\(\int \frac {a c+b c x^2}{x^3 (a+b x^2)^3} \, dx\) [140]

Optimal result

Integrand size = 23, antiderivative size = 53 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {c}{2 a^2 x^2}-\frac {b c}{2 a^2 \left (a+b x^2\right )}-\frac {2 b c \log (x)}{a^3}+\frac {b c \log \left (a+b x^2\right )}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, 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} \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=\frac {b c \log \left (a+b x^2\right )}{a^3}-\frac {2 b c \log (x)}{a^3}-\frac {b c}{2 a^2 \left (a+b x^2\right )}-\frac {c}{2 a^2 x^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {c \left (a \left (\frac {1}{x^2}+\frac {b}{a+b x^2}\right )+4 b \log (x)-2 b \log \left (a+b x^2\right )\right )}{2 a^3} \]

[In]

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

[Out]

-1/2*(c*(a*(x^(-2) + b/(a + b*x^2)) + 4*b*Log[x] - 2*b*Log[a + b*x^2]))/a^3

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08

method	result	size
default	\(c \left (-\frac {1}{2 a^{2} x^{2}}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {b^{2} \left (\frac {2 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{3}}\right )\)	\(57\)
risch	\(\frac {-\frac {b c \,x^{2}}{a^{2}}-\frac {c}{2 a}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {2 b c \ln \left (x \right )}{a^{3}}+\frac {b c \ln \left (-b \,x^{2}-a \right )}{a^{3}}\)	\(58\)
norman	\(\frac {-\frac {c}{2}+\frac {2 b^{2} c \,x^{4}}{a^{2}}+\frac {3 b^{3} c \,x^{6}}{2 a^{3}}}{x^{2} \left (b \,x^{2}+a \right )^{2}}+\frac {b c \ln \left (b \,x^{2}+a \right )}{a^{3}}-\frac {2 b c \ln \left (x \right )}{a^{3}}\)	\(66\)
parallelrisch	\(-\frac {4 \ln \left (x \right ) x^{4} b^{2} c -2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{2} c -2 b^{2} c \,x^{4}+4 \ln \left (x \right ) x^{2} a b c -2 \ln \left (b \,x^{2}+a \right ) x^{2} a b c +a^{2} c}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )}\)	\(87\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {2 \, a b c x^{2} + a^{2} c - 2 \, {\left (b^{2} c x^{4} + a b c x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{2} c x^{4} + a b c x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=c \left (\frac {- a - 2 b x^{2}}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} - \frac {2 b \log {\left (x \right )}}{a^{3}} + \frac {b \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{3}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {2 \, b c x^{2} + a c}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac {b c \log \left (b x^{2} + a\right )}{a^{3}} - \frac {b c \log \left (x^{2}\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {b c \log \left (x^{2}\right )}{a^{3}} + \frac {b c \log \left ({\left | b x^{2} + a \right |}\right )}{a^{3}} - \frac {2 \, b c x^{2} + a c}{2 \, {\left (b x^{4} + a x^{2}\right )} a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {a c+b c x^2}{x^3 \left (a+b x^2\right )^3} \, dx=\frac {b\,c\,\ln \left (b\,x^2+a\right )}{a^3}-\frac {\frac {c}{2\,a}+\frac {b\,c\,x^2}{a^2}}{b\,x^4+a\,x^2}-\frac {2\,b\,c\,\ln \left (x\right )}{a^3} \]

[In]

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

[Out]

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