3.214 \(\int \frac{x^5}{a x^2+b x^3} \, dx\)

Optimal. Leaf size=44 \[ \frac{a^2 x}{b^3}-\frac{a^3 \log (a+b x)}{b^4}-\frac{a x^2}{2 b^2}+\frac{x^3}{3 b} \]

[Out]

(a^2*x)/b^3 - (a*x^2)/(2*b^2) + x^3/(3*b) - (a^3*Log[a + b*x])/b^4

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Rubi [A]  time = 0.0270089, antiderivative size = 44, normalized size of antiderivative = 1., 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} \[ \frac{a^2 x}{b^3}-\frac{a^3 \log (a+b x)}{b^4}-\frac{a x^2}{2 b^2}+\frac{x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*x^2 + b*x^3),x]

[Out]

(a^2*x)/b^3 - (a*x^2)/(2*b^2) + x^3/(3*b) - (a^3*Log[a + b*x])/b^4

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]

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])

Rubi steps

\begin{align*} \int \frac{x^5}{a x^2+b x^3} \, dx &=\int \frac{x^3}{a+b x} \, dx\\ &=\int \left (\frac{a^2}{b^3}-\frac{a x}{b^2}+\frac{x^2}{b}-\frac{a^3}{b^3 (a+b x)}\right ) \, dx\\ &=\frac{a^2 x}{b^3}-\frac{a x^2}{2 b^2}+\frac{x^3}{3 b}-\frac{a^3 \log (a+b x)}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.003585, size = 44, normalized size = 1. \[ \frac{a^2 x}{b^3}-\frac{a^3 \log (a+b x)}{b^4}-\frac{a x^2}{2 b^2}+\frac{x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*x^2 + b*x^3),x]

[Out]

(a^2*x)/b^3 - (a*x^2)/(2*b^2) + x^3/(3*b) - (a^3*Log[a + b*x])/b^4

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Maple [A]  time = 0.002, size = 41, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}x}{{b}^{3}}}-{\frac{a{x}^{2}}{2\,{b}^{2}}}+{\frac{{x}^{3}}{3\,b}}-{\frac{{a}^{3}\ln \left ( bx+a \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a*x^2),x)

[Out]

a^2*x/b^3-1/2*a*x^2/b^2+1/3*x^3/b-a^3*ln(b*x+a)/b^4

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Maxima [A]  time = 0.971426, size = 57, normalized size = 1.3 \begin{align*} -\frac{a^{3} \log \left (b x + a\right )}{b^{4}} + \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a*x^2),x, algorithm="maxima")

[Out]

-a^3*log(b*x + a)/b^4 + 1/6*(2*b^2*x^3 - 3*a*b*x^2 + 6*a^2*x)/b^3

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Fricas [A]  time = 0.872357, size = 92, normalized size = 2.09 \begin{align*} \frac{2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a*x^2),x, algorithm="fricas")

[Out]

1/6*(2*b^3*x^3 - 3*a*b^2*x^2 + 6*a^2*b*x - 6*a^3*log(b*x + a))/b^4

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Sympy [A]  time = 0.486927, size = 37, normalized size = 0.84 \begin{align*} - \frac{a^{3} \log{\left (a + b x \right )}}{b^{4}} + \frac{a^{2} x}{b^{3}} - \frac{a x^{2}}{2 b^{2}} + \frac{x^{3}}{3 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a*x**2),x)

[Out]

-a**3*log(a + b*x)/b**4 + a**2*x/b**3 - a*x**2/(2*b**2) + x**3/(3*b)

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Giac [A]  time = 1.12933, size = 58, normalized size = 1.32 \begin{align*} -\frac{a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a*x^2),x, algorithm="giac")

[Out]

-a^3*log(abs(b*x + a))/b^4 + 1/6*(2*b^2*x^3 - 3*a*b*x^2 + 6*a^2*x)/b^3