3.47 \(\int \frac {-\sqrt [3]{\frac {a}{b}} B+2 (\frac {a}{b})^{2/3} C+B x+C x^2}{a-b x^3} \, dx\)

Optimal. Leaf size=75 \[ -\frac {2 \left (\frac {a}{b}\right )^{2/3} \left (B-C \sqrt [3]{\frac {a}{b}}\right ) \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{\frac {a}{b}}}+1}{\sqrt {3}}\right )}{\sqrt {3} a}-\frac {C \log \left (\sqrt [3]{\frac {a}{b}}-x\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1869, 31, 617, 204} \[ -\frac {2 \left (\frac {a}{b}\right )^{2/3} \left (B-C \sqrt [3]{\frac {a}{b}}\right ) \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{\frac {a}{b}}}+1}{\sqrt {3}}\right )}{\sqrt {3} a}-\frac {C \log \left (\sqrt [3]{\frac {a}{b}}-x\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(-((a/b)^(1/3)*B) + 2*(a/b)^(2/3)*C + B*x + C*x^2)/(a - b*x^3),x]

[Out]

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

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 1869

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/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 + (-(a/b))^(1/3)*B - 2*(-(a/b))^(2/3)*C, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rubi steps

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

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Mathematica [B]  time = 0.32, size = 244, normalized size = 3.25 \[ \frac {\sqrt [3]{b} \left (a^{2/3} B+\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (2 C \sqrt [3]{\frac {a}{b}}-B\right )\right ) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-2 \sqrt [3]{b} \left (a^{2/3} B+\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (2 C \sqrt [3]{\frac {a}{b}}-B\right )\right ) \log \left (\sqrt [3]{a}-\sqrt [3]{b} x\right )-2 \sqrt {3} \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (B-2 C \sqrt [3]{\frac {a}{b}}\right )+\sqrt [3]{a} B\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )-2 a C \log \left (a-b x^3\right )}{6 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[(-((a/b)^(1/3)*B) + 2*(a/b)^(2/3)*C + B*x + C*x^2)/(a - b*x^3),x]

[Out]

(-2*Sqrt[3]*a^(1/3)*b^(1/3)*(a^(1/3)*B + (a/b)^(1/3)*b^(1/3)*(B - 2*(a/b)^(1/3)*C))*ArcTan[(1 + (2*b^(1/3)*x)/
a^(1/3))/Sqrt[3]] - 2*b^(1/3)*(a^(2/3)*B + a^(1/3)*(a/b)^(1/3)*b^(1/3)*(-B + 2*(a/b)^(1/3)*C))*Log[a^(1/3) - b
^(1/3)*x] + b^(1/3)*(a^(2/3)*B + a^(1/3)*(a/b)^(1/3)*b^(1/3)*(-B + 2*(a/b)^(1/3)*C))*Log[a^(2/3) + a^(1/3)*b^(
1/3)*x + b^(2/3)*x^2] - 2*a*C*Log[a - b*x^3])/(6*a*b)

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fricas [B]  time = 3.20, size = 450, normalized size = 6.00 \[ \left [-\frac {C \log \left (x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) - \sqrt {\frac {1}{3}} \sqrt {\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} - B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} - C^{2} a}{a}} \log \left (-\frac {C^{3} a^{2} - B^{3} a b + 2 \, {\left (C^{3} a b - B^{3} b^{2}\right )} x^{3} - 3 \, {\left (C^{3} a b - B^{3} b^{2}\right )} x \left (\frac {a}{b}\right )^{\frac {2}{3}} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, B C a b x^{2} - B^{2} a b x - C^{2} a^{2} + {\left (2 \, B^{2} b^{2} x^{2} - C^{2} a b x - B C a b\right )} \left (\frac {a}{b}\right )^{\frac {2}{3}} + {\left (2 \, C^{2} a b x^{2} - B C a b x - B^{2} a b\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \sqrt {\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} - B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} - C^{2} a}{a}}}{b x^{3} - a}\right )}{b}, -\frac {2 \, \sqrt {\frac {1}{3}} \sqrt {-\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} - B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} - C^{2} a}{a}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (2 \, B^{2} b x + C^{2} a + {\left (2 \, C^{2} b x + B C b\right )} \left (\frac {a}{b}\right )^{\frac {2}{3}} + {\left (2 \, B C b x + B^{2} b\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} - B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} - C^{2} a}{a}}}{C^{3} a - B^{3} b}\right ) + C \log \left (x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(a/b)^(1/3)*B+2*(a/b)^(2/3)*C+B*x+C*x^2)/(-b*x^3+a),x, algorithm="fricas")

[Out]

[-(C*log(x - (a/b)^(1/3)) - sqrt(1/3)*sqrt((2*B*C*b*(a/b)^(2/3) - B^2*b*(a/b)^(1/3) - C^2*a)/a)*log(-(C^3*a^2
- B^3*a*b + 2*(C^3*a*b - B^3*b^2)*x^3 - 3*(C^3*a*b - B^3*b^2)*x*(a/b)^(2/3) + 3*sqrt(1/3)*(2*B*C*a*b*x^2 - B^2
*a*b*x - C^2*a^2 + (2*B^2*b^2*x^2 - C^2*a*b*x - B*C*a*b)*(a/b)^(2/3) + (2*C^2*a*b*x^2 - B*C*a*b*x - B^2*a*b)*(
a/b)^(1/3))*sqrt((2*B*C*b*(a/b)^(2/3) - B^2*b*(a/b)^(1/3) - C^2*a)/a))/(b*x^3 - a)))/b, -(2*sqrt(1/3)*sqrt(-(2
*B*C*b*(a/b)^(2/3) - B^2*b*(a/b)^(1/3) - C^2*a)/a)*arctan(-sqrt(1/3)*(2*B^2*b*x + C^2*a + (2*C^2*b*x + B*C*b)*
(a/b)^(2/3) + (2*B*C*b*x + B^2*b)*(a/b)^(1/3))*sqrt(-(2*B*C*b*(a/b)^(2/3) - B^2*b*(a/b)^(1/3) - C^2*a)/a)/(C^3
*a - B^3*b)) + C*log(x - (a/b)^(1/3)))/b]

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giac [A]  time = 0.18, size = 125, normalized size = 1.67 \[ \frac {2 \, \sqrt {3} {\left (C a b - \left (a b^{2}\right )^{\frac {2}{3}} B\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b^{2}} - \frac {{\left (C b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + B b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} - \left (a b^{2}\right )^{\frac {1}{3}} B b + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} C\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(a/b)^(1/3)*B+2*(a/b)^(2/3)*C+B*x+C*x^2)/(-b*x^3+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 124, normalized size = 1.65 \[ -\frac {2 \sqrt {3}\, B \arctan \left (\frac {\left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}+1\right ) \sqrt {3}}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {2 \sqrt {3}\, C \arctan \left (\frac {\left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}+1\right ) \sqrt {3}}{3}\right )}{3 b}-\frac {2 C \ln \left (x -\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b}+\frac {C \ln \left (x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 b}-\frac {C \ln \left (b \,x^{3}-a \right )}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-(a/b)^(1/3)*B+2*(a/b)^(2/3)*C+B*x+C*x^2)/(-b*x^3+a),x)

[Out]

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

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maxima [A]  time = 3.14, size = 78, normalized size = 1.04 \[ -\frac {C \log \left (x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b} - \frac {2 \, \sqrt {3} {\left (C a + {\left (3 \, B \left (\frac {a}{b}\right )^{\frac {2}{3}} - \frac {4 \, C a}{b}\right )} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(a/b)^(1/3)*B+2*(a/b)^(2/3)*C+B*x+C*x^2)/(-b*x^3+a),x, algorithm="maxima")

[Out]

-C*log(x - (a/b)^(1/3))/b - 2/9*sqrt(3)*(C*a + (3*B*(a/b)^(2/3) - 4*C*a/b)*b)*arctan(1/3*sqrt(3)*(2*x + (a/b)^
(1/3))/(a/b)^(1/3))/(a*b)

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mupad [B]  time = 6.36, size = 435, normalized size = 5.80 \[ \sum _{k=1}^3\ln \left (-\frac {C^2\,a+B^2\,b\,{\left (\frac {a}{b}\right )}^{1/3}-2\,B\,C\,b\,{\left (\frac {a}{b}\right )}^{2/3}}{b^3}-\frac {\mathrm {root}\left (27\,a^2\,b^3\,z^3+27\,C\,a^2\,b^2\,z^2-18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^3\,a^2,z,k\right )\,\left (6\,C\,a+\mathrm {root}\left (27\,a^2\,b^3\,z^3+27\,C\,a^2\,b^2\,z^2-18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^3\,a^2,z,k\right )\,a\,b\,9-3\,B\,b\,x\,{\left (\frac {a}{b}\right )}^{1/3}+6\,C\,b\,x\,{\left (\frac {a}{b}\right )}^{2/3}\right )}{b^2}+\frac {x\,\left (B^2-2\,C^2\,{\left (\frac {a}{b}\right )}^{2/3}+B\,C\,{\left (\frac {a}{b}\right )}^{1/3}\right )}{b^2}\right )\,\mathrm {root}\left (27\,a^2\,b^3\,z^3+27\,C\,a^2\,b^2\,z^2-18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^3\,a^2,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x + C*x^2 - B*(a/b)^(1/3) + 2*C*(a/b)^(2/3))/(a - b*x^3),x)

[Out]

symsum(log((x*(B^2 - 2*C^2*(a/b)^(2/3) + B*C*(a/b)^(1/3)))/b^2 - (root(27*a^2*b^3*z^3 + 27*C*a^2*b^2*z^2 - 18*
B*C*a*b^2*z*(a/b)^(2/3) + 9*B^2*a*b^2*z*(a/b)^(1/3) + 9*C^2*a^2*b*z - 18*B*C^2*a*b*(a/b)^(2/3) + 9*B^2*C*a*b*(
a/b)^(1/3) + 9*C^3*a^2, z, k)*(6*C*a + 9*root(27*a^2*b^3*z^3 + 27*C*a^2*b^2*z^2 - 18*B*C*a*b^2*z*(a/b)^(2/3) +
9*B^2*a*b^2*z*(a/b)^(1/3) + 9*C^2*a^2*b*z - 18*B*C^2*a*b*(a/b)^(2/3) + 9*B^2*C*a*b*(a/b)^(1/3) + 9*C^3*a^2, z
, k)*a*b - 3*B*b*x*(a/b)^(1/3) + 6*C*b*x*(a/b)^(2/3)))/b^2 - (C^2*a + B^2*b*(a/b)^(1/3) - 2*B*C*b*(a/b)^(2/3))
/b^3)*root(27*a^2*b^3*z^3 + 27*C*a^2*b^2*z^2 - 18*B*C*a*b^2*z*(a/b)^(2/3) + 9*B^2*a*b^2*z*(a/b)^(1/3) + 9*C^2*
a^2*b*z - 18*B*C^2*a*b*(a/b)^(2/3) + 9*B^2*C*a*b*(a/b)^(1/3) + 9*C^3*a^2, z, k), k, 1, 3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(a/b)**(1/3)*B+2*(a/b)**(2/3)*C+B*x+C*x**2)/(-b*x**3+a),x)

[Out]

Timed out

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