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

Optimal. Leaf size=39 \[ \frac {(A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {B x}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {388, 205} \begin {gather*} \frac {(A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {B x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(a + b*x^2),x]

[Out]

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 40, normalized size = 1.03 \begin {gather*} \frac {B x}{b}-\frac {(a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(a + b*x^2),x]

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 99, normalized size = 2.54 \begin {gather*} \left [\frac {2 \, B a b x + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2 \, a b^{2}}, \frac {B a b x - {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 34, normalized size = 0.87 \begin {gather*} \frac {B x}{b} - \frac {{\left (B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 45, normalized size = 1.15 \begin {gather*} \frac {A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}-\frac {B a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {B x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.50, size = 34, normalized size = 0.87 \begin {gather*} \frac {B x}{b} - \frac {{\left (B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 31, normalized size = 0.79 \begin {gather*} \frac {B\,x}{b}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{\sqrt {a}\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(a + b*x^2),x)

[Out]

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

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sympy [B]  time = 0.28, size = 82, normalized size = 2.10 \begin {gather*} \frac {B x}{b} + \frac {\sqrt {- \frac {1}{a b^{3}}} \left (- A b + B a\right ) \log {\left (- a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{2} - \frac {\sqrt {- \frac {1}{a b^{3}}} \left (- A b + B a\right ) \log {\left (a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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