3.2.37 \(\int \frac {a c+b c x^2}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=47 \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {c x}{2 a \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {21, 199, 205} \begin {gather*} \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {c x}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*c + b*c*x^2)/(a + b*x^2)^3,x]

[Out]

(c*x)/(2*a*(a + b*x^2)) + (c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.00 \begin {gather*} c \left (\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*c + b*c*x^2)/(a + b*x^2)^3,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a c+b c x^2}{\left (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a*c + b*c*x^2)/(a + b*x^2)^3,x]

[Out]

IntegrateAlgebraic[(a*c + b*c*x^2)/(a + b*x^2)^3, x]

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fricas [A]  time = 0.58, size = 128, normalized size = 2.72 \begin {gather*} \left [\frac {2 \, a b c x - {\left (b c x^{2} + a c\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{4 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac {a b c x + {\left (b c x^{2} + a c\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*c*x^2+a*c)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.38, size = 37, normalized size = 0.79 \begin {gather*} \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} + \frac {c x}{2 \, {\left (b x^{2} + a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*c*x^2+a*c)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 38, normalized size = 0.81 \begin {gather*} \frac {c x}{2 \left (b \,x^{2}+a \right ) a}+\frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*c*x^2+a*c)/(b*x^2+a)^3,x)

[Out]

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

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maxima [A]  time = 2.40, size = 37, normalized size = 0.79 \begin {gather*} \frac {c x}{2 \, {\left (a b x^{2} + a^{2}\right )}} + \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*c*x^2+a*c)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 35, normalized size = 0.74 \begin {gather*} \frac {c\,x}{2\,a\,\left (b\,x^2+a\right )}+\frac {c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c + b*c*x^2)/(a + b*x^2)^3,x)

[Out]

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

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sympy [B]  time = 0.23, size = 80, normalized size = 1.70 \begin {gather*} c \left (\frac {x}{2 a^{2} + 2 a b x^{2}} - \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (- a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*c*x**2+a*c)/(b*x**2+a)**3,x)

[Out]

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

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