/3*A/a/x^3+1/a^2/x*A*b-1/a/x*B+b^2/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A-b/a/(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^4/(b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.30634, size = 296, normalized size = 5.02 \begin{align*} \left [-\frac{3 \,{\left (B a - A b\right )} x^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) + 6 \,{\left (B a - A b\right )} x^{2} + 2 \, A a}{6 \, a^{2} x^{3}}, -\frac{3 \,{\left (B a - A b\right )} x^{3} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 3 \,{\left (B a - A b\right )} x^{2} + A a}{3 \, a^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.562504, size = 129, normalized size = 2.19 \begin{align*} \frac{\sqrt{- \frac{b}{a^{5}}} \left (- A b + B a\right ) \log{\left (- \frac{a^{3} \sqrt{- \frac{b}{a^{5}}} \left (- A b + B a\right )}{- A b^{2} + B a b} + x \right )}}{2} - \frac{\sqrt{- \frac{b}{a^{5}}} \left (- A b + B a\right ) \log{\left (\frac{a^{3} \sqrt{- \frac{b}{a^{5}}} \left (- A b + B a\right )}{- A b^{2} + B a b} + x \right )}}{2} - \frac{A a + x^{2} \left (- 3 A b + 3 B a\right )}{3 a^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**4/(b*x**2+a),x)

[Out]

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

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Giac [A]  time = 1.14362, size = 77, normalized size = 1.31 \begin{align*} -\frac{{\left (B a b - A b^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2}} - \frac{3 \, B a x^{2} - 3 \, A b x^{2} + A a}{3 \, a^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a),x, algorithm="giac")

[Out]

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