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

Optimal. Leaf size=60 \[ -\frac{3 \sqrt{b} c \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{3 c}{2 a^2 x}+\frac{c}{2 a x \left (a+b x^2\right )} \]

[Out]

(-3*c)/(2*a^2*x) + c/(2*a*x*(a + b*x^2)) - (3*Sqrt[b]*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2))

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Rubi [A]  time = 0.0220548, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {21, 290, 325, 205} \[ -\frac{3 \sqrt{b} c \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{3 c}{2 a^2 x}+\frac{c}{2 a x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c)/(2*a^2*x) + c/(2*a*x*(a + b*x^2)) - (3*Sqrt[b]*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2))

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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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}{x^2 \left (a+b x^2\right )^3} \, dx &=c \int \frac{1}{x^2 \left (a+b x^2\right )^2} \, dx\\ &=\frac{c}{2 a x \left (a+b x^2\right )}+\frac{(3 c) \int \frac{1}{x^2 \left (a+b x^2\right )} \, dx}{2 a}\\ &=-\frac{3 c}{2 a^2 x}+\frac{c}{2 a x \left (a+b x^2\right )}-\frac{(3 b c) \int \frac{1}{a+b x^2} \, dx}{2 a^2}\\ &=-\frac{3 c}{2 a^2 x}+\frac{c}{2 a x \left (a+b x^2\right )}-\frac{3 \sqrt{b} c \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0363686, size = 56, normalized size = 0.93 \[ c \left (-\frac{b x}{2 a^2 \left (a+b x^2\right )}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{1}{a^2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c*(-(1/(a^2*x)) - (b*x)/(2*a^2*(a + b*x^2)) - (3*Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)))

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Maple [A]  time = 0.007, size = 49, normalized size = 0.8 \begin{align*} -{\frac{c}{{a}^{2}x}}-{\frac{bcx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{3\,bc}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27035, size = 309, normalized size = 5.15 \begin{align*} \left [-\frac{6 \, b c x^{2} - 3 \,{\left (b c x^{3} + a c x\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) + 4 \, a c}{4 \,{\left (a^{2} b x^{3} + a^{3} x\right )}}, -\frac{3 \, b c x^{2} + 3 \,{\left (b c x^{3} + a c x\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 2 \, a c}{2 \,{\left (a^{2} b x^{3} + a^{3} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*b*c*x^2 - 3*(b*c*x^3 + a*c*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 4*a*c)/(a^
2*b*x^3 + a^3*x), -1/2*(3*b*c*x^2 + 3*(b*c*x^3 + a*c*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 2*a*c)/(a^2*b*x^3 + a^
3*x)]

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Sympy [A]  time = 0.44569, size = 92, normalized size = 1.53 \begin{align*} c \left (\frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (- \frac{a^{3} \sqrt{- \frac{b}{a^{5}}}}{b} + x \right )}}{4} - \frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (\frac{a^{3} \sqrt{- \frac{b}{a^{5}}}}{b} + x \right )}}{4} - \frac{2 a + 3 b x^{2}}{2 a^{3} x + 2 a^{2} b x^{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(3*sqrt(-b/a**5)*log(-a**3*sqrt(-b/a**5)/b + x)/4 - 3*sqrt(-b/a**5)*log(a**3*sqrt(-b/a**5)/b + x)/4 - (2*a +
3*b*x**2)/(2*a**3*x + 2*a**2*b*x**3))

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Giac [A]  time = 1.1715, size = 68, normalized size = 1.13 \begin{align*} -\frac{3 \, b c \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{2}} - \frac{3 \, b c x^{2} + 2 \, a c}{2 \,{\left (b x^{3} + a x\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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