3.31 \(\int \frac {c+d x^2}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=63 \[ \frac {(a d+b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}+\frac {x (b c-a d)}{2 a b \left (a+b x^2\right )} \]

[Out]

1/2*(-a*d+b*c)*x/a/b/(b*x^2+a)+1/2*(a*d+b*c)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^(3/2)

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Rubi [A]  time = 0.02, antiderivative size = 63, 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 = {385, 205} \[ \frac {(a d+b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}+\frac {x (b c-a d)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/(a + b*x^2)^2,x]

[Out]

((b*c - a*d)*x)/(2*a*b*(a + b*x^2)) + ((b*c + a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*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 385

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 63, normalized size = 1.00 \[ \frac {(a d+b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}-\frac {x (a d-b c)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/(a + b*x^2)^2,x]

[Out]

-1/2*((-(b*c) + a*d)*x)/(a*b*(a + b*x^2)) + ((b*c + a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*b^(3/2))

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fricas [A]  time = 0.72, size = 181, normalized size = 2.87 \[ \left [-\frac {{\left (a b c + a^{2} d + {\left (b^{2} c + a b d\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (a b^{2} c - a^{2} b d\right )} x}{4 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}, \frac {{\left (a b c + a^{2} d + {\left (b^{2} c + a b d\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (a b^{2} c - a^{2} b d\right )} x}{2 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((a*b*c + a^2*d + (b^2*c + a*b*d)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(a*b
^2*c - a^2*b*d)*x)/(a^2*b^3*x^2 + a^3*b^2), 1/2*((a*b*c + a^2*d + (b^2*c + a*b*d)*x^2)*sqrt(a*b)*arctan(sqrt(a
*b)*x/a) + (a*b^2*c - a^2*b*d)*x)/(a^2*b^3*x^2 + a^3*b^2)]

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giac [A]  time = 0.57, size = 57, normalized size = 0.90 \[ \frac {{\left (b c + a d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b} + \frac {b c x - a d x}{2 \, {\left (b x^{2} + a\right )} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c + a*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/2*(b*c*x - a*d*x)/((b*x^2 + a)*a*b)

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maple [A]  time = 0.01, size = 68, normalized size = 1.08 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a}+\frac {d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b}-\frac {\left (a d -b c \right ) x}{2 \left (b \,x^{2}+a \right ) a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(b*x^2+a)^2,x)

[Out]

-1/2*(a*d-b*c)/a/b*x/(b*x^2+a)+1/2/b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*d+1/2/a/(a*b)^(1/2)*arctan(1/(a*b)^
(1/2)*b*x)*c

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maxima [A]  time = 3.11, size = 57, normalized size = 0.90 \[ \frac {{\left (b c - a d\right )} x}{2 \, {\left (a b^{2} x^{2} + a^{2} b\right )}} + \frac {{\left (b c + a d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c - a*d)*x/(a*b^2*x^2 + a^2*b) + 1/2*(b*c + a*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b)

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mupad [B]  time = 5.04, size = 51, normalized size = 0.81 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (a\,d+b\,c\right )}{2\,a^{3/2}\,b^{3/2}}-\frac {x\,\left (a\,d-b\,c\right )}{2\,a\,b\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/(a + b*x^2)^2,x)

[Out]

(atan((b^(1/2)*x)/a^(1/2))*(a*d + b*c))/(2*a^(3/2)*b^(3/2)) - (x*(a*d - b*c))/(2*a*b*(a + b*x^2))

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sympy [B]  time = 0.40, size = 112, normalized size = 1.78 \[ \frac {x \left (- a d + b c\right )}{2 a^{2} b + 2 a b^{2} x^{2}} - \frac {\sqrt {- \frac {1}{a^{3} b^{3}}} \left (a d + b c\right ) \log {\left (- a^{2} b \sqrt {- \frac {1}{a^{3} b^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{3} b^{3}}} \left (a d + b c\right ) \log {\left (a^{2} b \sqrt {- \frac {1}{a^{3} b^{3}}} + x \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(b*x**2+a)**2,x)

[Out]

x*(-a*d + b*c)/(2*a**2*b + 2*a*b**2*x**2) - sqrt(-1/(a**3*b**3))*(a*d + b*c)*log(-a**2*b*sqrt(-1/(a**3*b**3))
+ x)/4 + sqrt(-1/(a**3*b**3))*(a*d + b*c)*log(a**2*b*sqrt(-1/(a**3*b**3)) + x)/4

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