3.29.10 \(\int \frac {1}{\sqrt {(3+5 x)^2}} \, dx\) [2810]

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

[Out]

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

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Rubi [A]
time = 0.00, 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 = {253, 15, 29} \begin {gather*} \frac {(5 x+3) \log (5 x+3)}{5 \sqrt {(5 x+3)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(3 + 5*x)^2],x]

[Out]

((3 + 5*x)*Log[3 + 5*x])/(5*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 253

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

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Mathematica [A]
time = 0.01, size = 26, normalized size = 1.00 \begin {gather*} \frac {(3+5 x) \log (3+5 x)}{5 \sqrt {(3+5 x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(3 + 5*x)^2],x]

[Out]

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

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Maple [A]
time = 0.22, size = 23, normalized size = 0.88

method	result	size
default	\(\frac {\left (3+5 x \right ) \ln \left (3+5 x \right )}{5 \sqrt {\left (3+5 x \right )^{2}}}\)	\(23\)
risch	\(\frac {\sqrt {\left (3+5 x \right )^{2}}\, \ln \left (3+5 x \right )}{15+25 x}\)	\(25\)
meijerg	\(\frac {3 \ln \left (1+\frac {5 x}{3}\right )}{5 \sqrt {\left (3+5 x \right )^{2}}}+\frac {x \ln \left (1+\frac {5 x}{3}\right )}{\sqrt {\left (3+5 x \right )^{2}}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((3+5*x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 6, normalized size = 0.23 \begin {gather*} \frac {1}{5} \, \log \left (x + \frac {3}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((3+5*x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/5*log(x + 3/5)

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Fricas [A]
time = 0.44, size = 8, normalized size = 0.31 \begin {gather*} \frac {1}{5} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((3+5*x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*log(5*x + 3)

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Sympy [A]
time = 0.01, size = 7, normalized size = 0.27 \begin {gather*} \frac {\log {\left (5 x + 3 \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((3+5*x)**2)**(1/2),x)

[Out]

log(5*x + 3)/5

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Giac [A]
time = 1.39, size = 15, normalized size = 0.58 \begin {gather*} \frac {1}{5} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \mathrm {sgn}\left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((3+5*x)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*log(abs(5*x + 3))*sgn(5*x + 3)

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Mupad [B]
time = 1.38, size = 14, normalized size = 0.54 \begin {gather*} \frac {\ln \left (5\,x+3\right )\,\mathrm {sign}\left (5\,x+3\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((5*x + 3)^2)^(1/2),x)

[Out]

(log(5*x + 3)*sign(5*x + 3))/5

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