3.1.40 \(\int \frac {1}{\sqrt [3]{-5+7 x-3 x^2+x^3}} \, dx\) [40]

Optimal. Leaf size=81 \[ \frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 (-1+x)}{\sqrt {3} \sqrt [3]{-5+7 x-3 x^2+x^3}}\right )+\frac {1}{4} \log (1-x)-\frac {3}{4} \log \left (1-x+\sqrt [3]{-5+7 x-3 x^2+x^3}\right ) \]

[Out]

1/4*ln(1-x)-3/4*ln(1-x+(x^3-3*x^2+7*x-5)^(1/3))+1/2*arctan(1/3*3^(1/2)+2/3*(-1+x)/(x^3-3*x^2+7*x-5)^(1/3)*3^(1
/2))*3^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 131, normalized size of antiderivative = 1.62, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2092, 2036, 335, 281, 245} \begin {gather*} \frac {\sqrt {3} \sqrt [3]{(x-1)^2+4} \sqrt [3]{x-1} \tan ^{-1}\left (\frac {\frac {2 (x-1)^{2/3}}{\sqrt [3]{(x-1)^2+4}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{(x-1)^3+4 (x-1)}}-\frac {3 \sqrt [3]{(x-1)^2+4} \sqrt [3]{x-1} \log \left ((x-1)^{2/3}-\sqrt [3]{(x-1)^2+4}\right )}{4 \sqrt [3]{(x-1)^3+4 (x-1)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 7*x - 3*x^2 + x^3)^(-1/3),x]

[Out]

(Sqrt[3]*(4 + (-1 + x)^2)^(1/3)*(-1 + x)^(1/3)*ArcTan[(1 + (2*(-1 + x)^(2/3))/(4 + (-1 + x)^2)^(1/3))/Sqrt[3]]
)/(2*(4*(-1 + x) + (-1 + x)^3)^(1/3)) - (3*(4 + (-1 + x)^2)^(1/3)*(-1 + x)^(1/3)*Log[-(4 + (-1 + x)^2)^(1/3) +
(-1 + x)^(2/3)])/(4*(4*(-1 + x) + (-1 + x)^3)^(1/3))

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2092

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]{-5+7 x-3 x^2+x^3}} \, dx &=\text {Subst}\left (\int \frac {1}{\sqrt [3]{4 x+x^3}} \, dx,x,-1+x\right )\\ &=\frac {\left (\sqrt [3]{4+(-1+x)^2} \sqrt [3]{-1+x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \sqrt [3]{4+x^2}} \, dx,x,-1+x\right )}{\sqrt [3]{4 (-1+x)+(-1+x)^3}}\\ &=\frac {\left (3 \sqrt [3]{4+(-1+x)^2} \sqrt [3]{-1+x}\right ) \text {Subst}\left (\int \frac {x}{\sqrt [3]{4+x^6}} \, dx,x,\sqrt [3]{-1+x}\right )}{\sqrt [3]{4 (-1+x)+(-1+x)^3}}\\ &=\frac {\left (3 \sqrt [3]{4+(-1+x)^2} \sqrt [3]{-1+x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{4+x^3}} \, dx,x,(-1+x)^{2/3}\right )}{2 \sqrt [3]{4 (-1+x)+(-1+x)^3}}\\ &=\frac {\sqrt {3} \sqrt [3]{4+(-1+x)^2} \sqrt [3]{-1+x} \tan ^{-1}\left (\frac {1+\frac {2 (-1+x)^{2/3}}{\sqrt [3]{4+(-1+x)^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{-4 (1-x)+(-1+x)^3}}-\frac {3 \sqrt [3]{4+(-1+x)^2} \sqrt [3]{-1+x} \log \left (\sqrt [3]{4+(-1+x)^2}-(-1+x)^{2/3}\right )}{4 \sqrt [3]{-4 (1-x)+(-1+x)^3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.01, size = 85, normalized size = 1.05 \begin {gather*} \frac {3 \sqrt [3]{(2-i)+i x} \sqrt [3]{i (-1+x)} ((-1+2 i)+x) F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};-\frac {1}{4} i ((-1+2 i)+x),-\frac {1}{2} i ((-1+2 i)+x)\right )}{4 \sqrt [3]{-5+7 x-3 x^2+x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 7*x - 3*x^2 + x^3)^(-1/3),x]

[Out]

(3*((2 - I) + I*x)^(1/3)*(I*(-1 + x))^(1/3)*((-1 + 2*I) + x)*AppellF1[2/3, 1/3, 1/3, 5/3, (-1/4*I)*((-1 + 2*I)
+ x), (-1/2*I)*((-1 + 2*I) + x)])/(4*(-5 + 7*x - 3*x^2 + x^3)^(1/3))

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[1/(x^3 - 3*x^2 + 7*x - 5)^(1/3),x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.90, size = 433, normalized size = 5.35

method	result	size
trager	\(-\frac {\ln \left (92 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+624 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {2}{3}}-675 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}} x -184 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -41 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+51 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {2}{3}}+675 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}}+624 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}} x +82 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -583 x^{2}-624 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}}-713 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1166 x -1643\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (212 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-624 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {2}{3}}-51 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}} x -424 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +463 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+675 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {2}{3}}+51 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}}-624 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}} x -926 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +161 x^{2}+624 \left (x^{3}-3 x^{2}+7 x -5\right )^{\frac {1}{3}}+1643 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-322 x +713\right )}{2}\)	\(433\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-3*x^2+7*x-5)^(1/3),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(92*RootOf(_Z^2-_Z+1)^2*x^2+624*RootOf(_Z^2-_Z+1)*(x^3-3*x^2+7*x-5)^(2/3)-675*RootOf(_Z^2-_Z+1)*(x^3-3*
x^2+7*x-5)^(1/3)*x-184*RootOf(_Z^2-_Z+1)^2*x-41*RootOf(_Z^2-_Z+1)*x^2+51*(x^3-3*x^2+7*x-5)^(2/3)+675*RootOf(_Z
^2-_Z+1)*(x^3-3*x^2+7*x-5)^(1/3)+624*(x^3-3*x^2+7*x-5)^(1/3)*x+82*RootOf(_Z^2-_Z+1)*x-583*x^2-624*(x^3-3*x^2+7
*x-5)^(1/3)-713*RootOf(_Z^2-_Z+1)+1166*x-1643)+1/2*RootOf(_Z^2-_Z+1)*ln(212*RootOf(_Z^2-_Z+1)^2*x^2-624*RootOf
(_Z^2-_Z+1)*(x^3-3*x^2+7*x-5)^(2/3)-51*RootOf(_Z^2-_Z+1)*(x^3-3*x^2+7*x-5)^(1/3)*x-424*RootOf(_Z^2-_Z+1)^2*x+4
63*RootOf(_Z^2-_Z+1)*x^2+675*(x^3-3*x^2+7*x-5)^(2/3)+51*RootOf(_Z^2-_Z+1)*(x^3-3*x^2+7*x-5)^(1/3)-624*(x^3-3*x
^2+7*x-5)^(1/3)*x-926*RootOf(_Z^2-_Z+1)*x+161*x^2+624*(x^3-3*x^2+7*x-5)^(1/3)+1643*RootOf(_Z^2-_Z+1)-322*x+713
)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-3*x^2+7*x-5)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^3 - 3*x^2 + 7*x - 5)^(-1/3), x)

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Fricas [A]
time = 0.46, size = 120, normalized size = 1.48 \begin {gather*} -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {22791076 \, \sqrt {3} {\left (x^{3} - 3 \, x^{2} + 7 \, x - 5\right )}^{\frac {1}{3}} {\left (x - 1\right )} + \sqrt {3} {\left (20389537 \, x^{2} - 40779074 \, x + 53222437\right )} + 17987998 \, \sqrt {3} {\left (x^{3} - 3 \, x^{2} + 7 \, x - 5\right )}^{\frac {2}{3}}}{7204617 \, x^{2} - 14409234 \, x - 20666867}\right ) - \frac {1}{4} \, \log \left (3 \, {\left (x^{3} - 3 \, x^{2} + 7 \, x - 5\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 3 \, {\left (x^{3} - 3 \, x^{2} + 7 \, x - 5\right )}^{\frac {2}{3}} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-3*x^2+7*x-5)^(1/3),x, algorithm="fricas")

[Out]

-1/2*sqrt(3)*arctan((22791076*sqrt(3)*(x^3 - 3*x^2 + 7*x - 5)^(1/3)*(x - 1) + sqrt(3)*(20389537*x^2 - 40779074
*x + 53222437) + 17987998*sqrt(3)*(x^3 - 3*x^2 + 7*x - 5)^(2/3))/(7204617*x^2 - 14409234*x - 20666867)) - 1/4*
log(3*(x^3 - 3*x^2 + 7*x - 5)^(1/3)*(x - 1) - 3*(x^3 - 3*x^2 + 7*x - 5)^(2/3) + 4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{x^{3} - 3 x^{2} + 7 x - 5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-3*x**2+7*x-5)**(1/3),x)

[Out]

Integral((x**3 - 3*x**2 + 7*x - 5)**(-1/3), x)

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-3*x^2+7*x-5)^(1/3),x)

[Out]

Could not integrate

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (x^3-3\,x^2+7\,x-5\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(7*x - 3*x^2 + x^3 - 5)^(1/3),x)

[Out]

int(1/(7*x - 3*x^2 + x^3 - 5)^(1/3), x)

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