3.1.47 \(\int \frac {(a+b x^2)^3}{x^4} \, dx\) [47]

Optimal. Leaf size=37 \[ -\frac {a^3}{3 x^3}-\frac {3 a^2 b}{x}+3 a b^2 x+\frac {b^3 x^3}{3} \]

[Out]

-1/3*a^3/x^3-3*a^2*b/x+3*a*b^2*x+1/3*b^3*x^3

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Rubi [A]
time = 0.01, antiderivative size = 37, 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^3}{3 x^3}-\frac {3 a^2 b}{x}+3 a b^2 x+\frac {b^3 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/x^4,x]

[Out]

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

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

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Mathematica [A]
time = 0.00, size = 37, normalized size = 1.00 \begin {gather*} -\frac {a^3}{3 x^3}-\frac {3 a^2 b}{x}+3 a b^2 x+\frac {b^3 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^3/x^4,x]

[Out]

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

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Maple [A]
time = 0.01, size = 34, normalized size = 0.92

method	result	size
default	\(-\frac {a^{3}}{3 x^{3}}-\frac {3 a^{2} b}{x}+3 a \,b^{2} x +\frac {b^{3} x^{3}}{3}\)	\(34\)
gosper	\(-\frac {-b^{3} x^{6}-9 a \,b^{2} x^{4}+9 a^{2} b \,x^{2}+a^{3}}{3 x^{3}}\)	\(36\)
risch	\(\frac {b^{3} x^{3}}{3}+3 a \,b^{2} x +\frac {-3 a^{2} b \,x^{2}-\frac {1}{3} a^{3}}{x^{3}}\)	\(36\)
norman	\(\frac {\frac {1}{3} b^{3} x^{6}+3 a \,b^{2} x^{4}-3 a^{2} b \,x^{2}-\frac {1}{3} a^{3}}{x^{3}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a^3/x^3-3*a^2*b/x+3*a*b^2*x+1/3*b^3*x^3

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Maxima [A]
time = 0.29, size = 34, normalized size = 0.92 \begin {gather*} \frac {1}{3} \, b^{3} x^{3} + 3 \, a b^{2} x - \frac {9 \, a^{2} b x^{2} + a^{3}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^4,x, algorithm="maxima")

[Out]

1/3*b^3*x^3 + 3*a*b^2*x - 1/3*(9*a^2*b*x^2 + a^3)/x^3

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Fricas [A]
time = 0.92, size = 36, normalized size = 0.97 \begin {gather*} \frac {b^{3} x^{6} + 9 \, a b^{2} x^{4} - 9 \, a^{2} b x^{2} - a^{3}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^4,x, algorithm="fricas")

[Out]

1/3*(b^3*x^6 + 9*a*b^2*x^4 - 9*a^2*b*x^2 - a^3)/x^3

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Sympy [A]
time = 0.05, size = 36, normalized size = 0.97 \begin {gather*} 3 a b^{2} x + \frac {b^{3} x^{3}}{3} + \frac {- a^{3} - 9 a^{2} b x^{2}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3/x**4,x)

[Out]

3*a*b**2*x + b**3*x**3/3 + (-a**3 - 9*a**2*b*x**2)/(3*x**3)

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Giac [A]
time = 0.55, size = 34, normalized size = 0.92 \begin {gather*} \frac {1}{3} \, b^{3} x^{3} + 3 \, a b^{2} x - \frac {9 \, a^{2} b x^{2} + a^{3}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^4,x, algorithm="giac")

[Out]

1/3*b^3*x^3 + 3*a*b^2*x - 1/3*(9*a^2*b*x^2 + a^3)/x^3

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Mupad [B]
time = 4.80, size = 36, normalized size = 0.97 \begin {gather*} \frac {b^3\,x^3}{3}-\frac {\frac {a^3}{3}+3\,b\,a^2\,x^2}{x^3}+3\,a\,b^2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/x^4,x)

[Out]

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

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