3.232 \(\int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx\)

Optimal. Leaf size=75 \[ -\frac {3}{2} \log \left (1-\frac {x-3}{\sqrt [3]{x^3-5 x^2+3 x+9}}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 (x-3)}{\sqrt [3]{x^3-5 x^2+3 x+9}}+1}{\sqrt {3}}\right )-\frac {1}{2} \log (x+1) \]

________________________________________________________________________________________

Rubi [B]  time = 0.12, antiderivative size = 188, normalized size of antiderivative = 2.51, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2067, 2064, 60} \[ -\frac {(9-3 x)^{2/3} \sqrt [3]{x+1} \log \left (-\frac {32}{3} (x-3)\right )}{2\ 3^{2/3} \sqrt [3]{x^3-5 x^2+3 x+9}}-\frac {\sqrt [3]{3} (9-3 x)^{2/3} \sqrt [3]{x+1} \log \left (\frac {\sqrt [3]{3} \sqrt [3]{x+1}}{\sqrt [3]{9-3 x}}+1\right )}{2 \sqrt [3]{x^3-5 x^2+3 x+9}}-\frac {(9-3 x)^{2/3} \sqrt [3]{x+1} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x+1}}{\sqrt [6]{3} \sqrt [3]{9-3 x}}\right )}{\sqrt [6]{3} \sqrt [3]{x^3-5 x^2+3 x+9}} \]

Antiderivative was successfully verified.

[In]

Int[(9 + 3*x - 5*x^2 + x^3)^(-1/3),x]

[Out]

-(((9 - 3*x)^(2/3)*(1 + x)^(1/3)*ArcTan[1/Sqrt[3] - (2*(1 + x)^(1/3))/(3^(1/6)*(9 - 3*x)^(1/3))])/(3^(1/6)*(9
+ 3*x - 5*x^2 + x^3)^(1/3))) - ((9 - 3*x)^(2/3)*(1 + x)^(1/3)*Log[(-32*(-3 + x))/3])/(2*3^(2/3)*(9 + 3*x - 5*x
^2 + x^3)^(1/3)) - (3^(1/3)*(9 - 3*x)^(2/3)*(1 + x)^(1/3)*Log[1 + (3^(1/3)*(1 + x)^(1/3))/(9 - 3*x)^(1/3)])/(2
*(9 + 3*x - 5*x^2 + x^3)^(1/3))

Rule 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

Rule 2064

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{27}-\frac {16 x}{3}+x^3}} \, dx,x,-\frac {5}{3}+x\right )\\ &=\frac {\left (16\ 2^{2/3} (3-x)^{2/3} \sqrt [3]{1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right )^{2/3} \sqrt [3]{\frac {128}{9}+\frac {16 x}{3}}} \, dx,x,-\frac {5}{3}+x\right )}{3 \sqrt [3]{9+3 x-5 x^2+x^3}}\\ &=-\frac {\sqrt {3} (3-x)^{2/3} \sqrt [3]{1+x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{3-x}}\right )}{\sqrt [3]{9+3 x-5 x^2+x^3}}-\frac {(3-x)^{2/3} \sqrt [3]{1+x} \log (3-x)}{2 \sqrt [3]{9+3 x-5 x^2+x^3}}-\frac {3 (3-x)^{2/3} \sqrt [3]{1+x} \log \left (\frac {3 \left (\sqrt [3]{3-x}+\sqrt [3]{1+x}\right )}{\sqrt [3]{3-x}}\right )}{2 \sqrt [3]{9+3 x-5 x^2+x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 49, normalized size = 0.65 \[ \frac {3 (x-3) \sqrt [3]{x+1} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {3-x}{4}\right )}{2^{2/3} \sqrt [3]{(x-3)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(9 + 3*x - 5*x^2 + x^3)^(-1/3),x]

[Out]

(3*(-3 + x)*(1 + x)^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, (3 - x)/4])/(2^(2/3)*((-3 + x)^2*(1 + x))^(1/3))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.26, size = 132, normalized size = 1.76 \[ -\log \left (\sqrt [3]{x^3-5 x^2+3 x+9}-x+3\right )+\frac {1}{2} \log \left (x^2+\left (x^3-5 x^2+3 x+9\right )^{2/3}+(x-3) \sqrt [3]{x^3-5 x^2+3 x+9}-6 x+9\right )-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^3-5 x^2+3 x+9}}{\sqrt [3]{x^3-5 x^2+3 x+9}+2 x-6}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(9 + 3*x - 5*x^2 + x^3)^(-1/3),x]

[Out]

-(Sqrt[3]*ArcTan[(Sqrt[3]*(9 + 3*x - 5*x^2 + x^3)^(1/3))/(-6 + 2*x + (9 + 3*x - 5*x^2 + x^3)^(1/3))]) - Log[3
- x + (9 + 3*x - 5*x^2 + x^3)^(1/3)] + Log[9 - 6*x + x^2 + (-3 + x)*(9 + 3*x - 5*x^2 + x^3)^(1/3) + (9 + 3*x -
5*x^2 + x^3)^(2/3)]/2

________________________________________________________________________________________

fricas [A]  time = 1.34, size = 128, normalized size = 1.71 \[ -\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 3\right )} + 2 \, \sqrt {3} {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}}{3 \, {\left (x - 3\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}} {\left (x - 3\right )} - 6 \, x + {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}} + 9}{x^{2} - 6 \, x + 9}\right ) - \log \left (-\frac {x - {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}} - 3}{x - 3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/3),x, algorithm="fricas")

[Out]

-sqrt(3)*arctan(1/3*(sqrt(3)*(x - 3) + 2*sqrt(3)*(x^3 - 5*x^2 + 3*x + 9)^(1/3))/(x - 3)) + 1/2*log((x^2 + (x^3
- 5*x^2 + 3*x + 9)^(1/3)*(x - 3) - 6*x + (x^3 - 5*x^2 + 3*x + 9)^(2/3) + 9)/(x^2 - 6*x + 9)) - log(-(x - (x^3
- 5*x^2 + 3*x + 9)^(1/3) - 3)/(x - 3))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/3),x, algorithm="giac")

[Out]

integrate((x^3 - 5*x^2 + 3*x + 9)^(-1/3), x)

________________________________________________________________________________________

maple [C]  time = 0.47, size = 672, normalized size = 8.96

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \ln \left (-\frac {20 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}-60 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x +27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}+27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -33 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}-81 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}-6 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -216 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-216 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -36 x^{2}+315 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )+648 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}+360 x -756}{-3+x}\right )}{3}-\frac {\ln \left (-\frac {20 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}-60 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x -27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -87 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}+81 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}+366 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -135 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-135 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x +45 x^{2}-315 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )+405 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}-198 x +189}{-3+x}\right ) \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )}{3}+\ln \left (-\frac {20 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}-60 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x -27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -87 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}+81 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}+366 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -135 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-135 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x +45 x^{2}-315 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )+405 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}-198 x +189}{-3+x}\right )\)	\(672\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-5*x^2+3*x+9)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/3*RootOf(_Z^2-3*_Z+9)*ln(-(20*RootOf(_Z^2-3*_Z+9)^2*x^2-60*RootOf(_Z^2-3*_Z+9)^2*x+27*RootOf(_Z^2-3*_Z+9)*(x
^3-5*x^2+3*x+9)^(2/3)+27*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(1/3)*x-33*RootOf(_Z^2-3*_Z+9)*x^2-81*RootOf(_Z
^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(1/3)-6*RootOf(_Z^2-3*_Z+9)*x-216*(x^3-5*x^2+3*x+9)^(2/3)-216*(x^3-5*x^2+3*x+9)^(
1/3)*x-36*x^2+315*RootOf(_Z^2-3*_Z+9)+648*(x^3-5*x^2+3*x+9)^(1/3)+360*x-756)/(-3+x))-1/3*ln(-(20*RootOf(_Z^2-3
*_Z+9)^2*x^2-60*RootOf(_Z^2-3*_Z+9)^2*x-27*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(2/3)-27*RootOf(_Z^2-3*_Z+9)*
(x^3-5*x^2+3*x+9)^(1/3)*x-87*RootOf(_Z^2-3*_Z+9)*x^2+81*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(1/3)+366*RootOf
(_Z^2-3*_Z+9)*x-135*(x^3-5*x^2+3*x+9)^(2/3)-135*(x^3-5*x^2+3*x+9)^(1/3)*x+45*x^2-315*RootOf(_Z^2-3*_Z+9)+405*(
x^3-5*x^2+3*x+9)^(1/3)-198*x+189)/(-3+x))*RootOf(_Z^2-3*_Z+9)+ln(-(20*RootOf(_Z^2-3*_Z+9)^2*x^2-60*RootOf(_Z^2
-3*_Z+9)^2*x-27*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(2/3)-27*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(1/3)*x-8
7*RootOf(_Z^2-3*_Z+9)*x^2+81*RootOf(_Z^2-3*_Z+9)*(x^3-5*x^2+3*x+9)^(1/3)+366*RootOf(_Z^2-3*_Z+9)*x-135*(x^3-5*
x^2+3*x+9)^(2/3)-135*(x^3-5*x^2+3*x+9)^(1/3)*x+45*x^2-315*RootOf(_Z^2-3*_Z+9)+405*(x^3-5*x^2+3*x+9)^(1/3)-198*
x+189)/(-3+x))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^3 - 5*x^2 + 3*x + 9)^(-1/3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (x^3-5\,x^2+3\,x+9\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x - 5*x^2 + x^3 + 9)^(1/3),x)

[Out]

int(1/(3*x - 5*x^2 + x^3 + 9)^(1/3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt [3]{x^{3} - 5 x^{2} + 3 x + 9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-5*x**2+3*x+9)**(1/3),x)

[Out]

Integral((x**3 - 5*x**2 + 3*x + 9)**(-1/3), x)

________________________________________________________________________________________