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

Optimal. Leaf size=56 \[ \frac {\tan ^{-1}\left (\sqrt [4]{2} \sqrt [4]{2 x^2+2 x+1}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\sqrt [4]{2} \sqrt [4]{2 x^2+2 x+1}\right )}{2^{3/4}} \]

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Rubi [A]  time = 0.04, antiderivative size = 42, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {694, 266, 63, 298, 203, 206} \begin {gather*} \frac {\tan ^{-1}\left (\sqrt [4]{(2 x+1)^2+1}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\sqrt [4]{(2 x+1)^2+1}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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 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+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{\frac {1}{2}+\frac {x^2}{2}}} \, dx,x,1+2 x\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{\frac {1}{2}+\frac {x}{2}} x} \, dx,x,(1+2 x)^2\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^2}{-1+2 x^4} \, dx,x,\sqrt [4]{1+2 x+2 x^2}\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+2 x+2 x^2}\right )}{\sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+2 x+2 x^2}\right )}{\sqrt {2}}\\ &=\frac {\tan ^{-1}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.70 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt [4]{4 x^2+4 x+2}\right )-\tanh ^{-1}\left (\sqrt [4]{4 x^2+4 x+2}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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)

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IntegrateAlgebraic [A]  time = 0.11, size = 56, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.64, size = 102, normalized size = 1.82 \begin {gather*} -\frac {1}{4} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} + 4 \, \sqrt {2 \, x^{2} + 2 \, x + 1}} - \frac {1}{4} \cdot 8^{\frac {3}{4}} {\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 ) + \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (1+2 x \right ) \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\left (2\,x+1\right )\,{\left (2\,x^2+2\,x+1\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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