3.2.83 \(\int \frac {x}{b x^2+c x^4} \, dx\) [183]

Optimal. Leaf size=22 \[ \frac {\log (x)}{b}-\frac {\log \left (b+c x^2\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1598, 272, 36, 29, 31} \begin {gather*} \frac {\log (x)}{b}-\frac {\log \left (b+c x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(b*x^2 + c*x^4),x]

[Out]

Log[x]/b - Log[b + c*x^2]/(2*b)

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]]

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

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Mathematica [A]
time = 0.00, size = 22, normalized size = 1.00 \begin {gather*} \frac {\log (x)}{b}-\frac {\log \left (b+c x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(b*x^2 + c*x^4),x]

[Out]

Log[x]/b - Log[b + c*x^2]/(2*b)

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Maple [A]
time = 0.09, size = 21, normalized size = 0.95

method	result	size
default	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b}\)	\(21\)
norman	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b}\)	\(21\)
risch	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (c \,x^{2}+b \right )}{2 b}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 23, normalized size = 1.05 \begin {gather*} -\frac {\log \left (c x^{2} + b\right )}{2 \, b} + \frac {\log \left (x^{2}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 18, normalized size = 0.82 \begin {gather*} -\frac {\log \left (c x^{2} + b\right ) - 2 \, \log \left (x\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(c*x^2 + b) - 2*log(x))/b

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Sympy [A]
time = 0.09, size = 15, normalized size = 0.68 \begin {gather*} \frac {\log {\left (x \right )}}{b} - \frac {\log {\left (\frac {b}{c} + x^{2} \right )}}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/b - log(b/c + x**2)/(2*b)

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Giac [A]
time = 4.68, size = 22, normalized size = 1.00 \begin {gather*} -\frac {\log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b} + \frac {\log \left ({\left | x \right |}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 18, normalized size = 0.82 \begin {gather*} -\frac {\ln \left (c\,x^2+b\right )-2\,\ln \left (x\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^2 + c*x^4),x)

[Out]

-(log(b + c*x^2) - 2*log(x))/(2*b)

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