3.3.24 \(\int \frac {x^6}{(a x^2+b x^3)^2} \, dx\) [224]

Optimal. Leaf size=33 \[ \frac {x}{b^2}-\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, 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, 45} \begin {gather*} -\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3}+\frac {x}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(a*x^2 + b*x^3)^2,x]

[Out]

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

Rule 45

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 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^6}{\left (a x^2+b x^3\right )^2} \, dx &=\int \frac {x^2}{(a+b x)^2} \, dx\\ &=\int \left (\frac {1}{b^2}+\frac {a^2}{b^2 (a+b x)^2}-\frac {2 a}{b^2 (a+b x)}\right ) \, dx\\ &=\frac {x}{b^2}-\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.88 \begin {gather*} \frac {b x-\frac {a^2}{a+b x}-2 a \log (a+b x)}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a*x^2 + b*x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.36, size = 34, normalized size = 1.03

method	result	size
default	\(\frac {x}{b^{2}}-\frac {a^{2}}{b^{3} \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(34\)
risch	\(\frac {x}{b^{2}}-\frac {a^{2}}{b^{3} \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(34\)
norman	\(\frac {\frac {x^{5}}{b}-\frac {2 a^{2} x^{3}}{b^{3}}}{x^{3} \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 36, normalized size = 1.09 \begin {gather*} -\frac {a^{2}}{b^{4} x + a b^{3}} + \frac {x}{b^{2}} - \frac {2 \, a \log \left (b x + a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a*x^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.47, size = 47, normalized size = 1.42 \begin {gather*} \frac {b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.07, size = 31, normalized size = 0.94 \begin {gather*} - \frac {a^{2}}{a b^{3} + b^{4} x} - \frac {2 a \log {\left (a + b x \right )}}{b^{3}} + \frac {x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**3+a*x**2)**2,x)

[Out]

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

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Giac [A]
time = 1.67, size = 34, normalized size = 1.03 \begin {gather*} \frac {x}{b^{2}} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {a^{2}}{{\left (b x + a\right )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.04, size = 36, normalized size = 1.09 \begin {gather*} \frac {x}{b^2}-\frac {a^2}{x\,b^4+a\,b^3}-\frac {2\,a\,\ln \left (a+b\,x\right )}{b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a*x^2 + b*x^3)^2,x)

[Out]

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

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