3.3.27 \(\int \frac {x^3}{(a x^2+b x^3)^2} \, dx\) [227]

Optimal. Leaf size=29 \[ \frac {1}{a (a+b x)}+\frac {\log (x)}{a^2}-\frac {\log (a+b x)}{a^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 29, 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 = {1598, 46} \begin {gather*} -\frac {\log (a+b x)}{a^2}+\frac {\log (x)}{a^2}+\frac {1}{a (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a*x^2 + b*x^3)^2,x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

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

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Mathematica [A]
time = 0.01, size = 24, normalized size = 0.83 \begin {gather*} \frac {\frac {a}{a+b x}+\log (x)-\log (a+b x)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a*x^2 + b*x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.36, size = 30, normalized size = 1.03

method	result	size
default	\(\frac {1}{a \left (b x +a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (b x +a \right )}{a^{2}}\)	\(30\)
risch	\(\frac {1}{a \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a^{2}}+\frac {\ln \left (-x \right )}{a^{2}}\)	\(32\)
norman	\(-\frac {b x}{a^{2} \left (b x +a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (b x +a \right )}{a^{2}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 28, normalized size = 0.97 \begin {gather*} \frac {1}{a b x + a^{2}} - \frac {\log \left (b x + a\right )}{a^{2}} + \frac {\log \left (x\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.08, size = 39, normalized size = 1.34 \begin {gather*} -\frac {{\left (b x + a\right )} \log \left (b x + a\right ) - {\left (b x + a\right )} \log \left (x\right ) - a}{a^{2} b x + a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.09, size = 22, normalized size = 0.76 \begin {gather*} \frac {1}{a^{2} + a b x} + \frac {\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.28, size = 31, normalized size = 1.07 \begin {gather*} -\frac {\log \left ({\left | b x + a \right |}\right )}{a^{2}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2}} + \frac {1}{{\left (b x + a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 26, normalized size = 0.90 \begin {gather*} \frac {1}{a^2+b\,x\,a}-\frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^3)^2,x)

[Out]

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

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