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

Optimal. Leaf size=48 \[ \frac {C \log \left (\sqrt [3]{b} x+2\right )}{\sqrt [3]{b}}-\frac {2 C \tan ^{-1}\left (\frac {1-\sqrt [3]{b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}} \]

[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 = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1863, 31, 617, 204} \[ \frac {C \log \left (\sqrt [3]{b} x+2\right )}{\sqrt [3]{b}}-\frac {2 C \tan ^{-1}\left (\frac {1-\sqrt [3]{b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}} \]

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 1863

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] && PolyQ[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 = 0.02, size = 76, normalized size = 1.58 \[ \frac {C \left (-\log \left (b^{2/3} x^2-2 \sqrt [3]{b} x+4\right )+\log \left (b x^3+8\right )+2 \log \left (\sqrt [3]{b} x+2\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{b} x-1}{\sqrt {3}}\right )\right )}{3 \sqrt [3]{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 134, normalized size = 2.79 \[ \left [\frac {\sqrt {\frac {1}{3}} C b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {b x^{3} + 6 \, \sqrt {\frac {1}{3}} {\left (b x^{2} + b^{\frac {2}{3}} x - 2 \, b^{\frac {1}{3}}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} - 6 \, b^{\frac {1}{3}} x - 4}{b x^{3} + 8}\right ) + C b^{\frac {2}{3}} \log \left (b x + 2 \, b^{\frac {2}{3}}\right )}{b}, \frac {2 \, \sqrt {\frac {1}{3}} C b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {2}{3}} x - b^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}}}\right ) + C b^{\frac {2}{3}} \log \left (b x + 2 \, b^{\frac {2}{3}}\right )}{b}\right ] \]

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

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giac [B]  time = 0.42, size = 115, normalized size = 2.40 \[ \frac {2}{3} \, \sqrt {3} C \left (-\frac {1}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (x + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {1}{b}\right )^{\frac {1}{3}}}\right ) - \frac {1}{3} \, {\left (C b^{\frac {2}{3}} \left (-\frac {1}{b}\right )^{\frac {2}{3}} + 2 \, C\right )} \left (-\frac {1}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - 2 \, \left (-\frac {1}{b}\right )^{\frac {1}{3}} \right |}\right ) + \frac {1}{3} \, {\left (C \left (-\frac {1}{b}\right )^{\frac {1}{3}} + \frac {C}{b^{\frac {1}{3}}}\right )} \log \left (x^{2} + 2 \, x \left (-\frac {1}{b}\right )^{\frac {1}{3}} + 4 \, \left (-\frac {1}{b}\right )^{\frac {2}{3}}\right ) \]

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="giac")

[Out]

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

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maple [B]  time = 0.06, size = 117, normalized size = 2.44 \[ \frac {C \ln \left (b \,x^{3}+8\right )}{3 b^{\frac {1}{3}}}+\frac {8^{\frac {1}{3}} \sqrt {3}\, C \arctan \left (\frac {\sqrt {3}\, \left (\frac {8^{\frac {2}{3}} x}{4 \left (\frac {1}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}+\frac {8^{\frac {1}{3}} C \ln \left (x +8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}-\frac {8^{\frac {1}{3}} C \ln \left (x^{2}-8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}} x +8^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {1}{b}\right )^{\frac {2}{3}} b} \]

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^(1/3)*ln(b*x^3+8)

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maxima [A]  time = 2.99, size = 47, normalized size = 0.98 \[ \frac {2 \, \sqrt {3} C \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {2}{3}} x - b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {1}{3}}} + \frac {C \log \left (\frac {b^{\frac {1}{3}} x + 2}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} \]

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="maxima")

[Out]

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

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mupad [B]  time = 5.14, size = 147, normalized size = 3.06 \[ \sum _{k=1}^3\ln \left (-\frac {\left (C-\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,b^{1/3}\,3\right )\,\left (-C+\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,b^{1/3}\,3+C\,b^{1/3}\,x\right )\,8}{b^{5/3}}\right )\,\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right ) \]

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 - 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)*b^(1/3))*(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)*b^(1/3) - C + C*b^(1/3)*x))/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), k, 1, 3)

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sympy [A]  time = 0.63, size = 58, normalized size = 1.21 \[ \operatorname {RootSum} {\left (3 t^{3} b^{\frac {5}{3}} - 3 t^{2} C b^{\frac {4}{3}} + t C^{2} b - C^{3} b^{\frac {2}{3}}, \left (t \mapsto t \log {\left (x + \frac {3 t \sqrt [3]{b} - C}{C \sqrt [3]{b}} \right )} \right )\right )} \]

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**(5/3) - 3*_t**2*C*b**(4/3) + _t*C**2*b - C**3*b**(2/3), Lambda(_t, _t*log(x + (3*_t*b**(1/3
) - C)/(C*b**(1/3)))))

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