\(\int \frac {a c+b c x^2}{x (a+b x^2)^2} \, dx\) [131]

Optimal result

Integrand size = 23, antiderivative size = 24 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=\frac {c \log (x)}{a}-\frac {c \log \left (a+b x^2\right )}{2 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {21, 272, 36, 29, 31} \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=\frac {c \log (x)}{a}-\frac {c \log \left (a+b x^2\right )}{2 a} \]

[In]

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

[Out]

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

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 \left (a+b x^2\right )} \, dx \\ & = \frac {1}{2} c \text {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {c \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 a}-\frac {(b c) \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,x^2\right )}{2 a} \\ & = \frac {c \log (x)}{a}-\frac {c \log \left (a+b x^2\right )}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=c \left (\frac {\log (x)}{a}-\frac {\log \left (a+b x^2\right )}{2 a}\right ) \]

[In]

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

[Out]

c*(Log[x]/a - Log[a + b*x^2]/(2*a))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
default	\(c \left (\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a}\right )\)	\(23\)
norman	\(\frac {c \ln \left (x \right )}{a}-\frac {c \ln \left (b \,x^{2}+a \right )}{2 a}\)	\(23\)
risch	\(\frac {c \ln \left (x \right )}{a}-\frac {c \ln \left (b \,x^{2}+a \right )}{2 a}\)	\(23\)
parallelrisch	\(\frac {2 c \ln \left (x \right )-c \ln \left (b \,x^{2}+a \right )}{2 a}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=-\frac {c \log \left (b x^{2} + a\right ) - 2 \, c \log \left (x\right )}{2 \, a} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=c \left (\frac {\log {\left (x \right )}}{a} - \frac {\log {\left (\frac {a}{b} + x^{2} \right )}}{2 a}\right ) \]

[In]

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

[Out]

c*(log(x)/a - log(a/b + x**2)/(2*a))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=-\frac {c \log \left (b x^{2} + a\right )}{2 \, a} + \frac {c \log \left (x^{2}\right )}{2 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=\frac {c \log \left (x^{2}\right )}{2 \, a} - \frac {c \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx=-\frac {c\,\left (\ln \left (b\,x^2+a\right )-2\,\ln \left (x\right )\right )}{2\,a} \]

[In]

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

[Out]

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