3.23.49 \(\int \frac {-3+x}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\)

Optimal. Leaf size=168 \[ -\frac {\log \left (2 \sqrt [3]{1-x^2}+2^{2/3} x+2^{2/3}\right )}{2^{2/3}}+\frac {\log \left (-\sqrt [3]{2} x^2-2 \left (1-x^2\right )^{2/3}+\left (2^{2/3} x+2^{2/3}\right ) \sqrt [3]{1-x^2}-2 \sqrt [3]{2} x-\sqrt [3]{2}\right )}{2\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^2}}{\sqrt [3]{1-x^2}-2^{2/3} x-2^{2/3}}\right )}{2^{2/3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 95, normalized size of antiderivative = 0.57, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1008} \begin {gather*} \frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}-\frac {3 \log \left ((x+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{1-x}\right )}{2\ 2^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} (x+1)^{2/3}}{\sqrt {3} \sqrt [3]{1-x}}\right )}{2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + x)/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

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

Rule 1008

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)^(1/3)*((d_) + (f_.)*(x_)^2)), x_Symbol] :> Simp[(Sqrt[3]*h*ArcT
an[1/Sqrt[3] - (2^(2/3)*(1 - (3*h*x)/g)^(2/3))/(Sqrt[3]*(1 + (3*h*x)/g)^(1/3))])/(2^(2/3)*a^(1/3)*f), x] + (-S
imp[(3*h*Log[(1 - (3*h*x)/g)^(2/3) + 2^(1/3)*(1 + (3*h*x)/g)^(1/3)])/(2^(5/3)*a^(1/3)*f), x] + Simp[(h*Log[d +
f*x^2])/(2^(5/3)*a^(1/3)*f), x]) /; FreeQ[{a, c, d, f, g, h}, x] && EqQ[c*d + 3*a*f, 0] && EqQ[c*g^2 + 9*a*h^
2, 0] && GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.22, size = 143, normalized size = 0.85 \begin {gather*} \frac {1}{6} x^2 F_1\left (1;\frac {1}{3},1;2;x^2,-\frac {x^2}{3}\right )+\frac {27 x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{1-x^2} \left (x^2+3\right ) \left (2 x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};x^2,-\frac {x^2}{3}\right )-F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};x^2,-\frac {x^2}{3}\right )\right )-9 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};x^2,-\frac {x^2}{3}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-3 + x)/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

(x^2*AppellF1[1, 1/3, 1, 2, x^2, -1/3*x^2])/6 + (27*x*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2])/((1 - x^2)^(1
/3)*(3 + x^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, x^2, -1/3*x^2]
- AppellF1[3/2, 4/3, 1, 5/2, x^2, -1/3*x^2])))

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IntegrateAlgebraic [A]  time = 0.25, size = 168, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^2}}{-2^{2/3}-2^{2/3} x+\sqrt [3]{1-x^2}}\right )}{2^{2/3}}-\frac {\log \left (2^{2/3}+2^{2/3} x+2 \sqrt [3]{1-x^2}\right )}{2^{2/3}}+\frac {\log \left (-\sqrt [3]{2}-2 \sqrt [3]{2} x-\sqrt [3]{2} x^2+\left (2^{2/3}+2^{2/3} x\right ) \sqrt [3]{1-x^2}-2 \left (1-x^2\right )^{2/3}\right )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-3 + x)/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

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

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fricas [B]  time = 6.50, size = 315, normalized size = 1.88 \begin {gather*} -\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {4^{\frac {1}{6}} \sqrt {3} {\left (12 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 9 \, x\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - 12 \, \left (-1\right )^{\frac {1}{3}} {\left (x^{5} + 19 \, x^{4} + 42 \, x^{3} + 6 \, x^{2} - 27 \, x - 9\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}} {\left (x^{6} - 18 \, x^{5} - 117 \, x^{4} - 36 \, x^{3} + 207 \, x^{2} + 54 \, x - 27\right )}\right )}}{6 \, {\left (x^{6} + 54 \, x^{5} + 171 \, x^{4} + 108 \, x^{3} - 81 \, x^{2} - 162 \, x - 27\right )}}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {6 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3 \, x\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} + 18 \, x^{3} + 24 \, x^{2} - 18 \, x - 9\right )} + 6 \, {\left (x^{3} + 7 \, x^{2} + 3 \, x - 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}}{x^{4} + 6 \, x^{2} + 9}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (\frac {6 \cdot 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} + 12 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{x^{2} + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="fricas")

[Out]

-1/6*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*sqrt(3)*(12*4^(2/3)*(-1)^(2/3)*(x^4 + 3*x^3 + 3*x^2 + 9*x)*
(-x^2 + 1)^(2/3) - 12*(-1)^(1/3)*(x^5 + 19*x^4 + 42*x^3 + 6*x^2 - 27*x - 9)*(-x^2 + 1)^(1/3) + 4^(1/3)*(x^6 -
18*x^5 - 117*x^4 - 36*x^3 + 207*x^2 + 54*x - 27))/(x^6 + 54*x^5 + 171*x^4 + 108*x^3 - 81*x^2 - 162*x - 27)) -
1/24*4^(2/3)*(-1)^(1/3)*log(-(6*4^(2/3)*(-1)^(1/3)*(x^2 + 3*x)*(-x^2 + 1)^(2/3) - 4^(1/3)*(-1)^(2/3)*(x^4 + 18
*x^3 + 24*x^2 - 18*x - 9) + 6*(x^3 + 7*x^2 + 3*x - 3)*(-x^2 + 1)^(1/3))/(x^4 + 6*x^2 + 9)) + 1/12*4^(2/3)*(-1)
^(1/3)*log((6*4^(1/3)*(-1)^(2/3)*(-x^2 + 1)^(1/3)*(x + 1) - 4^(2/3)*(-1)^(1/3)*(x^2 + 3) + 12*(-x^2 + 1)^(2/3)
)/(x^2 + 3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="giac")

[Out]

integrate((x - 3)/((x^2 + 3)*(-x^2 + 1)^(1/3)), x)

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maple [C]  time = 8.26, size = 1552, normalized size = 9.24

method	result	size
trager	\(\text {Expression too large to display}\)	\(1552\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3+x)/(-x^2+1)^(1/3)/(x^2+3),x,method=_RETURNVERBOSE)

[Out]

RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*ln((-6*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)
*RootOf(_Z^3+2)^3*x^2-8*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x^2+18*(-x^2+1)
^(2/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2-18*RootOf(RootOf(_Z^3+2)^2+2*_Z*Ro
otOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x-24*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2
)^2*x+3*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2*x+18*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2
)*RootOf(_Z^3+2)*x+3*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2+18*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^
3+2)+4*_Z^2)*RootOf(_Z^3+2)+3*RootOf(_Z^3+2)*x^2+4*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2-6*(
-x^2+1)^(2/3)+18*RootOf(_Z^3+2)*x+24*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x-9*RootOf(_Z^3+2)-12
*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2))/(2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*R
ootOf(_Z^3+2)^2*x-x+3)/(2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2*x-x-3))-1/2*ln(
-(-2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x^2+8*RootOf(RootOf(_Z^3+2)^2+2*_Z*R
ootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x^2+18*(-x^2+1)^(2/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_
Z^2)*RootOf(_Z^3+2)^2-6*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x+24*RootOf(RootO
f(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x+6*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2*x+18*(-x^2+1)^(
1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)*x+6*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2+18
*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)+RootOf(_Z^3+2)*x^2-4*RootOf
(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2-12*(-x^2+1)^(2/3)+3*RootOf(_Z^3+2)-12*RootOf(RootOf(_Z^3+2)^
2+2*_Z*RootOf(_Z^3+2)+4*_Z^2))/(2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2*x-x+3)/
(2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2*x-x-3))*RootOf(_Z^3+2)-ln(-(-2*RootOf(
RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x^2+8*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2
)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x^2+18*(-x^2+1)^(2/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(
_Z^3+2)^2-6*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x+24*RootOf(RootOf(_Z^3+2)^2+
2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x+6*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2*x+18*(-x^2+1)^(1/3)*RootOf(
RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)*x+6*(-x^2+1)^(1/3)*RootOf(_Z^3+2)^2+18*(-x^2+1)^(1
/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)+RootOf(_Z^3+2)*x^2-4*RootOf(RootOf(_Z^3
+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2-12*(-x^2+1)^(2/3)+3*RootOf(_Z^3+2)-12*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootO
f(_Z^3+2)+4*_Z^2))/(2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2*x-x+3)/(2*RootOf(Ro
otOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2*x-x-3))*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2
)+4*_Z^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="maxima")

[Out]

integrate((x - 3)/((x^2 + 3)*(-x^2 + 1)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x-3}{{\left (1-x^2\right )}^{1/3}\,\left (x^2+3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 3)/((1 - x^2)^(1/3)*(x^2 + 3)),x)

[Out]

int((x - 3)/((1 - x^2)^(1/3)*(x^2 + 3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(-x**2+1)**(1/3)/(x**2+3),x)

[Out]

Integral((x - 3)/((-(x - 1)*(x + 1))**(1/3)*(x**2 + 3)), x)

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