3.2.22 \(\int \frac {x (b+2 c x^2)}{b x^2+c x^4} \, dx\) [122]

Optimal. Leaf size=16 \[ \frac {1}{2} \log \left (b x^2+c x^4\right ) \]

[Out]

1/2*ln(c*x^4+b*x^2)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1598, 457, 78} \begin {gather*} \frac {1}{2} \log \left (b+c x^2\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(b + 2*c*x^2))/(b*x^2 + c*x^4),x]

[Out]

Log[x] + Log[b + c*x^2]/2

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x \left (b+2 c x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {b+2 c x^2}{x \left (b+c x^2\right )} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {b+2 c x}{x (b+c x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x}+\frac {c}{b+c x}\right ) \, dx,x,x^2\right )\\ &=\log (x)+\frac {1}{2} \log \left (b+c x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 15, normalized size = 0.94 \begin {gather*} \log (x)+\frac {1}{2} \log \left (b+c x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(b + 2*c*x^2))/(b*x^2 + c*x^4),x]

[Out]

Log[x] + Log[b + c*x^2]/2

________________________________________________________________________________________

Maple [A]
time = 0.15, size = 14, normalized size = 0.88

method	result	size
default	\(\ln \left (x \right )+\frac {\ln \left (c \,x^{2}+b \right )}{2}\)	\(14\)
norman	\(\ln \left (x \right )+\frac {\ln \left (c \,x^{2}+b \right )}{2}\)	\(14\)
risch	\(\ln \left (x \right )+\frac {\ln \left (c \,x^{2}+b \right )}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*c*x^2+b)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{2} \, \log \left (c x^{2} + b\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*c*x^2+b)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/2*log(c*x^2 + b) + 1/2*log(x^2)

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \log \left (c x^{2} + b\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*c*x^2+b)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/2*log(c*x^2 + b) + log(x)

________________________________________________________________________________________

Sympy [A]
time = 0.10, size = 12, normalized size = 0.75 \begin {gather*} \log {\left (x \right )} + \frac {\log {\left (\frac {b}{c} + x^{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*c*x**2+b)/(c*x**4+b*x**2),x)

[Out]

log(x) + log(b/c + x**2)/2

________________________________________________________________________________________

Giac [A]
time = 4.41, size = 15, normalized size = 0.94 \begin {gather*} \frac {1}{2} \, \log \left ({\left | c x^{4} + b x^{2} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*c*x^2+b)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*log(abs(c*x^4 + b*x^2))

________________________________________________________________________________________

Mupad [B]
time = 0.06, size = 13, normalized size = 0.81 \begin {gather*} \frac {\ln \left (c\,x^2+b\right )}{2}+\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(b + 2*c*x^2))/(b*x^2 + c*x^4),x)

[Out]

log(b + c*x^2)/2 + log(x)

________________________________________________________________________________________