\(\int \frac {a c+b c x^2}{x^2 (a+b x^2)^3} \, dx\) [139]

Optimal result

Integrand size = 23, antiderivative size = 60 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {3 c}{2 a^2 x}+\frac {c}{2 a x \left (a+b x^2\right )}-\frac {3 \sqrt {b} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

-3/2*c/a^2/x+1/2*c/a/x/(b*x^2+a)-3/2*c*arctan(x*b^(1/2)/a^(1/2))*b^(1/2)/a^(5/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {21, 296, 331, 211} \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {3 \sqrt {b} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3 c}{2 a^2 x}+\frac {c}{2 a x \left (a+b x^2\right )} \]

[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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]

Rubi steps \begin{align*} \text {integral}& = 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] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=c \left (-\frac {1}{a^2 x}-\frac {b x}{2 a^2 \left (a+b x^2\right )}-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}\right ) \]

[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)))

Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78

method	result	size
default	\(c \left (-\frac {1}{a^{2} x}-\frac {b \left (\frac {x}{2 b \,x^{2}+2 a}+\frac {3 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}\right )\)	\(47\)
risch	\(\frac {-\frac {3 b c \,x^{2}}{2 a^{2}}-\frac {c}{a}}{x \left (b \,x^{2}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{2}+b \,c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{5}+2 b \,c^{2}\right ) x +a^{3} c \textit {\_R} \right )\right )}{4}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.40 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\left [-\frac {6 \, b c x^{2} - 3 \, {\left (b c x^{3} + a c x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 4 \, a c}{4 \, {\left (a^{2} b x^{3} + a^{3} x\right )}}, -\frac {3 \, b c x^{2} + 3 \, {\left (b c x^{3} + a c x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 2 \, a c}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.57 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=c \left (\frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x \right )}}{4} - \frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x \right )}}{4} + \frac {- 2 a - 3 b x^{2}}{2 a^{3} x + 2 a^{2} b x^{3}}\right ) \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {3 \, b c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {3 \, b c x^{2} + 2 \, a c}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}} \]

[In]

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

[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)/(a^2*b*x^3 + a^3*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {3 \, b c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {3 \, b c x^{2} + 2 \, a c}{2 \, {\left (b x^{3} + a x\right )} a^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.46 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {c}{a}+\frac {3\,b\,c\,x^2}{2\,a^2}}{b\,x^3+a\,x}-\frac {3\,\sqrt {b}\,c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}} \]

[In]

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

[Out]

- (c/a + (3*b*c*x^2)/(2*a^2))/(a*x + b*x^3) - (3*b^(1/2)*c*atan((b^(1/2)*x)/a^(1/2)))/(2*a^(5/2))