3.31 \(\int \frac {8 C+(-b)^{2/3} C x^2}{-8+b x^3} \, dx\)

Optimal. Leaf size=57 \[ \frac {2 C \tan ^{-1}\left (\frac {1-\sqrt [3]{-b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{-b}}-\frac {C \log \left (\sqrt [3]{-b} x+2\right )}{\sqrt [3]{-b}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1864, 31, 617, 204} \[ \frac {2 C \tan ^{-1}\left (\frac {1-\sqrt [3]{-b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{-b}}-\frac {C \log \left (\sqrt [3]{-b} x+2\right )}{\sqrt [3]{-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 1864

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, With[{q = (-a)^(1/3)/(-b)^(1/3)}, Dist[C/b, Int[1/(q + x), x], x] + Dist[(B + C*q)/b, Int[1/(q^2 - q*x +
x^2), x], x]] /; EqQ[A*(-b)^(2/3) - (-a)^(1/3)*(-b)^(1/3)*B - 2*(-a)^(2/3)*C, 0]] /; FreeQ[{a, b}, x] && Poly
Q[P2, x, 2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 99, normalized size = 1.74 \[ \frac {C \left (-b^{2/3} \log \left (b^{2/3} x^2+2 \sqrt [3]{b} x+4\right )+2 b^{2/3} \log \left (2-\sqrt [3]{b} x\right )-2 \sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{b} x+1}{\sqrt {3}}\right )+(-b)^{2/3} \log \left (8-b x^3\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 182, normalized size = 3.19 \[ \left [\frac {\sqrt {\frac {1}{3}} C b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {b x^{3} - 6 \, \sqrt {\frac {1}{3}} {\left (b x^{2} - \left (-b\right )^{\frac {2}{3}} x + 2 \, \left (-b\right )^{\frac {1}{3}}\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 6 \, \left (-b\right )^{\frac {1}{3}} x + 4}{b x^{3} - 8}\right ) + C \left (-b\right )^{\frac {2}{3}} \log \left (b x - 2 \, \left (-b\right )^{\frac {2}{3}}\right )}{b}, -\frac {2 \, \sqrt {\frac {1}{3}} C b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {2}{3}} x - \left (-b\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}\right ) - C \left (-b\right )^{\frac {2}{3}} \log \left (b x - 2 \, \left (-b\right )^{\frac {2}{3}}\right )}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(1/3)*C*b*sqrt((-b)^(1/3)/b)*log((b*x^3 - 6*sqrt(1/3)*(b*x^2 - (-b)^(2/3)*x + 2*(-b)^(1/3))*sqrt((-b)^(1
/3)/b) + 6*(-b)^(1/3)*x + 4)/(b*x^3 - 8)) + C*(-b)^(2/3)*log(b*x - 2*(-b)^(2/3)))/b, -(2*sqrt(1/3)*C*b*sqrt(-(
-b)^(1/3)/b)*arctan(sqrt(1/3)*((-b)^(2/3)*x - (-b)^(1/3))*sqrt(-(-b)^(1/3)/b)) - C*(-b)^(2/3)*log(b*x - 2*(-b)
^(2/3)))/b]

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giac [B]  time = 0.31, size = 91, normalized size = 1.60 \[ -\frac {2 \, \sqrt {3} C {\left | b \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (x + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{3 \, b} + \frac {1}{3} \, {\left (\frac {C \left (-b\right )^{\frac {2}{3}}}{b} - \frac {C}{b^{\frac {1}{3}}}\right )} \log \left (x^{2} + \frac {2 \, x}{b^{\frac {1}{3}}} + \frac {4}{b^{\frac {2}{3}}}\right ) + \frac {{\left (2 \, C + \frac {C \left (-b\right )^{\frac {2}{3}}}{b^{\frac {2}{3}}}\right )} \log \left ({\left | x - \frac {2}{b^{\frac {1}{3}}} \right |}\right )}{3 \, b^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(3)*C*abs(b)^(2/3)*arctan(1/3*sqrt(3)*b^(1/3)*(x + 1/b^(1/3)))/b + 1/3*(C*(-b)^(2/3)/b - C/b^(1/3))*l
og(x^2 + 2*x/b^(1/3) + 4/b^(2/3)) + 1/3*(2*C + C*(-b)^(2/3)/b^(2/3))*log(abs(x - 2/b^(1/3)))/b^(1/3)

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maple [B]  time = 0.05, size = 122, normalized size = 2.14 \[ -\frac {8^{\frac {1}{3}} \sqrt {3}\, C \arctan \left (\frac {\sqrt {3}\, \left (\frac {8^{\frac {2}{3}} x}{4 \left (\frac {1}{b}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{3 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}+\frac {8^{\frac {1}{3}} C \ln \left (x -8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}-\frac {8^{\frac {1}{3}} C \ln \left (x^{2}+8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}} x +8^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}+\frac {\left (-b \right )^{\frac {2}{3}} C \ln \left (b \,x^{3}-8\right )}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*C/b*8^(1/3)/(1/b)^(2/3)*ln(x-8^(1/3)*(1/b)^(1/3))-1/6*C/b*8^(1/3)/(1/b)^(2/3)*ln(x^2+8^(1/3)*(1/b)^(1/3)*x
+8^(2/3)*(1/b)^(2/3))-1/3*C/b*8^(1/3)/(1/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(1/4*8^(2/3)/(1/b)^(1/3)*x+1))+1/
3*C*(-b)^(2/3)/b*ln(b*x^3-8)

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maxima [B]  time = 2.94, size = 122, normalized size = 2.14 \[ \frac {{\left (C \left (-b\right )^{\frac {2}{3}} - C b^{\frac {2}{3}}\right )} \log \left (b^{\frac {2}{3}} x^{2} + 2 \, b^{\frac {1}{3}} x + 4\right )}{3 \, b} + \frac {{\left (C \left (-b\right )^{\frac {2}{3}} + 2 \, C b^{\frac {2}{3}}\right )} \log \left (\frac {b^{\frac {1}{3}} x - 2}{b^{\frac {1}{3}}}\right )}{3 \, b} + \frac {2 \, \sqrt {3} {\left (C \left (-b\right )^{\frac {2}{3}} b^{\frac {4}{3}} - {\left (C \left (-b\right )^{\frac {2}{3}} b^{\frac {1}{3}} + 3 \, C b\right )} b\right )} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {2}{3}} x + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{9 \, b^{\frac {7}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(C*(-b)^(2/3) - C*b^(2/3))*log(b^(2/3)*x^2 + 2*b^(1/3)*x + 4)/b + 1/3*(C*(-b)^(2/3) + 2*C*b^(2/3))*log((b^
(1/3)*x - 2)/b^(1/3))/b + 2/9*sqrt(3)*(C*(-b)^(2/3)*b^(4/3) - (C*(-b)^(2/3)*b^(1/3) + 3*C*b)*b)*arctan(1/3*sqr
t(3)*(b^(2/3)*x + b^(1/3))/b^(1/3))/b^(7/3)

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mupad [B]  time = 5.27, size = 176, normalized size = 3.09 \[ \sum _{k=1}^3\ln \left (\frac {8\,C^2}{{\left (-b\right )}^{5/3}}+\mathrm {root}\left (27\,b^3\,z^3-27\,C\,{\left (-b\right )}^{8/3}\,z^2-9\,C^2\,{\left (-b\right )}^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,\left (-\frac {\mathrm {root}\left (27\,b^3\,z^3-27\,C\,{\left (-b\right )}^{8/3}\,z^2-9\,C^2\,{\left (-b\right )}^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,72}{b}+\frac {48\,C}{{\left (-b\right )}^{4/3}}+\frac {24\,C\,x}{b}\right )-\frac {8\,C^2\,x}{{\left (-b\right )}^{4/3}}\right )\,\mathrm {root}\left (27\,b^3\,z^3-27\,C\,{\left (-b\right )}^{8/3}\,z^2-9\,C^2\,{\left (-b\right )}^{7/3}\,z-9\,C^3\,b^2,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((8*C^2)/(-b)^(5/3) + root(27*b^3*z^3 - 27*C*(-b)^(8/3)*z^2 - 9*C^2*(-b)^(7/3)*z - 9*C^3*b^2, z, k)*
((48*C)/(-b)^(4/3) - (72*root(27*b^3*z^3 - 27*C*(-b)^(8/3)*z^2 - 9*C^2*(-b)^(7/3)*z - 9*C^3*b^2, z, k))/b + (2
4*C*x)/b) - (8*C^2*x)/(-b)^(4/3))*root(27*b^3*z^3 - 27*C*(-b)^(8/3)*z^2 - 9*C^2*(-b)^(7/3)*z - 9*C^3*b^2, z, k
), k, 1, 3)

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sympy [A]  time = 0.99, size = 58, normalized size = 1.02 \[ \operatorname {RootSum} {\left (3 t^{3} b^{2} - 3 t^{2} C b \left (- b\right )^{\frac {2}{3}} + t C^{2} \left (- b\right )^{\frac {4}{3}} - C^{3} b, \left (t \mapsto t \log {\left (- \frac {3 t}{C} + x + \frac {\left (- b\right )^{\frac {2}{3}}}{b} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(3*_t**3*b**2 - 3*_t**2*C*b*(-b)**(2/3) + _t*C**2*(-b)**(4/3) - C**3*b, Lambda(_t, _t*log(-3*_t/C + x +
(-b)**(2/3)/b)))

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