\(\int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx\) [722]

Optimal result

Integrand size = 22, antiderivative size = 56 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {\arctan \left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}} \]

[Out]

1/2*arctan(2^(1/4)*(2*x^2+2*x+1)^(1/4))*2^(1/4)-1/2*arctanh(2^(1/4)*(2*x^2+2*x+1)^(1/4))*2^(1/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {708, 272, 65, 304, 209, 212} \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {\arctan \left (\sqrt [4]{(2 x+1)^2+1}\right )}{2^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{(2 x+1)^2+1}\right )}{2^{3/4}} \]

[In]

Int[1/((1 + 2*x)*(1 + 2*x + 2*x^2)^(1/4)),x]

[Out]

ArcTan[(1 + (1 + 2*x)^2)^(1/4)]/2^(3/4) - ArcTanh[(1 + (1 + 2*x)^2)^(1/4)]/2^(3/4)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {\arctan \left (\sqrt [4]{2+4 x+4 x^2}\right )-\text {arctanh}\left (\sqrt [4]{2+4 x+4 x^2}\right )}{2^{3/4}} \]

[In]

Integrate[1/((1 + 2*x)*(1 + 2*x + 2*x^2)^(1/4)),x]

[Out]

(ArcTan[(2 + 4*x + 4*x^2)^(1/4)] - ArcTanh[(2 + 4*x + 4*x^2)^(1/4)])/2^(3/4)

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{4}} \left (2 \arctan \left (2^{\frac {1}{4}} \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}\right )-\ln \left (\frac {2^{\frac {3}{4}}+2 \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}{-2^{\frac {3}{4}}+2 \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}\right )\right )}{4}\)	\(70\)

[In]

int(1/(2*x+1)/(2*x^2+2*x+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.84 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=-\frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} i \cdot 8^{\frac {3}{4}} \log \left (i \cdot 8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} i \cdot 8^{\frac {3}{4}} \log \left (-i \cdot 8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (-8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) \]

[In]

integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="fricas")

[Out]

-1/16*8^(3/4)*log(8^(1/4) + 2*(2*x^2 + 2*x + 1)^(1/4)) + 1/16*I*8^(3/4)*log(I*8^(1/4) + 2*(2*x^2 + 2*x + 1)^(1
/4)) - 1/16*I*8^(3/4)*log(-I*8^(1/4) + 2*(2*x^2 + 2*x + 1)^(1/4)) + 1/16*8^(3/4)*log(-8^(1/4) + 2*(2*x^2 + 2*x
+ 1)^(1/4))

Sympy [F]

\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int \frac {1}{\left (2 x + 1\right ) \sqrt [4]{2 x^{2} + 2 x + 1}}\, dx \]

[In]

integrate(1/(1+2*x)/(2*x**2+2*x+1)**(1/4),x)

[Out]

Integral(1/((2*x + 1)*(2*x**2 + 2*x + 1)**(1/4)), x)

Maxima [F]

\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}} {\left (2 \, x + 1\right )}} \,d x } \]

[In]

integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((2*x^2 + 2*x + 1)^(1/4)*(2*x + 1)), x)

Giac [F]

\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}} {\left (2 \, x + 1\right )}} \,d x } \]

[In]

integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((2*x^2 + 2*x + 1)^(1/4)*(2*x + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int \frac {1}{\left (2\,x+1\right )\,{\left (2\,x^2+2\,x+1\right )}^{1/4}} \,d x \]

[In]

int(1/((2*x + 1)*(2*x + 2*x^2 + 1)^(1/4)),x)

[Out]

int(1/((2*x + 1)*(2*x + 2*x^2 + 1)^(1/4)), x)