3.2.96 \(\int \frac {-1-2 x+2 x^2}{(-1+2 x) \sqrt {x+x^4}} \, dx\)

Optimal. Leaf size=21 \[ \tanh ^{-1}\left (\frac {2 \sqrt {x^4+x}}{2 x^2+1}\right ) \]

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Rubi [F]  time = 0.69, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-2 x+2 x^2}{(-1+2 x) \sqrt {x+x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[1 + x^3]*ArcSinh[x^(3/2)])/(3*Sqrt[x + x^4]) - (x*(1 + x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3]
)*x)^2]*EllipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3])/4])/(2*3^(1/4)*Sqrt[(x*(1
+ x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[x + x^4]) + (3*Sqrt[x]*Sqrt[1 + x^3]*Defer[Subst][Defer[Int][1/((1 - Sqrt
[2]*x)*Sqrt[1 + x^6]), x], x, Sqrt[x]])/(2*Sqrt[x + x^4]) + (3*Sqrt[x]*Sqrt[1 + x^3]*Defer[Subst][Defer[Int][1
/((1 + Sqrt[2]*x)*Sqrt[1 + x^6]), x], x, Sqrt[x]])/(2*Sqrt[x + x^4])

Rubi steps

\begin {align*} \int \frac {-1-2 x+2 x^2}{(-1+2 x) \sqrt {x+x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {-1-2 x+2 x^2}{\sqrt {x} (-1+2 x) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {-1-2 x^2+2 x^4}{\left (-1+2 x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{2 \sqrt {1+x^6}}+\frac {x^2}{\sqrt {1+x^6}}-\frac {3}{2 \left (-1+2 x^2\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}\\ &=-\frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}+\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}-\frac {\left (3 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+2 x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}\\ &=-\frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} F\left (\cos ^{-1}\left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}+\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{3/2}\right )}{3 \sqrt {x+x^4}}-\frac {\left (3 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{2 \left (1-\sqrt {2} x\right ) \sqrt {1+x^6}}-\frac {1}{2 \left (1+\sqrt {2} x\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}\\ &=\frac {2 \sqrt {x} \sqrt {1+x^3} \sinh ^{-1}\left (x^{3/2}\right )}{3 \sqrt {x+x^4}}-\frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} F\left (\cos ^{-1}\left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}+\frac {\left (3 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}}+\frac {\left (3 \sqrt {x} \sqrt {1+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}}\\ \end {align*}

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Mathematica [C]  time = 1.13, size = 250, normalized size = 11.90 \begin {gather*} \frac {-\sqrt {6} \left (1+\sqrt [3]{-1}\right ) \sqrt {\frac {x-\sqrt [3]{-1}}{\left (1+\sqrt [3]{-1}\right ) x}} \sqrt {\frac {(x+1) \left (2 x+i \sqrt {3}-1\right )}{x^2}} x^2 F\left (\sin ^{-1}\left (\sqrt {\frac {(-1)^{2/3} (x+1)}{\left (-1+(-1)^{2/3}\right ) x}}\right )|1+(-1)^{2/3}\right )+\sqrt {6} \left (1+\sqrt [3]{-1}\right ) \sqrt {\frac {x-\sqrt [3]{-1}}{\left (1+\sqrt [3]{-1}\right ) x}} \sqrt {\frac {(x+1) \left (2 x+i \sqrt {3}-1\right )}{x^2}} x^2 \Pi \left (\frac {1}{3} \left (1+\sqrt [3]{-1}\right );\sin ^{-1}\left (\sqrt {\frac {(-1)^{2/3} (x+1)}{\left (-1+(-1)^{2/3}\right ) x}}\right )|1+(-1)^{2/3}\right )+2 \sqrt {x^3+1} \sqrt {x} \sinh ^{-1}\left (x^{3/2}\right )}{3 \sqrt {x^4+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[1 + x^3]*ArcSinh[x^(3/2)] - Sqrt[6]*(1 + (-1)^(1/3))*x^2*Sqrt[(-(-1)^(1/3) + x)/((1 + (-1)^(1/
3))*x)]*Sqrt[((1 + x)*(-1 + I*Sqrt[3] + 2*x))/x^2]*EllipticF[ArcSin[Sqrt[((-1)^(2/3)*(1 + x))/((-1 + (-1)^(2/3
))*x)]], 1 + (-1)^(2/3)] + Sqrt[6]*(1 + (-1)^(1/3))*x^2*Sqrt[(-(-1)^(1/3) + x)/((1 + (-1)^(1/3))*x)]*Sqrt[((1
+ x)*(-1 + I*Sqrt[3] + 2*x))/x^2]*EllipticPi[(1 + (-1)^(1/3))/3, ArcSin[Sqrt[((-1)^(2/3)*(1 + x))/((-1 + (-1)^
(2/3))*x)]], 1 + (-1)^(2/3)])/(3*Sqrt[x + x^4])

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IntegrateAlgebraic [A]  time = 1.21, size = 21, normalized size = 1.00 \begin {gather*} \tanh ^{-1}\left (\frac {2 \sqrt {x^4+x}}{2 x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 25, normalized size = 1.19 \begin {gather*} \log \left (\frac {2 \, x^{2} + 2 \, \sqrt {x^{4} + x} + 1}{2 \, x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 780, normalized size = 37.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-1/2-1/2*I*3^(1/2))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*(-(x-1/2+1/2*I*3^(1/2)
)/(1/2-1/2*I*3^(1/2))/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)/(3/2+1/2*I*3^(1/2)
)/(x*(1+x)*(x-1/2+1/2*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*(-EllipticF(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^
(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2
))+EllipticPi(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),(1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1/2)),
((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)))+(-1/2-1/2*I*3^(1
/2))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*(-(x-1/2+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2
))/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)/(3/2+1/2*I*3^(1/2))/(x*(1+x)*(x-1/2+1
/2*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*EllipticF(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),(
(-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))-(-1/2-1/2*I*3^(1/2
))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*(-(x-1/2+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))
/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)/(3/2+1/2*I*3^(1/2))/(x*(1+x)*(x-1/2+1/2
*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*(EllipticF(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),((
-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+2*EllipticPi(((3/2+
1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),3*(1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1/2)),((-3/2+1/2*I*3^(1/
2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {-2\,x^2+2\,x+1}{\left (2\,x-1\right )\,\sqrt {x^4+x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((2*x**2 - 2*x - 1)/(sqrt(x*(x + 1)*(x**2 - x + 1))*(2*x - 1)), x)

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