3.2811 \(\int \frac {1}{\sqrt {(6+10 x)^2}} \, dx\)

Optimal. Leaf size=26 \[ \frac {(5 x+3) \log (5 x+3)}{10 \sqrt {(5 x+3)^2}} \]

[Out]

1/10*(3+5*x)*ln(3+5*x)/((3+5*x)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {247, 15, 29} \[ \frac {(5 x+3) \log (10 x+6)}{10 \sqrt {(5 x+3)^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(6 + 10*x)^2],x]

[Out]

((3 + 5*x)*Log[6 + 10*x])/(10*Sqrt[(3 + 5*x)^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {(6+10 x)^2}} \, dx &=\frac {1}{10} \operatorname {Subst}\left (\int \frac {1}{\sqrt {x^2}} \, dx,x,6+10 x\right )\\ &=\frac {(6+10 x) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,6+10 x\right )}{10 \sqrt {(6+10 x)^2}}\\ &=\frac {(3+5 x) \log (3+5 x)}{10 \sqrt {(3+5 x)^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.00 \[ \frac {(10 x+6) \log (10 x+6)}{10 \sqrt {(10 x+6)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(6 + 10*x)^2],x]

[Out]

((6 + 10*x)*Log[6 + 10*x])/(10*Sqrt[(6 + 10*x)^2])

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fricas [A]  time = 0.64, size = 8, normalized size = 0.31 \[ \frac {1}{10} \, \log \left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((6+10*x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/10*log(5*x + 3)

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giac [A]  time = 0.17, size = 15, normalized size = 0.58 \[ \frac {1}{10} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \mathrm {sgn}\left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((6+10*x)^2)^(1/2),x, algorithm="giac")

[Out]

1/10*log(abs(5*x + 3))*sgn(5*x + 3)

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maple [A]  time = 0.00, size = 26, normalized size = 1.00 \[ \frac {\left (5 x +3\right ) \sqrt {4}\, \ln \left (5 x +3\right )}{20 \sqrt {\left (5 x +3\right )^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((6+10*x)^2)^(1/2),x)

[Out]

1/20/((5*x+3)^2)^(1/2)*(5*x+3)*4^(1/2)*ln(5*x+3)

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maxima [A]  time = 1.13, size = 6, normalized size = 0.23 \[ \frac {1}{10} \, \log \left (x + \frac {3}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((6+10*x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/10*log(x + 3/5)

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mupad [B]  time = 1.26, size = 14, normalized size = 0.54 \[ \frac {\ln \left (10\,x+6\right )\,\mathrm {sign}\left (10\,x+6\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((10*x + 6)^2)^(1/2),x)

[Out]

(log(10*x + 6)*sign(10*x + 6))/10

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sympy [A]  time = 0.08, size = 7, normalized size = 0.27 \[ \frac {\log {\left (10 x + 6 \right )}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((6+10*x)**2)**(1/2),x)

[Out]

log(10*x + 6)/10

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