3.215 \(\int \frac{x^2}{(a+b x)^7} \, dx\)

Optimal. Leaf size=47 \[ -\frac{a^2}{6 b^3 (a+b x)^6}+\frac{2 a}{5 b^3 (a+b x)^5}-\frac{1}{4 b^3 (a+b x)^4} \]

[Out]

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

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Rubi [A]  time = 0.0209128, antiderivative size = 47, 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} \[ -\frac{a^2}{6 b^3 (a+b x)^6}+\frac{2 a}{5 b^3 (a+b x)^5}-\frac{1}{4 b^3 (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x)^7,x]

[Out]

-a^2/(6*b^3*(a + b*x)^6) + (2*a)/(5*b^3*(a + b*x)^5) - 1/(4*b^3*(a + b*x)^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^2}{(a+b x)^7} \, dx &=\int \left (\frac{a^2}{b^2 (a+b x)^7}-\frac{2 a}{b^2 (a+b x)^6}+\frac{1}{b^2 (a+b x)^5}\right ) \, dx\\ &=-\frac{a^2}{6 b^3 (a+b x)^6}+\frac{2 a}{5 b^3 (a+b x)^5}-\frac{1}{4 b^3 (a+b x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0144736, size = 31, normalized size = 0.66 \[ -\frac{a^2+6 a b x+15 b^2 x^2}{60 b^3 (a+b x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x)^7,x]

[Out]

-(a^2 + 6*a*b*x + 15*b^2*x^2)/(60*b^3*(a + b*x)^6)

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Maple [A]  time = 0.004, size = 42, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2}}{6\,{b}^{3} \left ( bx+a \right ) ^{6}}}+{\frac{2\,a}{5\,{b}^{3} \left ( bx+a \right ) ^{5}}}-{\frac{1}{4\,{b}^{3} \left ( bx+a \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x+a)^7,x)

[Out]

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

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Maxima [B]  time = 1.07224, size = 117, normalized size = 2.49 \begin{align*} -\frac{15 \, b^{2} x^{2} + 6 \, a b x + a^{2}}{60 \,{\left (b^{9} x^{6} + 6 \, a b^{8} x^{5} + 15 \, a^{2} b^{7} x^{4} + 20 \, a^{3} b^{6} x^{3} + 15 \, a^{4} b^{5} x^{2} + 6 \, a^{5} b^{4} x + a^{6} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^7,x, algorithm="maxima")

[Out]

-1/60*(15*b^2*x^2 + 6*a*b*x + a^2)/(b^9*x^6 + 6*a*b^8*x^5 + 15*a^2*b^7*x^4 + 20*a^3*b^6*x^3 + 15*a^4*b^5*x^2 +
 6*a^5*b^4*x + a^6*b^3)

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Fricas [B]  time = 1.50643, size = 182, normalized size = 3.87 \begin{align*} -\frac{15 \, b^{2} x^{2} + 6 \, a b x + a^{2}}{60 \,{\left (b^{9} x^{6} + 6 \, a b^{8} x^{5} + 15 \, a^{2} b^{7} x^{4} + 20 \, a^{3} b^{6} x^{3} + 15 \, a^{4} b^{5} x^{2} + 6 \, a^{5} b^{4} x + a^{6} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^7,x, algorithm="fricas")

[Out]

-1/60*(15*b^2*x^2 + 6*a*b*x + a^2)/(b^9*x^6 + 6*a*b^8*x^5 + 15*a^2*b^7*x^4 + 20*a^3*b^6*x^3 + 15*a^4*b^5*x^2 +
 6*a^5*b^4*x + a^6*b^3)

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Sympy [B]  time = 0.791499, size = 92, normalized size = 1.96 \begin{align*} - \frac{a^{2} + 6 a b x + 15 b^{2} x^{2}}{60 a^{6} b^{3} + 360 a^{5} b^{4} x + 900 a^{4} b^{5} x^{2} + 1200 a^{3} b^{6} x^{3} + 900 a^{2} b^{7} x^{4} + 360 a b^{8} x^{5} + 60 b^{9} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x+a)**7,x)

[Out]

-(a**2 + 6*a*b*x + 15*b**2*x**2)/(60*a**6*b**3 + 360*a**5*b**4*x + 900*a**4*b**5*x**2 + 1200*a**3*b**6*x**3 +
900*a**2*b**7*x**4 + 360*a*b**8*x**5 + 60*b**9*x**6)

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Giac [A]  time = 1.15681, size = 39, normalized size = 0.83 \begin{align*} -\frac{15 \, b^{2} x^{2} + 6 \, a b x + a^{2}}{60 \,{\left (b x + a\right )}^{6} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^7,x, algorithm="giac")

[Out]

-1/60*(15*b^2*x^2 + 6*a*b*x + a^2)/((b*x + a)^6*b^3)