\(\int \frac {a c+b c x^2}{x^2 (a+b x^2)} \, dx\) [125]

Optimal result

Integrand size = 23, antiderivative size = 6 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[Out]

-c/x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {21, 30} \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-(c/x)

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = c \int \frac {1}{x^2} \, dx \\ & = -\frac {c}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-(c/x)

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17

method	result	size
gosper	\(-\frac {c}{x}\)	\(7\)
default	\(-\frac {c}{x}\)	\(7\)
norman	\(-\frac {c}{x}\)	\(7\)
risch	\(-\frac {c}{x}\)	\(7\)
parallelrisch	\(-\frac {c}{x}\)	\(7\)

[In]

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

[Out]

-c/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-c/x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.50 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=- \frac {c}{x} \]

[In]

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

[Out]

-c/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-c/x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-c/x

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {a c+b c x^2}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c}{x} \]

[In]

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

[Out]

-c/x