3.3.31 \(\int \frac {(a+b x^3)^2}{x^9} \, dx\) [231]

Optimal. Leaf size=30 \[ -\frac {a^2}{8 x^8}-\frac {2 a b}{5 x^5}-\frac {b^2}{2 x^2} \]

[Out]

-1/8*a^2/x^8-2/5*a*b/x^5-1/2*b^2/x^2

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \begin {gather*} -\frac {a^2}{8 x^8}-\frac {2 a b}{5 x^5}-\frac {b^2}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2/x^9,x]

[Out]

-1/8*a^2/x^8 - (2*a*b)/(5*x^5) - b^2/(2*x^2)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^2}{x^9} \, dx &=\int \left (\frac {a^2}{x^9}+\frac {2 a b}{x^6}+\frac {b^2}{x^3}\right ) \, dx\\ &=-\frac {a^2}{8 x^8}-\frac {2 a b}{5 x^5}-\frac {b^2}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} -\frac {a^2}{8 x^8}-\frac {2 a b}{5 x^5}-\frac {b^2}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^2/x^9,x]

[Out]

-1/8*a^2/x^8 - (2*a*b)/(5*x^5) - b^2/(2*x^2)

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Maple [A]
time = 0.12, size = 25, normalized size = 0.83

method	result	size
default	\(-\frac {a^{2}}{8 x^{8}}-\frac {2 a b}{5 x^{5}}-\frac {b^{2}}{2 x^{2}}\)	\(25\)
norman	\(\frac {-\frac {1}{2} b^{2} x^{6}-\frac {2}{5} a b \,x^{3}-\frac {1}{8} a^{2}}{x^{8}}\)	\(26\)
risch	\(\frac {-\frac {1}{2} b^{2} x^{6}-\frac {2}{5} a b \,x^{3}-\frac {1}{8} a^{2}}{x^{8}}\)	\(26\)
gosper	\(-\frac {20 b^{2} x^{6}+16 a b \,x^{3}+5 a^{2}}{40 x^{8}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/8*a^2/x^8-2/5*a*b/x^5-1/2*b^2/x^2

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Maxima [A]
time = 0.30, size = 26, normalized size = 0.87 \begin {gather*} -\frac {20 \, b^{2} x^{6} + 16 \, a b x^{3} + 5 \, a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^9,x, algorithm="maxima")

[Out]

-1/40*(20*b^2*x^6 + 16*a*b*x^3 + 5*a^2)/x^8

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Fricas [A]
time = 0.34, size = 26, normalized size = 0.87 \begin {gather*} -\frac {20 \, b^{2} x^{6} + 16 \, a b x^{3} + 5 \, a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^9,x, algorithm="fricas")

[Out]

-1/40*(20*b^2*x^6 + 16*a*b*x^3 + 5*a^2)/x^8

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Sympy [A]
time = 0.07, size = 27, normalized size = 0.90 \begin {gather*} \frac {- 5 a^{2} - 16 a b x^{3} - 20 b^{2} x^{6}}{40 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2/x**9,x)

[Out]

(-5*a**2 - 16*a*b*x**3 - 20*b**2*x**6)/(40*x**8)

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Giac [A]
time = 2.44, size = 26, normalized size = 0.87 \begin {gather*} -\frac {20 \, b^{2} x^{6} + 16 \, a b x^{3} + 5 \, a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^9,x, algorithm="giac")

[Out]

-1/40*(20*b^2*x^6 + 16*a*b*x^3 + 5*a^2)/x^8

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Mupad [B]
time = 0.03, size = 26, normalized size = 0.87 \begin {gather*} -\frac {\frac {a^2}{8}+\frac {2\,a\,b\,x^3}{5}+\frac {b^2\,x^6}{2}}{x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^2/x^9,x)

[Out]

-(a^2/8 + (b^2*x^6)/2 + (2*a*b*x^3)/5)/x^8

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