\(\int \frac {1}{(1+x) \sqrt [4]{2+2 x+x^2}} \, dx\) [345]

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {1}{(1+x) \sqrt [4]{2+2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2+2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2+2 x+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {708, 272, 65, 304, 209, 212} \[ \int \frac {1}{(1+x) \sqrt [4]{2+2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{(x+1)^2+1}\right )-\text {arctanh}\left (\sqrt [4]{(x+1)^2+1}\right ) \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1+x) \sqrt [4]{2+2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2+2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2+2 x+x^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48

method	result	size
pseudoelliptic	\(\frac {\ln \left (\left (x^{2}+2 x +2\right )^{\frac {1}{4}}-1\right )}{2}-\frac {\ln \left (\left (x^{2}+2 x +2\right )^{\frac {1}{4}}+1\right )}{2}+\arctan \left (\left (x^{2}+2 x +2\right )^{\frac {1}{4}}\right )\)	\(43\)
trager	\(\frac {\ln \left (-\frac {2 \left (x^{2}+2 x +2\right )^{\frac {3}{4}}-2 \sqrt {x^{2}+2 x +2}-x^{2}+2 \left (x^{2}+2 x +2\right )^{\frac {1}{4}}-2 x -3}{\left (1+x \right )^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \left (x^{2}+2 x +2\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+2 x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{2}+2 x +2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (1+x \right )^{2}}\right )}{2}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(1+x) \sqrt [4]{2+2 x+x^2}} \, dx=\arctan \left ({\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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