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

Optimal. Leaf size=61 \[ -\frac {\sqrt {c} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{5/2}}-\frac {b B-A c}{b^2 x}-\frac {A}{3 b x^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 453, 325, 205} \begin {gather*} -\frac {b B-A c}{b^2 x}-\frac {\sqrt {c} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{5/2}}-\frac {A}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^2*(b*x^2 + c*x^4)),x]

[Out]

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

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 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 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 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 60, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {c} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{5/2}}+\frac {A c-b B}{b^2 x}-\frac {A}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*A/(b*x^3) + (-(b*B) + A*c)/(b^2*x) - (Sqrt[c]*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(5/2)

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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 (b x^2+c x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^2*(b*x^2 + c*x^4)),x]

[Out]

IntegrateAlgebraic[(A + B*x^2)/(x^2*(b*x^2 + c*x^4)), x]

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fricas [A]  time = 0.41, size = 135, normalized size = 2.21 \begin {gather*} \left [-\frac {3 \, {\left (B b - A c\right )} x^{3} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) + 6 \, {\left (B b - A c\right )} x^{2} + 2 \, A b}{6 \, b^{2} x^{3}}, -\frac {3 \, {\left (B b - A c\right )} x^{3} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 3 \, {\left (B b - A c\right )} x^{2} + A b}{3 \, b^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 57, normalized size = 0.93 \begin {gather*} -\frac {{\left (B b c - A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{2}} - \frac {3 \, B b x^{2} - 3 \, A c x^{2} + A b}{3 \, b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 72, normalized size = 1.18 \begin {gather*} \frac {A \,c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{2}}-\frac {B c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b}+\frac {A c}{b^{2} x}-\frac {B}{b x}-\frac {A}{3 b \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^2/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 2.82, size = 56, normalized size = 0.92 \begin {gather*} -\frac {{\left (B b c - A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{2}} - \frac {3 \, {\left (B b - A c\right )} x^{2} + A b}{3 \, b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 53, normalized size = 0.87 \begin {gather*} \frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{5/2}}-\frac {\frac {A}{3\,b}-\frac {x^2\,\left (A\,c-B\,b\right )}{b^2}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^2*(b*x^2 + c*x^4)),x)

[Out]

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

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sympy [B]  time = 0.44, size = 129, normalized size = 2.11 \begin {gather*} \frac {\sqrt {- \frac {c}{b^{5}}} \left (- A c + B b\right ) \log {\left (- \frac {b^{3} \sqrt {- \frac {c}{b^{5}}} \left (- A c + B b\right )}{- A c^{2} + B b c} + x \right )}}{2} - \frac {\sqrt {- \frac {c}{b^{5}}} \left (- A c + B b\right ) \log {\left (\frac {b^{3} \sqrt {- \frac {c}{b^{5}}} \left (- A c + B b\right )}{- A c^{2} + B b c} + x \right )}}{2} + \frac {- A b + x^{2} \left (3 A c - 3 B b\right )}{3 b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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