3.1.57 \(\int \frac {(a+b x)^2}{x^2} \, dx\) [57]

Optimal. Leaf size=20 \[ -\frac {a^2}{x}+b^2 x+2 a b \log (x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {45} \begin {gather*} -\frac {a^2}{x}+2 a b \log (x)+b^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/x^2,x]

[Out]

-(a^2/x) + b^2*x + 2*a*b*Log[x]

Rule 45

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

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Mathematica [A]
time = 0.00, size = 20, normalized size = 1.00 \begin {gather*} -\frac {a^2}{x}+b^2 x+2 a b \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/x^2,x]

[Out]

-(a^2/x) + b^2*x + 2*a*b*Log[x]

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Mathics [A]
time = 1.73, size = 22, normalized size = 1.10 \begin {gather*} \frac {-a^2+b x \left (2 a \text {Log}\left [x\right ]+b x\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(a + b*x)^2/x^2,x]')

[Out]

(-a ^ 2 + b x (2 a Log[x] + b x)) / x

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Maple [A]
time = 0.07, size = 21, normalized size = 1.05

method	result	size
default	\(-\frac {a^{2}}{x}+b^{2} x +2 a b \ln \left (x \right )\)	\(21\)
risch	\(-\frac {a^{2}}{x}+b^{2} x +2 a b \ln \left (x \right )\)	\(21\)
norman	\(\frac {x^{2} b^{2}-a^{2}}{x}+2 a b \ln \left (x \right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 20, normalized size = 1.00 \begin {gather*} b^{2} x + 2 \, a b \log \left (x\right ) - \frac {a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + 2*a*b*log(x) - a^2/x

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Fricas [A]
time = 0.31, size = 24, normalized size = 1.20 \begin {gather*} \frac {b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*x^2 + 2*a*b*x*log(x) - a^2)/x

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Sympy [A]
time = 0.06, size = 17, normalized size = 0.85 \begin {gather*} - \frac {a^{2}}{x} + 2 a b \log {\left (x \right )} + b^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2/x + 2*a*b*log(x) + b**2*x

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Giac [A]
time = 0.00, size = 20, normalized size = 1.00 \begin {gather*} x b^{2}-\frac {a^{2}}{x}+2 b a \ln \left |x\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + 2*a*b*log(abs(x)) - a^2/x

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Mupad [B]
time = 0.07, size = 20, normalized size = 1.00 \begin {gather*} b^2\,x-\frac {a^2}{x}+2\,a\,b\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/x^2,x)

[Out]

b^2*x - a^2/x + 2*a*b*log(x)

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