3.3.11 \(\int \frac {x}{a x^3+b x^4} \, dx\)

Optimal. Leaf size=28 \[ -\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2}-\frac {1}{a x} \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1584, 44} \begin {gather*} -\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2}-\frac {1}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a*x^3 + b*x^4),x]

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 1584

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}{a x^3+b x^4} \, dx &=\int \frac {1}{x^2 (a+b x)} \, dx\\ &=\int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac {1}{a x}-\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \begin {gather*} -\frac {b \log (x)}{a^2}+\frac {b \log (a+b x)}{a^2}-\frac {1}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x^3 + b*x^4),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{a x^3+b x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/(a*x^3 + b*x^4),x]

[Out]

IntegrateAlgebraic[x/(a*x^3 + b*x^4), x]

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fricas [A]  time = 0.38, size = 26, normalized size = 0.93 \begin {gather*} \frac {b x \log \left (b x + a\right ) - b x \log \relax (x) - a}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 30, normalized size = 1.07 \begin {gather*} \frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 29, normalized size = 1.04 \begin {gather*} -\frac {b \ln \relax (x )}{a^{2}}+\frac {b \ln \left (b x +a \right )}{a^{2}}-\frac {1}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^4+a*x^3),x)

[Out]

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

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maxima [A]  time = 1.36, size = 28, normalized size = 1.00 \begin {gather*} \frac {b \log \left (b x + a\right )}{a^{2}} - \frac {b \log \relax (x)}{a^{2}} - \frac {1}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^4+a*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.21, size = 25, normalized size = 0.89 \begin {gather*} \frac {2\,b\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^2}-\frac {1}{a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x^3 + b*x^4),x)

[Out]

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

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sympy [A]  time = 0.21, size = 19, normalized size = 0.68 \begin {gather*} - \frac {1}{a x} + \frac {b \left (- \log {\relax (x )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**4+a*x**3),x)

[Out]

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

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