3.3.25 \(\int \frac {(a+b x^3)^2}{x^3} \, dx\) [225]

Optimal. Leaf size=26 \[ -\frac {a^2}{2 x^2}+2 a b x+\frac {b^2 x^4}{4} \]

[Out]

-1/2*a^2/x^2+2*a*b*x+1/4*b^2*x^4

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Rubi [A]
time = 0.01, antiderivative size = 26, 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}{2 x^2}+2 a b x+\frac {b^2 x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2/x^3,x]

[Out]

-1/2*a^2/x^2 + 2*a*b*x + (b^2*x^4)/4

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

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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} -\frac {a^2}{2 x^2}+2 a b x+\frac {b^2 x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^2/x^3,x]

[Out]

-1/2*a^2/x^2 + 2*a*b*x + (b^2*x^4)/4

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Maple [A]
time = 0.11, size = 23, normalized size = 0.88

method	result	size
default	\(-\frac {a^{2}}{2 x^{2}}+2 a b x +\frac {b^{2} x^{4}}{4}\)	\(23\)
risch	\(-\frac {a^{2}}{2 x^{2}}+2 a b x +\frac {b^{2} x^{4}}{4}\)	\(23\)
norman	\(\frac {\frac {1}{4} b^{2} x^{6}+2 a b \,x^{3}-\frac {1}{2} a^{2}}{x^{2}}\)	\(26\)
gosper	\(-\frac {-b^{2} x^{6}-8 a b \,x^{3}+2 a^{2}}{4 x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2/x^2+2*a*b*x+1/4*b^2*x^4

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Maxima [A]
time = 0.30, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} + 2 \, a b x - \frac {a^{2}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2*a*b*x - 1/2*a^2/x^2

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Fricas [A]
time = 0.33, size = 25, normalized size = 0.96 \begin {gather*} \frac {b^{2} x^{6} + 8 \, a b x^{3} - 2 \, a^{2}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^6 + 8*a*b*x^3 - 2*a^2)/x^2

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Sympy [A]
time = 0.03, size = 22, normalized size = 0.85 \begin {gather*} - \frac {a^{2}}{2 x^{2}} + 2 a b x + \frac {b^{2} x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.71, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} + 2 \, a b x - \frac {a^{2}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2*a*b*x - 1/2*a^2/x^2

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Mupad [B]
time = 0.03, size = 22, normalized size = 0.85 \begin {gather*} \frac {b^2\,x^4}{4}-\frac {a^2}{2\,x^2}+2\,a\,b\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^2/x^3,x)

[Out]

(b^2*x^4)/4 - a^2/(2*x^2) + 2*a*b*x

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