3.2.42 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx\) [142]

Optimal. Leaf size=62 \[ \frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {a \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]

[Out]

b*x*((b*x+a)^2)^(1/2)/(b*x+a)+a*ln(x)*((b*x+a)^2)^(1/2)/(b*x+a)

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Rubi [A]
time = 0.01, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \begin {gather*} \frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {a \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x,x]

[Out]

(b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (a*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x])/(a + b*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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.44 \begin {gather*} \frac {\sqrt {(a+b x)^2} (b x+a \log (x))}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x,x]

[Out]

(Sqrt[(a + b*x)^2]*(b*x + a*Log[x]))/(a + b*x)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.14, size = 20, normalized size = 0.32

method	result	size
default	\(\mathrm {csgn}\left (b x +a \right ) \left (b x +a +a \ln \left (-b x \right )\right )\)	\(20\)
risch	\(\frac {b x \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {a \ln \left (x \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x+a)^2)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

csgn(b*x+a)*(b*x+a+a*ln(-b*x))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (40) = 80\).
time = 0.28, size = 82, normalized size = 1.32 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/x,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*a*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*a*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) +
sqrt(b^2*x^2 + 2*a*b*x + a^2)

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Fricas [A]
time = 1.29, size = 8, normalized size = 0.13 \begin {gather*} b x + a \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/x,x, algorithm="fricas")

[Out]

b*x + a*log(x)

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Sympy [A]
time = 0.02, size = 7, normalized size = 0.11 \begin {gather*} a \log {\left (x \right )} + b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)**2)**(1/2)/x,x)

[Out]

a*log(x) + b*x

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Giac [A]
time = 1.81, size = 21, normalized size = 0.34 \begin {gather*} b x \mathrm {sgn}\left (b x + a\right ) + a \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/x,x, algorithm="giac")

[Out]

b*x*sgn(b*x + a) + a*log(abs(x))*sgn(b*x + a)

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Mupad [B]
time = 0.28, size = 98, normalized size = 1.58 \begin {gather*} \sqrt {a^2+2\,a\,b\,x+b^2\,x^2}-\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )\,\sqrt {a^2}+\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )}{\sqrt {b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)/x,x)

[Out]

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

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