3.7.2 \(\int \frac {(-1+6 x^4) \sqrt {x+2 x^5}}{(1+2 x^4) (1-x^2+4 x^4+4 x^8)} \, dx\)

Optimal. Leaf size=47 \[ \tan ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right )-\tanh ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right ) \]

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Rubi [F]  time = 1.80, 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 {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)),x]

[Out]

(-4*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][1/(Sqrt[1 + 2*x^8]*(1 - x^2 + 2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*S
qrt[1 + 2*x^4]) + (3*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][x^2/(Sqrt[1 + 2*x^8]*(1 - x^2 + 2*x^8)), x], x, S
qrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4]) + (4*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][1/(Sqrt[1 + 2*x^8]*(1 + x^2 +
2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4]) + (3*Sqrt[x + 2*x^5]*Defer[Subst][Defer[Int][x^2/(Sqrt[1 +
2*x^8]*(1 + x^2 + 2*x^8)), x], x, Sqrt[x]])/(Sqrt[x]*Sqrt[1 + 2*x^4])

Rubi steps

\begin {align*} \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx &=\frac {\sqrt {x+2 x^5} \int \frac {\sqrt {x} \left (-1+6 x^4\right )}{\sqrt {1+2 x^4} \left (1-x^2+4 x^4+4 x^8\right )} \, dx}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (2 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1+6 x^8\right )}{\sqrt {1+2 x^8} \left (1-x^4+4 x^8+4 x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (2 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \left (\frac {-4+3 x^2}{2 \sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {4+3 x^2}{2 \sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \frac {-4+3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \frac {4+3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \left (-\frac {4}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \left (\frac {4}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (3 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (3 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}-\frac {\left (4 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (4 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)),x]

[Out]

Integrate[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)), x]

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IntegrateAlgebraic [A]  time = 0.43, size = 47, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {\sqrt {x+2 x^5}}{1+2 x^4}\right )-\tanh ^{-1}\left (\frac {\sqrt {x+2 x^5}}{1+2 x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + 6*x^4)*Sqrt[x + 2*x^5])/((1 + 2*x^4)*(1 - x^2 + 4*x^4 + 4*x^8)),x]

[Out]

ArcTan[Sqrt[x + 2*x^5]/(1 + 2*x^4)] - ArcTanh[Sqrt[x + 2*x^5]/(1 + 2*x^4)]

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fricas [A]  time = 0.50, size = 60, normalized size = 1.28 \begin {gather*} -\frac {1}{2} \, \arctan \left (\frac {2 \, x^{4} - x + 1}{2 \, \sqrt {2 \, x^{5} + x}}\right ) + \frac {1}{2} \, \log \left (\frac {2 \, x^{4} + x - 2 \, \sqrt {2 \, x^{5} + x} + 1}{2 \, x^{4} - x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="fricas")

[Out]

-1/2*arctan(1/2*(2*x^4 - x + 1)/sqrt(2*x^5 + x)) + 1/2*log((2*x^4 + x - 2*sqrt(2*x^5 + x) + 1)/(2*x^4 - x + 1)
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="giac")

[Out]

integrate(sqrt(2*x^5 + x)*(6*x^4 - 1)/((4*x^8 + 4*x^4 - x^2 + 1)*(2*x^4 + 1)), x)

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maple [C]  time = 0.67, size = 100, normalized size = 2.13

method	result	size
trager	\(\frac {\ln \left (-\frac {-2 x^{4}+2 \sqrt {2 x^{5}+x}-x -1}{2 x^{4}-x +1}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {2 x^{5}+x}-\RootOf \left (\textit {\_Z}^{2}+1\right )}{2 x^{4}+x +1}\right )}{2}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(-(-2*x^4+2*(2*x^5+x)^(1/2)-x-1)/(2*x^4-x+1))-1/2*RootOf(_Z^2+1)*ln(-(-2*RootOf(_Z^2+1)*x^4+RootOf(_Z^2+
1)*x+2*(2*x^5+x)^(1/2)-RootOf(_Z^2+1))/(2*x^4+x+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-1)*(2*x^5+x)^(1/2)/(2*x^4+1)/(4*x^8+4*x^4-x^2+1),x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^5 + x)*(6*x^4 - 1)/((4*x^8 + 4*x^4 - x^2 + 1)*(2*x^4 + 1)), x)

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mupad [B]  time = 1.34, size = 72, normalized size = 1.53 \begin {gather*} \frac {\ln \left (\frac {x}{2}-\sqrt {2\,x^5+x}+x^4+\frac {1}{2}\right )}{2}-\frac {\ln \left (2\,x^4-x+1\right )}{2}+\frac {\ln \left (x^4-\frac {x}{2}+\frac {1}{2}-\sqrt {2\,x^5+x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {\ln \left (2\,x^4+x+1\right )\,1{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x + 2*x^5)^(1/2)*(6*x^4 - 1))/((2*x^4 + 1)*(4*x^4 - x^2 + 4*x^8 + 1)),x)

[Out]

log(x/2 - (x + 2*x^5)^(1/2) + x^4 + 1/2)/2 + (log(x^4 - (x + 2*x^5)^(1/2)*1i - x/2 + 1/2)*1i)/2 - (log(x + 2*x
^4 + 1)*1i)/2 - log(2*x^4 - x + 1)/2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**4-1)*(2*x**5+x)**(1/2)/(2*x**4+1)/(4*x**8+4*x**4-x**2+1),x)

[Out]

Timed out

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