3.7.31 \(\int \frac {\sqrt {-1+x^5} (2+3 x^5)}{1-a x^4-2 x^5+x^{10}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt {x^5-1}}\right )}{\sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt {x^5-1}}\right )}{\sqrt [4]{a}} \]

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Rubi [F]  time = 0.54, 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 {\sqrt {-1+x^5} \left (2+3 x^5\right )}{1-a x^4-2 x^5+x^{10}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Sqrt[-1 + x^5]*(2 + 3*x^5))/(1 - a*x^4 - 2*x^5 + x^10),x]

[Out]

-2*Defer[Int][Sqrt[-1 + x^5]/(-1 + a*x^4 + 2*x^5 - x^10), x] + 3*Defer[Int][(x^5*Sqrt[-1 + x^5])/(1 - a*x^4 -
2*x^5 + x^10), x]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x^5} \left (2+3 x^5\right )}{1-a x^4-2 x^5+x^{10}} \, dx &=\int \left (-\frac {2 \sqrt {-1+x^5}}{-1+a x^4+2 x^5-x^{10}}+\frac {3 x^5 \sqrt {-1+x^5}}{1-a x^4-2 x^5+x^{10}}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt {-1+x^5}}{-1+a x^4+2 x^5-x^{10}} \, dx\right )+3 \int \frac {x^5 \sqrt {-1+x^5}}{1-a x^4-2 x^5+x^{10}} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-1+x^5} \left (2+3 x^5\right )}{1-a x^4-2 x^5+x^{10}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(Sqrt[-1 + x^5]*(2 + 3*x^5))/(1 - a*x^4 - 2*x^5 + x^10),x]

[Out]

Integrate[(Sqrt[-1 + x^5]*(2 + 3*x^5))/(1 - a*x^4 - 2*x^5 + x^10), x]

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IntegrateAlgebraic [A]  time = 3.59, size = 49, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt {-1+x^5}}\right )}{\sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt {-1+x^5}}\right )}{\sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-1 + x^5]*(2 + 3*x^5))/(1 - a*x^4 - 2*x^5 + x^10),x]

[Out]

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

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fricas [B]  time = 0.69, size = 195, normalized size = 3.98 \begin {gather*} -\frac {\arctan \left (\frac {a^{\frac {1}{4}} x}{\sqrt {x^{5} - 1}}\right )}{a^{\frac {1}{4}}} - \frac {\log \left (\frac {x^{10} + a x^{4} - 2 \, x^{5} + 2 \, \sqrt {x^{5} - 1} {\left (a^{\frac {3}{4}} x^{3} + \frac {a x^{6} - a x}{a^{\frac {3}{4}}}\right )} + \frac {2 \, {\left (a x^{7} - a x^{2}\right )}}{\sqrt {a}} + 1}{x^{10} - a x^{4} - 2 \, x^{5} + 1}\right )}{4 \, a^{\frac {1}{4}}} + \frac {\log \left (\frac {x^{10} + a x^{4} - 2 \, x^{5} - 2 \, \sqrt {x^{5} - 1} {\left (a^{\frac {3}{4}} x^{3} + \frac {a x^{6} - a x}{a^{\frac {3}{4}}}\right )} + \frac {2 \, {\left (a x^{7} - a x^{2}\right )}}{\sqrt {a}} + 1}{x^{10} - a x^{4} - 2 \, x^{5} + 1}\right )}{4 \, a^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-1)^(1/2)*(3*x^5+2)/(x^10-a*x^4-2*x^5+1),x, algorithm="fricas")

[Out]

-arctan(a^(1/4)*x/sqrt(x^5 - 1))/a^(1/4) - 1/4*log((x^10 + a*x^4 - 2*x^5 + 2*sqrt(x^5 - 1)*(a^(3/4)*x^3 + (a*x
^6 - a*x)/a^(3/4)) + 2*(a*x^7 - a*x^2)/sqrt(a) + 1)/(x^10 - a*x^4 - 2*x^5 + 1))/a^(1/4) + 1/4*log((x^10 + a*x^
4 - 2*x^5 - 2*sqrt(x^5 - 1)*(a^(3/4)*x^3 + (a*x^6 - a*x)/a^(3/4)) + 2*(a*x^7 - a*x^2)/sqrt(a) + 1)/(x^10 - a*x
^4 - 2*x^5 + 1))/a^(1/4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, x^{5} + 2\right )} \sqrt {x^{5} - 1}}{x^{10} - a x^{4} - 2 \, x^{5} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-1)^(1/2)*(3*x^5+2)/(x^10-a*x^4-2*x^5+1),x, algorithm="giac")

[Out]

integrate((3*x^5 + 2)*sqrt(x^5 - 1)/(x^10 - a*x^4 - 2*x^5 + 1), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{5}-1}\, \left (3 x^{5}+2\right )}{x^{10}-a \,x^{4}-2 x^{5}+1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5-1)^(1/2)*(3*x^5+2)/(x^10-a*x^4-2*x^5+1),x)

[Out]

int((x^5-1)^(1/2)*(3*x^5+2)/(x^10-a*x^4-2*x^5+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, x^{5} + 2\right )} \sqrt {x^{5} - 1}}{x^{10} - a x^{4} - 2 \, x^{5} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5-1)^(1/2)*(3*x^5+2)/(x^10-a*x^4-2*x^5+1),x, algorithm="maxima")

[Out]

integrate((3*x^5 + 2)*sqrt(x^5 - 1)/(x^10 - a*x^4 - 2*x^5 + 1), x)

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mupad [B]  time = 8.02, size = 98, normalized size = 2.00 \begin {gather*} \frac {\ln \left (\frac {x^5+\sqrt {a}\,x^2-2\,a^{1/4}\,x\,\sqrt {x^5-1}-1}{\sqrt {a}\,x^2-x^5+1}\right )}{2\,a^{1/4}}+\frac {\ln \left (\frac {x^5-\sqrt {a}\,x^2-1+a^{1/4}\,x\,\sqrt {x^5-1}\,2{}\mathrm {i}}{x^5+\sqrt {a}\,x^2-1}\right )\,1{}\mathrm {i}}{2\,a^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^5 - 1)^(1/2)*(3*x^5 + 2))/(a*x^4 + 2*x^5 - x^10 - 1),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (x - 1\right ) \left (x^{4} + x^{3} + x^{2} + x + 1\right )} \left (3 x^{5} + 2\right )}{- a x^{4} + x^{10} - 2 x^{5} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5-1)**(1/2)*(3*x**5+2)/(x**10-a*x**4-2*x**5+1),x)

[Out]

Integral(sqrt((x - 1)*(x**4 + x**3 + x**2 + x + 1))*(3*x**5 + 2)/(-a*x**4 + x**10 - 2*x**5 + 1), x)

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