3.4.44 \(\int (\frac {a}{x}+b x)^2 \, dx\) [344]

Optimal. Leaf size=24 \[ -\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1607, 276} \begin {gather*} -\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a/x + b*x)^2,x]

[Out]

-(a^2/x) + 2*a*b*x + (b^2*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]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 24, normalized size = 1.00 \begin {gather*} -\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a/x + b*x)^2,x]

[Out]

-(a^2/x) + 2*a*b*x + (b^2*x^3)/3

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Maple [A]
time = 0.03, size = 23, normalized size = 0.96

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} + 2 \, a b x - \frac {a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3 + 2*a*b*x - a^2/x

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Fricas [A]
time = 1.45, size = 25, normalized size = 1.04 \begin {gather*} \frac {b^{2} x^{4} + 6 \, a b x^{2} - 3 \, a^{2}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^2*x^4 + 6*a*b*x^2 - 3*a^2)/x

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Sympy [A]
time = 0.03, size = 19, normalized size = 0.79 \begin {gather*} - \frac {a^{2}}{x} + 2 a b x + \frac {b^{2} x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2/x + 2*a*b*x + b**2*x**3/3

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Giac [A]
time = 0.66, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} + 2 \, a b x - \frac {a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3 + 2*a*b*x - a^2/x

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Mupad [B]
time = 0.04, size = 22, normalized size = 0.92 \begin {gather*} \frac {b^2\,x^3}{3}-\frac {a^2}{x}+2\,a\,b\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + a/x)^2,x)

[Out]

(b^2*x^3)/3 - a^2/x + 2*a*b*x

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