3.2.37 \(\int \frac {x^3}{a x^2+b x^3} \, dx\)

Optimal. Leaf size=18 \[ \frac {x}{b}-\frac {a \log (a+b x)}{b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 43} \begin {gather*} \frac {x}{b}-\frac {a \log (a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a*x^2 + b*x^3),x]

[Out]

x/b - (a*Log[a + b*x])/b^2

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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^3}{a x^2+b x^3} \, dx &=\int \frac {x}{a+b x} \, dx\\ &=\int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx\\ &=\frac {x}{b}-\frac {a \log (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} \frac {x}{b}-\frac {a \log (a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a*x^2 + b*x^3),x]

[Out]

x/b - (a*Log[a + b*x])/b^2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3}{a x^2+b x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^3/(a*x^2 + b*x^3),x]

[Out]

IntegrateAlgebraic[x^3/(a*x^2 + b*x^3), x]

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fricas [A]  time = 0.38, size = 17, normalized size = 0.94 \begin {gather*} \frac {b x - a \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^3+a*x^2),x, algorithm="fricas")

[Out]

(b*x - a*log(b*x + a))/b^2

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giac [A]  time = 0.15, size = 19, normalized size = 1.06 \begin {gather*} \frac {x}{b} - \frac {a \log \left ({\left | b x + a \right |}\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^3+a*x^2),x, algorithm="giac")

[Out]

x/b - a*log(abs(b*x + a))/b^2

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maple [A]  time = 0.05, size = 19, normalized size = 1.06 \begin {gather*} -\frac {a \ln \left (b x +a \right )}{b^{2}}+\frac {x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^3+a*x^2),x)

[Out]

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

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maxima [A]  time = 1.30, size = 18, normalized size = 1.00 \begin {gather*} \frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^3+a*x^2),x, algorithm="maxima")

[Out]

x/b - a*log(b*x + a)/b^2

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mupad [B]  time = 0.04, size = 18, normalized size = 1.00 \begin {gather*} -\frac {a\,\ln \left (a+b\,x\right )-b\,x}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^3),x)

[Out]

-(a*log(a + b*x) - b*x)/b^2

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sympy [A]  time = 0.11, size = 14, normalized size = 0.78 \begin {gather*} - \frac {a \log {\left (a + b x \right )}}{b^{2}} + \frac {x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**3+a*x**2),x)

[Out]

-a*log(a + b*x)/b**2 + x/b

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