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

Optimal. Leaf size=23 \[ \frac {a}{b^2 (a+b x)}+\frac {\log (a+b x)}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 23, 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} \[ \frac {a}{b^2 (a+b x)}+\frac {\log (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*x^2 + b*x^3)^2,x]

[Out]

a/(b^2*(a + b*x)) + 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^5}{\left (a x^2+b x^3\right )^2} \, dx &=\int \frac {x}{(a+b x)^2} \, dx\\ &=\int \left (-\frac {a}{b (a+b x)^2}+\frac {1}{b (a+b x)}\right ) \, dx\\ &=\frac {a}{b^2 (a+b x)}+\frac {\log (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.87 \[ \frac {\frac {a}{a+b x}+\log (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*x^2 + b*x^3)^2,x]

[Out]

(a/(a + b*x) + Log[a + b*x])/b^2

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fricas [A]  time = 0.39, size = 28, normalized size = 1.22 \[ \frac {{\left (b x + a\right )} \log \left (b x + a\right ) + a}{b^{3} x + a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)*log(b*x + a) + a)/(b^3*x + a*b^2)

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giac [A]  time = 0.15, size = 24, normalized size = 1.04 \[ \frac {\log \left ({\left | b x + a \right |}\right )}{b^{2}} + \frac {a}{{\left (b x + a\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 24, normalized size = 1.04 \[ \frac {a}{\left (b x +a \right ) b^{2}}+\frac {\ln \left (b x +a \right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 26, normalized size = 1.13 \[ \frac {a}{b^{3} x + a b^{2}} + \frac {\log \left (b x + a\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(b^3*x + a*b^2) + log(b*x + a)/b^2

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mupad [B]  time = 0.04, size = 23, normalized size = 1.00 \[ \frac {\ln \left (a+b\,x\right )}{b^2}+\frac {a}{b^2\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*x^2 + b*x^3)^2,x)

[Out]

log(a + b*x)/b^2 + a/(b^2*(a + b*x))

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sympy [A]  time = 0.14, size = 20, normalized size = 0.87 \[ \frac {a}{a b^{2} + b^{3} x} + \frac {\log {\left (a + b x \right )}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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