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

Optimal. Leaf size=50 \[ \frac{a^3}{2 b^4 (a+b x)^2}-\frac{3 a^2}{b^4 (a+b x)}-\frac{3 a \log (a+b x)}{b^4}+\frac{x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.0262389, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{a^3}{2 b^4 (a+b x)^2}-\frac{3 a^2}{b^4 (a+b x)}-\frac{3 a \log (a+b x)}{b^4}+\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x)^3,x]

[Out]

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

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^3}{(a+b x)^3} \, dx &=\int \left (\frac{1}{b^3}-\frac{a^3}{b^3 (a+b x)^3}+\frac{3 a^2}{b^3 (a+b x)^2}-\frac{3 a}{b^3 (a+b x)}\right ) \, dx\\ &=\frac{x}{b^3}+\frac{a^3}{2 b^4 (a+b x)^2}-\frac{3 a^2}{b^4 (a+b x)}-\frac{3 a \log (a+b x)}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.0613447, size = 40, normalized size = 0.8 \[ -\frac{\frac{a^2 (5 a+6 b x)}{(a+b x)^2}+6 a \log (a+b x)-2 b x}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x)^3,x]

[Out]

-(-2*b*x + (a^2*(5*a + 6*b*x))/(a + b*x)^2 + 6*a*Log[a + b*x])/(2*b^4)

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Maple [A]  time = 0.007, size = 49, normalized size = 1. \begin{align*}{\frac{x}{{b}^{3}}}+{\frac{{a}^{3}}{2\,{b}^{4} \left ( bx+a \right ) ^{2}}}-3\,{\frac{{a}^{2}}{{b}^{4} \left ( bx+a \right ) }}-3\,{\frac{a\ln \left ( bx+a \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0621, size = 77, normalized size = 1.54 \begin{align*} -\frac{6 \, a^{2} b x + 5 \, a^{3}}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac{x}{b^{3}} - \frac{3 \, a \log \left (b x + a\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*a^2*b*x + 5*a^3)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) + x/b^3 - 3*a*log(b*x + a)/b^4

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Fricas [A]  time = 1.51677, size = 176, normalized size = 3.52 \begin{align*} \frac{2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*x^3 + 4*a*b^2*x^2 - 4*a^2*b*x - 5*a^3 - 6*(a*b^2*x^2 + 2*a^2*b*x + a^3)*log(b*x + a))/(b^6*x^2 + 2*
a*b^5*x + a^2*b^4)

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Sympy [A]  time = 0.600725, size = 56, normalized size = 1.12 \begin{align*} - \frac{3 a \log{\left (a + b x \right )}}{b^{4}} - \frac{5 a^{3} + 6 a^{2} b x}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} + \frac{x}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a*log(a + b*x)/b**4 - (5*a**3 + 6*a**2*b*x)/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + x/b**3

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Giac [A]  time = 1.15452, size = 59, normalized size = 1.18 \begin{align*} \frac{x}{b^{3}} - \frac{3 \, a \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac{6 \, a^{2} b x + 5 \, a^{3}}{2 \,{\left (b x + a\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b^3 - 3*a*log(abs(b*x + a))/b^4 - 1/2*(6*a^2*b*x + 5*a^3)/((b*x + a)^2*b^4)