3.709 \(\int \frac{1}{2 a+2 b+x^4} \, dx\)

Optimal. Leaf size=79 \[ -\frac{\tan ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2\ 2^{3/4} (-a-b)^{3/4}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2\ 2^{3/4} (-a-b)^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0604985, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {212, 206, 203} \[ -\frac{\tan ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2\ 2^{3/4} (-a-b)^{3/4}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{-a-b}}\right )}{2\ 2^{3/4} (-a-b)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(2*a + 2*b + x^4)^(-1),x]

[Out]

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

Rule 212

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

Rubi steps

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

Mathematica [A]  time = 0.0629929, size = 128, normalized size = 1.62 \[ \frac{-\log \left (-2 \sqrt [4]{2} x \sqrt [4]{a+b}+2 \sqrt{a+b}+\sqrt{2} x^2\right )+\log \left (2 \sqrt [4]{2} x \sqrt [4]{a+b}+2 \sqrt{a+b}+\sqrt{2} x^2\right )-2 \tan ^{-1}\left (1-\frac{\sqrt [4]{2} x}{\sqrt [4]{a+b}}\right )+2 \tan ^{-1}\left (\frac{\sqrt [4]{2} x}{\sqrt [4]{a+b}}+1\right )}{8 \sqrt [4]{2} (a+b)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2*a + 2*b + x^4)^(-1),x]

[Out]

(-2*ArcTan[1 - (2^(1/4)*x)/(a + b)^(1/4)] + 2*ArcTan[1 + (2^(1/4)*x)/(a + b)^(1/4)] - Log[2*Sqrt[a + b] - 2*2^
(1/4)*(a + b)^(1/4)*x + Sqrt[2]*x^2] + Log[2*Sqrt[a + b] + 2*2^(1/4)*(a + b)^(1/4)*x + Sqrt[2]*x^2])/(8*2^(1/4
)*(a + b)^(3/4))

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Maple [B]  time = 0.006, size = 137, normalized size = 1.7 \begin{align*}{\frac{\sqrt{2}}{8}\ln \left ({ \left ({x}^{2}+\sqrt [4]{2\,a+2\,b}x\sqrt{2}+\sqrt{2\,a+2\,b} \right ) \left ({x}^{2}-\sqrt [4]{2\,a+2\,b}x\sqrt{2}+\sqrt{2\,a+2\,b} \right ) ^{-1}} \right ) \left ( 2\,a+2\,b \right ) ^{-{\frac{3}{4}}}}+{\frac{\sqrt{2}}{4}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{2\,a+2\,b}}}}+1 \right ) \left ( 2\,a+2\,b \right ) ^{-{\frac{3}{4}}}}+{\frac{\sqrt{2}}{4}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{2\,a+2\,b}}}}-1 \right ) \left ( 2\,a+2\,b \right ) ^{-{\frac{3}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/(2*a+2*b)^(3/4)*2^(1/2)*ln((x^2+(2*a+2*b)^(1/4)*x*2^(1/2)+(2*a+2*b)^(1/2))/(x^2-(2*a+2*b)^(1/4)*x*2^(1/2)+
(2*a+2*b)^(1/2)))+1/4/(2*a+2*b)^(3/4)*2^(1/2)*arctan(2^(1/2)/(2*a+2*b)^(1/4)*x+1)+1/4/(2*a+2*b)^(3/4)*2^(1/2)*
arctan(2^(1/2)/(2*a+2*b)^(1/4)*x-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82359, size = 782, normalized size = 9.9 \begin{align*} \left (\frac{1}{8}\right )^{\frac{1}{4}} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{1}{4}} \arctan \left (-4 \, \left (\frac{1}{8}\right )^{\frac{3}{4}}{\left (a^{2} + 2 \, a b + b^{2}\right )} x \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{3}{4}} + 4 \, \left (\frac{1}{8}\right )^{\frac{3}{4}}{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{x^{2} + 2 \, \sqrt{\frac{1}{2}}{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{3}{4}}\right ) + \frac{1}{4} \, \left (\frac{1}{8}\right )^{\frac{1}{4}} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{1}{4}} \log \left (2 \, \left (\frac{1}{8}\right )^{\frac{1}{4}}{\left (a + b\right )} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{1}{4}} + x\right ) - \frac{1}{4} \, \left (\frac{1}{8}\right )^{\frac{1}{4}} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{1}{4}} \log \left (-2 \, \left (\frac{1}{8}\right )^{\frac{1}{4}}{\left (a + b\right )} \left (-\frac{1}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}\right )^{\frac{1}{4}} + x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.271048, size = 42, normalized size = 0.53 \begin{align*} \operatorname{RootSum}{\left (t^{4} \left (2048 a^{3} + 6144 a^{2} b + 6144 a b^{2} + 2048 b^{3}\right ) + 1, \left ( t \mapsto t \log{\left (8 t a + 8 t b + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_t**4*(2048*a**3 + 6144*a**2*b + 6144*a*b**2 + 2048*b**3) + 1, Lambda(_t, _t*log(8*_t*a + 8*_t*b + x))
)

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Giac [B]  time = 1.14887, size = 296, normalized size = 3.75 \begin{align*} \frac{{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2}{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}}}\right )}{4 \,{\left (\sqrt{2} a + \sqrt{2} b\right )}} + \frac{{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2}{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}}}\right )}{4 \,{\left (\sqrt{2} a + \sqrt{2} b\right )}} + \frac{{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} \log \left (x^{2} + \sqrt{2}{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} x + \sqrt{2 \, a + 2 \, b}\right )}{8 \,{\left (\sqrt{2} a + \sqrt{2} b\right )}} - \frac{{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} \log \left (x^{2} - \sqrt{2}{\left (2 \, a + 2 \, b\right )}^{\frac{1}{4}} x + \sqrt{2 \, a + 2 \, b}\right )}{8 \,{\left (\sqrt{2} a + \sqrt{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a + 2*b)^(1/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(sqrt(2)*a + sqr
t(2)*b) + 1/4*(2*a + 2*b)^(1/4)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(2*a + 2*b)^(1/4))/(2*a + 2*b)^(1/4))/(sqrt(
2)*a + sqrt(2)*b) + 1/8*(2*a + 2*b)^(1/4)*log(x^2 + sqrt(2)*(2*a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(sqrt(2)*a
+ sqrt(2)*b) - 1/8*(2*a + 2*b)^(1/4)*log(x^2 - sqrt(2)*(2*a + 2*b)^(1/4)*x + sqrt(2*a + 2*b))/(sqrt(2)*a + sqr
t(2)*b)