3.1.51 \(\int \frac {A+B x^2}{b x^2+c x^4} \, dx\) [51]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {1607, 464, 211} \begin {gather*} \frac {(b B-A c) \text {ArcTan}\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 211

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

Rule 464

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 1607

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 = 0.02, size = 42, normalized size = 1.00 \begin {gather*} -\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 {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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Maple [A]
time = 0.41, size = 37, normalized size = 0.88

method	result	size
default	\(\frac {\left (-A c +B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{b \sqrt {b c}}-\frac {A}{b x}\)	\(37\)
risch	\(-\frac {A}{b x}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (b^{3} c \,\textit {\_Z}^{2}+A^{2} c^{2}-2 A B b c +B^{2} b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{3} c +2 A^{2} c^{2}-4 A B b c +2 B^{2} b^{2}\right ) x +\left (A \,b^{2} c -B \,b^{3}\right ) \textit {\_R} \right )\right )}{2}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, 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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Fricas [A]
time = 2.03, 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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (34) = 68\).
time = 0.16, 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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Giac [A]
time = 0.64, 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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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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