3.28.21 \(\int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx\)

Optimal. Leaf size=250 \[ -\frac {\sqrt [3]{(1-2 x)^2} \left (7^{2/3} \sqrt [3]{2 x-1}+7\right ) \left (\sqrt [3]{7} (2 x-1)^{2/3}-7^{2/3} \sqrt [3]{2 x-1}+7\right )^2 \left (\frac {3 (2 x-1)^{2/3}}{2^{2/3}}+\sqrt [3]{2} 7^{2/3} \log \left (7^{2/3} \sqrt [3]{2 x-1}+7\right )-\left (\frac {7}{2}\right )^{2/3} \log \left (-\sqrt [3]{7} (2 x-1)^{2/3}+7^{2/3} \sqrt [3]{2 x-1}-7\right )+\sqrt [3]{2} \sqrt {3} 7^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{2 x-1}}{\sqrt {3} \sqrt [3]{7}}\right )\right )}{14 (x+3) \sqrt [3]{2 x-1} \left (-2 \sqrt [3]{7} x+(14 x-7)^{2/3}-7 \sqrt [3]{2 x-1}+\sqrt [3]{7}\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 179, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {646, 50, 56, 617, 204, 31} \begin {gather*} \frac {3 \sqrt [3]{4 x^2-4 x+1}}{2^{2/3}}-\frac {\left (\frac {7}{2}\right )^{2/3} \sqrt [3]{4 x^2-4 x+1} \log (x+3)}{(2 x-1)^{2/3}}+\frac {3 \left (\frac {7}{2}\right )^{2/3} \sqrt [3]{4 x^2-4 x+1} \log \left (\sqrt [3]{8 x-4}+2^{2/3} \sqrt [3]{7}\right )}{(2 x-1)^{2/3}}+\frac {\sqrt [3]{2} \sqrt {3} 7^{2/3} \sqrt [3]{4 x^2-4 x+1} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{2 x-1}}{\sqrt {3} \sqrt [3]{7}}\right )}{(2 x-1)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 8*x + 8*x^2)^(1/3)/(3 + x),x]

[Out]

(3*(1 - 4*x + 4*x^2)^(1/3))/2^(2/3) + (2^(1/3)*Sqrt[3]*7^(2/3)*(1 - 4*x + 4*x^2)^(1/3)*ArcTan[1/Sqrt[3] - (2*(
-1 + 2*x)^(1/3))/(Sqrt[3]*7^(1/3))])/(-1 + 2*x)^(2/3) - ((7/2)^(2/3)*(1 - 4*x + 4*x^2)^(1/3)*Log[3 + x])/(-1 +
2*x)^(2/3) + (3*(7/2)^(2/3)*(1 - 4*x + 4*x^2)^(1/3)*Log[2^(2/3)*7^(1/3) + (-4 + 8*x)^(1/3)])/(-1 + 2*x)^(2/3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx &=\frac {\sqrt [3]{2-8 x+8 x^2} \int \frac {(-4+8 x)^{2/3}}{3+x} \, dx}{(-4+8 x)^{2/3}}\\ &=\frac {3 \sqrt [3]{1-4 x+4 x^2}}{2^{2/3}}-\frac {\left (28 \sqrt [3]{2-8 x+8 x^2}\right ) \int \frac {1}{(3+x) \sqrt [3]{-4+8 x}} \, dx}{(-4+8 x)^{2/3}}\\ &=\frac {3 \sqrt [3]{1-4 x+4 x^2}}{2^{2/3}}-\frac {\left (\frac {7}{2}\right )^{2/3} \sqrt [3]{1-4 x+4 x^2} \log (3+x)}{(-1+2 x)^{2/3}}-\frac {\left (42 \sqrt [3]{2-8 x+8 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{2 \sqrt [3]{2} 7^{2/3}-2^{2/3} \sqrt [3]{7} x+x^2} \, dx,x,\sqrt [3]{-4+8 x}\right )}{(-4+8 x)^{2/3}}+\frac {\left (3 \sqrt [3]{2} 7^{2/3} \sqrt [3]{2-8 x+8 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{2^{2/3} \sqrt [3]{7}+x} \, dx,x,\sqrt [3]{-4+8 x}\right )}{(-4+8 x)^{2/3}}\\ &=\frac {3 \sqrt [3]{1-4 x+4 x^2}}{2^{2/3}}-\frac {\left (\frac {7}{2}\right )^{2/3} \sqrt [3]{1-4 x+4 x^2} \log (3+x)}{(-1+2 x)^{2/3}}+\frac {3 \left (\frac {7}{2}\right )^{2/3} \sqrt [3]{1-4 x+4 x^2} \log \left (2^{2/3} \sqrt [3]{7}+\sqrt [3]{-4+8 x}\right )}{(-1+2 x)^{2/3}}-\frac {\left (6 \sqrt [3]{2} 7^{2/3} \sqrt [3]{2-8 x+8 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\sqrt [3]{\frac {2}{7}} \sqrt [3]{-4+8 x}\right )}{(-4+8 x)^{2/3}}\\ &=\frac {3 \sqrt [3]{1-4 x+4 x^2}}{2^{2/3}}+\frac {\sqrt [3]{2} \sqrt {3} 7^{2/3} \sqrt [3]{1-4 x+4 x^2} \tan ^{-1}\left (\frac {7-2\ 7^{2/3} \sqrt [3]{-1+2 x}}{7 \sqrt {3}}\right )}{(-1+2 x)^{2/3}}-\frac {\left (\frac {7}{2}\right )^{2/3} \sqrt [3]{1-4 x+4 x^2} \log (3+x)}{(-1+2 x)^{2/3}}+\frac {3 \left (\frac {7}{2}\right )^{2/3} \sqrt [3]{1-4 x+4 x^2} \log \left (2^{2/3} \sqrt [3]{7}+\sqrt [3]{-4+8 x}\right )}{(-1+2 x)^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 37, normalized size = 0.15 \begin {gather*} -\frac {3 \sqrt [3]{(1-2 x)^2} \left (\, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {1}{7} (1-2 x)\right )-1\right )}{2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 8*x + 8*x^2)^(1/3)/(3 + x),x]

[Out]

(-3*((1 - 2*x)^2)^(1/3)*(-1 + Hypergeometric2F1[2/3, 1, 5/3, (1 - 2*x)/7]))/2^(2/3)

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IntegrateAlgebraic [A]  time = 5.42, size = 250, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [3]{(1-2 x)^2} \left (7+7^{2/3} \sqrt [3]{-1+2 x}\right ) \left (7-7^{2/3} \sqrt [3]{-1+2 x}+\sqrt [3]{7} (-1+2 x)^{2/3}\right )^2 \left (\frac {3 (-1+2 x)^{2/3}}{2^{2/3}}+\sqrt [3]{2} \sqrt {3} 7^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+2 x}}{\sqrt {3} \sqrt [3]{7}}\right )+\sqrt [3]{2} 7^{2/3} \log \left (7+7^{2/3} \sqrt [3]{-1+2 x}\right )-\left (\frac {7}{2}\right )^{2/3} \log \left (-7+7^{2/3} \sqrt [3]{-1+2 x}-\sqrt [3]{7} (-1+2 x)^{2/3}\right )\right )}{14 (3+x) \sqrt [3]{-1+2 x} \left (\sqrt [3]{7}-2 \sqrt [3]{7} x-7 \sqrt [3]{-1+2 x}+(-7+14 x)^{2/3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 - 8*x + 8*x^2)^(1/3)/(3 + x),x]

[Out]

-1/14*(((1 - 2*x)^2)^(1/3)*(7 + 7^(2/3)*(-1 + 2*x)^(1/3))*(7 - 7^(2/3)*(-1 + 2*x)^(1/3) + 7^(1/3)*(-1 + 2*x)^(
2/3))^2*((3*(-1 + 2*x)^(2/3))/2^(2/3) + 2^(1/3)*Sqrt[3]*7^(2/3)*ArcTan[1/Sqrt[3] - (2*(-1 + 2*x)^(1/3))/(Sqrt[
3]*7^(1/3))] + 2^(1/3)*7^(2/3)*Log[7 + 7^(2/3)*(-1 + 2*x)^(1/3)] - (7/2)^(2/3)*Log[-7 + 7^(2/3)*(-1 + 2*x)^(1/
3) - 7^(1/3)*(-1 + 2*x)^(2/3)]))/((3 + x)*(-1 + 2*x)^(1/3)*(7^(1/3) - 2*7^(1/3)*x - 7*(-1 + 2*x)^(1/3) + (-7 +
14*x)^(2/3)))

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fricas [A]  time = 0.55, size = 169, normalized size = 0.68 \begin {gather*} 98^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {98^{\frac {2}{3}} \sqrt {3} {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} - 7 \, \sqrt {3} {\left (2 \, x - 1\right )}}{21 \, {\left (2 \, x - 1\right )}}\right ) - \frac {1}{2} \cdot 98^{\frac {1}{3}} \log \left (\frac {98^{\frac {2}{3}} {\left (4 \, x^{2} - 4 \, x + 1\right )} - 7 \cdot 98^{\frac {1}{3}} {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} {\left (2 \, x - 1\right )} + 49 \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {2}{3}}}{4 \, x^{2} - 4 \, x + 1}\right ) + 98^{\frac {1}{3}} \log \left (\frac {98^{\frac {1}{3}} {\left (2 \, x - 1\right )} + 7 \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{2 \, x - 1}\right ) + \frac {3}{2} \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-8*x+2)^(1/3)/(3+x),x, algorithm="fricas")

[Out]

98^(1/3)*sqrt(3)*arctan(1/21*(98^(2/3)*sqrt(3)*(8*x^2 - 8*x + 2)^(1/3) - 7*sqrt(3)*(2*x - 1))/(2*x - 1)) - 1/2
*98^(1/3)*log((98^(2/3)*(4*x^2 - 4*x + 1) - 7*98^(1/3)*(8*x^2 - 8*x + 2)^(1/3)*(2*x - 1) + 49*(8*x^2 - 8*x + 2
)^(2/3))/(4*x^2 - 4*x + 1)) + 98^(1/3)*log((98^(1/3)*(2*x - 1) + 7*(8*x^2 - 8*x + 2)^(1/3))/(2*x - 1)) + 3/2*(
8*x^2 - 8*x + 2)^(1/3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{x + 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-8*x+2)^(1/3)/(3+x),x, algorithm="giac")

[Out]

integrate((8*x^2 - 8*x + 2)^(1/3)/(x + 3), x)

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maple [A]  time = 2.56, size = 110, normalized size = 0.44

method	result	size
risch	\(\frac {3 \,2^{\frac {1}{3}} \left (\left (-1+2 x \right )^{2}\right )^{\frac {1}{3}}}{2}+\frac {\left (7^{\frac {2}{3}} \ln \left (\left (-1+2 x \right )^{\frac {1}{3}}+7^{\frac {1}{3}}\right )-\frac {7^{\frac {2}{3}} \ln \left (\left (-1+2 x \right )^{\frac {2}{3}}-7^{\frac {1}{3}} \left (-1+2 x \right )^{\frac {1}{3}}+7^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, 7^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,7^{\frac {2}{3}} \left (-1+2 x \right )^{\frac {1}{3}}}{7}-1\right )}{3}\right )\right ) 2^{\frac {1}{3}} \left (\left (-1+2 x \right )^{2}\right )^{\frac {1}{3}}}{\left (-1+2 x \right )^{\frac {2}{3}}}\)	\(110\)
trager	\(\text {Expression too large to display}\)	\(1484\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2-8*x+2)^(1/3)/(3+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{x + 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-8*x+2)^(1/3)/(3+x),x, algorithm="maxima")

[Out]

integrate((8*x^2 - 8*x + 2)^(1/3)/(x + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (8\,x^2-8\,x+2\right )}^{1/3}}{x+3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2 - 8*x + 2)^(1/3)/(x + 3),x)

[Out]

int((8*x^2 - 8*x + 2)^(1/3)/(x + 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**2-8*x+2)**(1/3)/(3+x),x)

[Out]

2**(1/3)*Integral((4*x**2 - 4*x + 1)**(1/3)/(x + 3), x)

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