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

Optimal. Leaf size=25 \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 205} \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(a*c + b*c*x^2)/(a + b*x^2)^2,x]

[Out]

(c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b])

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fricas [A]  time = 0.55, size = 69, normalized size = 2.76 \[ \left [-\frac {\sqrt {-a b} c \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2 \, a b}, \frac {\sqrt {a b} c \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{a b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 16, normalized size = 0.64 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*arctan(b*x/sqrt(a*b))/sqrt(a*b)

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maple [A]  time = 0.00, size = 17, normalized size = 0.68 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.23, size = 16, normalized size = 0.64 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*arctan(b*x/sqrt(a*b))/sqrt(a*b)

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mupad [B]  time = 0.05, size = 17, normalized size = 0.68 \[ \frac {c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c + b*c*x^2)/(a + b*x^2)^2,x)

[Out]

(c*atan((b^(1/2)*x)/a^(1/2)))/(a^(1/2)*b^(1/2))

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sympy [B]  time = 0.15, size = 54, normalized size = 2.16 \[ c \left (- \frac {\sqrt {- \frac {1}{a b}} \log {\left (- a \sqrt {- \frac {1}{a b}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{a b}} \log {\left (a \sqrt {- \frac {1}{a b}} + x \right )}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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