\(\int \frac {a c+b c x^2}{x^3 (a+b x^2)^2} \, dx\) [133]

Optimal result

Integrand size = 23, antiderivative size = 38 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {c}{2 a x^2}-\frac {b c \log (x)}{a^2}+\frac {b c \log \left (a+b x^2\right )}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {21, 272, 46} \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {b c \log \left (a+b x^2\right )}{2 a^2}-\frac {b c \log (x)}{a^2}-\frac {c}{2 a x^2} \]

[In]

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

[Out]

-1/2*c/(a*x^2) - (b*c*Log[x])/a^2 + (b*c*Log[a + b*x^2])/(2*a^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 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = c \int \frac {1}{x^3 \left (a+b x^2\right )} \, dx \\ & = \frac {1}{2} c \text {Subst}\left (\int \frac {1}{x^2 (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} c \text {Subst}\left (\int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {c}{2 a x^2}-\frac {b c \log (x)}{a^2}+\frac {b c \log \left (a+b x^2\right )}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=c \left (-\frac {1}{2 a x^2}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^2\right )}{2 a^2}\right ) \]

[In]

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

[Out]

c*(-1/2*1/(a*x^2) - (b*Log[x])/a^2 + (b*Log[a + b*x^2])/(2*a^2))

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
default	\(c \left (-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 a^{2}}\right )\)	\(34\)
parallelrisch	\(-\frac {2 b c \ln \left (x \right ) x^{2}-b c \ln \left (b \,x^{2}+a \right ) x^{2}+a c}{2 a^{2} x^{2}}\)	\(37\)
risch	\(-\frac {c}{2 a \,x^{2}}-\frac {b c \ln \left (x \right )}{a^{2}}+\frac {b c \ln \left (-b \,x^{2}-a \right )}{2 a^{2}}\)	\(38\)
norman	\(\frac {-\frac {c}{2}+\frac {b^{2} c \,x^{4}}{2 a^{2}}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {b c \ln \left (x \right )}{a^{2}}+\frac {b c \ln \left (b \,x^{2}+a \right )}{2 a^{2}}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {b c x^{2} \log \left (b x^{2} + a\right ) - 2 \, b c x^{2} \log \left (x\right ) - a c}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=c \left (- \frac {1}{2 a x^{2}} - \frac {b \log {\left (x \right )}}{a^{2}} + \frac {b \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{2}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {b c \log \left (b x^{2} + a\right )}{2 \, a^{2}} - \frac {b c \log \left (x^{2}\right )}{2 \, a^{2}} - \frac {c}{2 \, a x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {b c \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {b c \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2}} + \frac {b c x^{2} - a c}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {b\,c\,\ln \left (b\,x^2+a\right )}{2\,a^2}-\frac {c}{2\,a\,x^2}-\frac {b\,c\,\ln \left (x\right )}{a^2} \]

[In]

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

[Out]

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