3.7.98 \(\int \frac {\sqrt [4]{1-x}}{1+x} \, dx\)

Optimal. Leaf size=58 \[ 4 \sqrt [4]{1-x}-2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {50, 63, 212, 206, 203} \begin {gather*} 4 \sqrt [4]{1-x}-2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(1/4)/(1 + x),x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{1-x}}{1+x} \, dx &=4 \sqrt [4]{1-x}+2 \int \frac {1}{(1-x)^{3/4} (1+x)} \, dx\\ &=4 \sqrt [4]{1-x}-8 \operatorname {Subst}\left (\int \frac {1}{2-x^4} \, dx,x,\sqrt [4]{1-x}\right )\\ &=4 \sqrt [4]{1-x}-\left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{1-x}\right )-\left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{1-x}\right )\\ &=4 \sqrt [4]{1-x}-2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 58, normalized size = 1.00 \begin {gather*} 4 \sqrt [4]{1-x}-2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(1/4)/(1 + x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 58, normalized size = 1.00 \begin {gather*} 4 \sqrt [4]{1-x}-2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{1-x}}{\sqrt [4]{2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 82, normalized size = 1.41 \begin {gather*} 4 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {\sqrt {2} + \sqrt {-x + 1}} - \frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-x + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left (-2^{\frac {1}{4}} + {\left (-x + 1\right )}^{\frac {1}{4}}\right ) + 4 \, {\left (-x + 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.17, size = 64, normalized size = 1.10 \begin {gather*} -2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-x + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-x + 1\right )}^{\frac {1}{4}} \right |}\right ) + 4 \, {\left (-x + 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*2^(1/4)*arctan(1/2*2^(3/4)*(-x + 1)^(1/4)) - 2^(1/4)*log(2^(1/4) + (-x + 1)^(1/4)) + 2^(1/4)*log(abs(-2^(1/
4) + (-x + 1)^(1/4))) + 4*(-x + 1)^(1/4)

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maple [A]  time = 0.01, size = 62, normalized size = 1.07 \begin {gather*} -2 \,2^{\frac {1}{4}} \arctan \left (\frac {\left (-x +1\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )-2^{\frac {1}{4}} \ln \left (\frac {\left (-x +1\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-x +1\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right )+4 \left (-x +1\right )^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.91, size = 61, normalized size = 1.05 \begin {gather*} -2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} - {\left (-x + 1\right )}^{\frac {1}{4}}}{2^{\frac {1}{4}} + {\left (-x + 1\right )}^{\frac {1}{4}}}\right ) + 4 \, {\left (-x + 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 46, normalized size = 0.79 \begin {gather*} 4\,{\left (1-x\right )}^{1/4}-2\,2^{1/4}\,\mathrm {atanh}\left (\frac {2^{3/4}\,{\left (1-x\right )}^{1/4}}{2}\right )-2\,2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (1-x\right )}^{1/4}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x)^(1/4)/(x + 1),x)

[Out]

4*(1 - x)^(1/4) - 2*2^(1/4)*atanh((2^(3/4)*(1 - x)^(1/4))/2) - 2*2^(1/4)*atan((2^(3/4)*(1 - x)^(1/4))/2)

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sympy [C]  time = 2.35, size = 243, normalized size = 4.19 \begin {gather*} \frac {5 \sqrt [4]{-1} \sqrt [4]{x - 1} \Gamma \left (\frac {5}{4}\right )}{\Gamma \left (\frac {9}{4}\right )} + \frac {5 \sqrt [4]{-2} e^{- \frac {i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \sqrt [4]{x - 1} e^{\frac {i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {5}{4}\right )}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {5 \left (-1\right )^{\frac {3}{4}} \sqrt [4]{2} e^{- \frac {i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \sqrt [4]{x - 1} e^{\frac {3 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {5}{4}\right )}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {5 \sqrt [4]{-2} e^{- \frac {i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \sqrt [4]{x - 1} e^{\frac {5 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {5}{4}\right )}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {5 \left (-1\right )^{\frac {3}{4}} \sqrt [4]{2} e^{- \frac {i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \sqrt [4]{x - 1} e^{\frac {7 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {5}{4}\right )}{4 \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(-1)**(1/4)*(x - 1)**(1/4)*gamma(5/4)/gamma(9/4) + 5*(-2)**(1/4)*exp(-I*pi/4)*log(-2**(3/4)*(x - 1)**(1/4)*e
xp_polar(I*pi/4)/2 + 1)*gamma(5/4)/(4*gamma(9/4)) - 5*(-1)**(3/4)*2**(1/4)*exp(-I*pi/4)*log(-2**(3/4)*(x - 1)*
*(1/4)*exp_polar(3*I*pi/4)/2 + 1)*gamma(5/4)/(4*gamma(9/4)) - 5*(-2)**(1/4)*exp(-I*pi/4)*log(-2**(3/4)*(x - 1)
**(1/4)*exp_polar(5*I*pi/4)/2 + 1)*gamma(5/4)/(4*gamma(9/4)) + 5*(-1)**(3/4)*2**(1/4)*exp(-I*pi/4)*log(-2**(3/
4)*(x - 1)**(1/4)*exp_polar(7*I*pi/4)/2 + 1)*gamma(5/4)/(4*gamma(9/4))

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