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

Optimal. Leaf size=29 \[ \tan ^{-1}\left (\sqrt [4]{x^2+2 x+2}\right )-\tanh ^{-1}\left (\sqrt [4]{x^2+2 x+2}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {694, 266, 63, 298, 203, 206} \begin {gather*} \tan ^{-1}\left (\sqrt [4]{(x+1)^2+1}\right )-\tanh ^{-1}\left (\sqrt [4]{(x+1)^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 63

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 203

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

Rule 206

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

Rule 266

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 298

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 694

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.93 \begin {gather*} \tan ^{-1}\left (\sqrt [4]{(x+1)^2+1}\right )-\tanh ^{-1}\left (\sqrt [4]{(x+1)^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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)]

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IntegrateAlgebraic [A]  time = 0.05, size = 29, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\sqrt [4]{2+2 x+x^2}\right )-\tanh ^{-1}\left (\sqrt [4]{2+2 x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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)]

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fricas [A]  time = 0.45, size = 42, normalized size = 1.45 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 1.15, size = 143, normalized size = 4.93

method	result	size
trager	\(-\frac {\ln \left (-\frac {2 \left (x^{2}+2 x +2\right )^{\frac {3}{4}}+2 \sqrt {x^{2}+2 x +2}+x^{2}+2 \left (x^{2}+2 x +2\right )^{\frac {1}{4}}+2 x +3}{\left (1+x \right )^{2}}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+2 x +2}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{2}+2 x +2\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x -3 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{2}+2 x +2\right )^{\frac {1}{4}}}{\left (1+x \right )^{2}}\right )}{2}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(-(2*(x^2+2*x+2)^(3/4)+2*(x^2+2*x+2)^(1/2)+x^2+2*(x^2+2*x+2)^(1/4)+2*x+3)/(1+x)^2)+1/2*RootOf(_Z^2+1)*l
n(-(2*RootOf(_Z^2+1)*(x^2+2*x+2)^(1/2)-RootOf(_Z^2+1)*x^2+2*(x^2+2*x+2)^(3/4)-2*RootOf(_Z^2+1)*x-3*RootOf(_Z^2
+1)-2*(x^2+2*x+2)^(1/4))/(1+x)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{\left (x+1\right )\,{\left (x^2+2\,x+2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x + 1\right ) \sqrt [4]{x^{2} + 2 x + 2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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