\(\int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx\) [609]

Optimal result

Integrand size = 27, antiderivative size = 33 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=-x+\frac {4}{3} \sqrt {2+3 x}-\frac {4}{3} \log \left (1+\sqrt {2+3 x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {442, 383, 78} \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=-x+\frac {4}{3} \sqrt {3 x+2}-\frac {4}{3} \log \left (\sqrt {3 x+2}+1\right ) \]

[In]

Int[(1 - Sqrt[2 + 3*x])/(1 + Sqrt[2 + 3*x]),x]

[Out]

-x + (4*Sqrt[2 + 3*x])/3 - (4*Log[1 + Sqrt[2 + 3*x]])/3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, Dis
t[g, Subst[Int[x^(g - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}
, x] && NeQ[b*c - a*d, 0] && FractionQ[n]

Rule 442

Int[((a_.) + (b_.)*(u_)^(n_))^(p_.)*((c_.) + (d_.)*(u_)^(n_))^(q_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1],
Subst[Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x, u], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && LinearQ[u, x] && N
eQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1-\sqrt {x}}{1+\sqrt {x}} \, dx,x,2+3 x\right ) \\ & = \frac {2}{3} \text {Subst}\left (\int \frac {(1-x) x}{1+x} \, dx,x,\sqrt {2+3 x}\right ) \\ & = \frac {2}{3} \text {Subst}\left (\int \left (2-x-\frac {2}{1+x}\right ) \, dx,x,\sqrt {2+3 x}\right ) \\ & = -x+\frac {4}{3} \sqrt {2+3 x}-\frac {4}{3} \log \left (1+\sqrt {2+3 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=\frac {1}{3} \left (-2-3 x+4 \sqrt {2+3 x}-4 \log \left (1+\sqrt {2+3 x}\right )\right ) \]

[In]

Integrate[(1 - Sqrt[2 + 3*x])/(1 + Sqrt[2 + 3*x]),x]

[Out]

(-2 - 3*x + 4*Sqrt[2 + 3*x] - 4*Log[1 + Sqrt[2 + 3*x]])/3

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-x -\frac {2}{3}+\frac {4 \sqrt {3 x +2}}{3}-\frac {4 \ln \left (1+\sqrt {3 x +2}\right )}{3}\)	\(27\)
default	\(-x -\frac {2}{3}+\frac {4 \sqrt {3 x +2}}{3}-\frac {4 \ln \left (1+\sqrt {3 x +2}\right )}{3}\)	\(27\)
trager	\(-x +\frac {4 \sqrt {3 x +2}}{3}-\frac {2 \ln \left (2 \sqrt {3 x +2}+3+3 x \right )}{3}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=-x + \frac {4}{3} \, \sqrt {3 \, x + 2} - \frac {4}{3} \, \log \left (\sqrt {3 \, x + 2} + 1\right ) \]

[In]

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

[Out]

-x + 4/3*sqrt(3*x + 2) - 4/3*log(sqrt(3*x + 2) + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=- x + \frac {4 \sqrt {3 x + 2}}{3} - \frac {4 \log {\left (\sqrt {3 x + 2} + 1 \right )}}{3} \]

[In]

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

[Out]

-x + 4*sqrt(3*x + 2)/3 - 4*log(sqrt(3*x + 2) + 1)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=-x + \frac {4}{3} \, \sqrt {3 \, x + 2} - \frac {4}{3} \, \log \left (\sqrt {3 \, x + 2} + 1\right ) - \frac {2}{3} \]

[In]

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

[Out]

-x + 4/3*sqrt(3*x + 2) - 4/3*log(sqrt(3*x + 2) + 1) - 2/3

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=-x + \frac {4}{3} \, \sqrt {3 \, x + 2} - \frac {4}{3} \, \log \left (\sqrt {3 \, x + 2} + 1\right ) - \frac {2}{3} \]

[In]

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

[Out]

-x + 4/3*sqrt(3*x + 2) - 4/3*log(sqrt(3*x + 2) + 1) - 2/3

Mupad [B] (verification not implemented)

Time = 17.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1-\sqrt {2+3 x}}{1+\sqrt {2+3 x}} \, dx=\frac {4\,\sqrt {3\,x+2}}{3}-\frac {4\,\ln \left (\sqrt {3\,x+2}+1\right )}{3}-x \]

[In]

int(-((3*x + 2)^(1/2) - 1)/((3*x + 2)^(1/2) + 1),x)

[Out]

(4*(3*x + 2)^(1/2))/3 - (4*log((3*x + 2)^(1/2) + 1))/3 - x