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

Optimal. Leaf size=33 \[ \frac {c x}{b}-\frac {\sqrt {a} c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {21, 321, 205} \[ \frac {c x}{b}-\frac {\sqrt {a} c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 321

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ c \left (\frac {x}{b}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 86, normalized size = 2.61 \[ \left [\frac {c \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 2 \, c x}{2 \, b}, -\frac {c \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - c x}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 28, normalized size = 0.85 \[ -\frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {c x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 0.88 \[ -\frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {c x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.58, size = 28, normalized size = 0.85 \[ -\frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {c x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 25, normalized size = 0.76 \[ \frac {c\,x}{b}-\frac {\sqrt {a}\,c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 58, normalized size = 1.76 \[ c \left (\frac {\sqrt {- \frac {a}{b^{3}}} \log {\left (- b \sqrt {- \frac {a}{b^{3}}} + x \right )}}{2} - \frac {\sqrt {- \frac {a}{b^{3}}} \log {\left (b \sqrt {- \frac {a}{b^{3}}} + x \right )}}{2} + \frac {x}{b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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