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

Optimal. Leaf size=42 \[ \frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {A}{b x} \]

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1593, 453, 205} \begin {gather*} \frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {A}{b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 + c*x^4),x]

[Out]

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

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 1593

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 42, normalized size = 1.00 \begin {gather*} \frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {A}{b x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{b x^2+c x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 105, normalized size = 2.50 \begin {gather*} \left [\frac {{\left (B b - A c\right )} \sqrt {-b c} x \log \left (\frac {c x^{2} + 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) - 2 \, A b c}{2 \, b^{2} c x}, \frac {{\left (B b - A c\right )} \sqrt {b c} x \arctan \left (\frac {\sqrt {b c} x}{b}\right ) - A b c}{b^{2} c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 36, normalized size = 0.86 \begin {gather*} \frac {{\left (B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {A}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b) - A/(b*x)

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maple [A]  time = 0.05, size = 48, normalized size = 1.14 \begin {gather*} -\frac {A c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b}+\frac {B \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}}-\frac {A}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 36, normalized size = 0.86 \begin {gather*} \frac {{\left (B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {A}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b) - A/(b*x)

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mupad [B]  time = 0.09, size = 35, normalized size = 0.83 \begin {gather*} -\frac {A}{b\,x}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{3/2}\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(b*x^2 + c*x^4),x)

[Out]

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

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sympy [B]  time = 0.42, size = 82, normalized size = 1.95 \begin {gather*} - \frac {A}{b x} - \frac {\sqrt {- \frac {1}{b^{3} c}} \left (- A c + B b\right ) \log {\left (- b^{2} \sqrt {- \frac {1}{b^{3} c}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{b^{3} c}} \left (- A c + B b\right ) \log {\left (b^{2} \sqrt {- \frac {1}{b^{3} c}} + x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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