\(\int \frac {A+B x^2}{a+b x^2} \, dx\) [62]

Optimal result

Integrand size = 17, antiderivative size = 39 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B x}{b}+\frac {(A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {396, 211} \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {(A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {B x}{b} \]

[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 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 396

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B x}{b}-\frac {(-A b+a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \]

[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))

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {B x}{b}+\frac {\left (A b -B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(34\)
risch	\(\frac {B x}{b}-\frac {\ln \left (b x +\sqrt {-a b}\right ) A}{2 \sqrt {-a b}}+\frac {\ln \left (b x +\sqrt {-a b}\right ) B a}{2 b \sqrt {-a b}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) A}{2 \sqrt {-a b}}-\frac {\ln \left (-b x +\sqrt {-a b}\right ) B a}{2 b \sqrt {-a b}}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.54 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\left [\frac {2 \, B a b x + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2 \, a b^{2}}, \frac {B a b x - {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{a b^{2}}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (34) = 68\).

Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.10 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B x}{b} + \frac {\sqrt {- \frac {1}{a b^{3}}} \left (- A b + B a\right ) \log {\left (- a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{2} - \frac {\sqrt {- \frac {1}{a b^{3}}} \left (- A b + B a\right ) \log {\left (a b \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{2} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B x}{b} - \frac {{\left (B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B x}{b} - \frac {{\left (B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

[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)

Mupad [B] (verification not implemented)

Time = 4.92 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2}{a+b x^2} \, dx=\frac {B\,x}{b}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{\sqrt {a}\,b^{3/2}} \]

[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))