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

Optimal. Leaf size=71 \[ \frac {C \log \left (\sqrt [3]{\frac {a}{b}}+x\right )}{b}-\frac {2 \left (\frac {a}{b}\right )^{2/3} \left (C \sqrt [3]{\frac {a}{b}}+B\right ) \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{\frac {a}{b}}}}{\sqrt {3}}\right )}{\sqrt {3} a} \]

[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 = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1867, 31, 617, 204} \[ \frac {C \log \left (\sqrt [3]{\frac {a}{b}}+x\right )}{b}-\frac {2 \left (\frac {a}{b}\right )^{2/3} \left (C \sqrt [3]{\frac {a}{b}}+B\right ) \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{\frac {a}{b}}}}{\sqrt {3}}\right )}{\sqrt {3} a} \]

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 1867

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 = 0.33, size = 247, normalized size = 3.48 \[ \frac {\sqrt [3]{b} \left (a^{2/3} B-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (2 C \sqrt [3]{\frac {a}{b}}+B\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \sqrt [3]{b} \left (\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (2 C \sqrt [3]{\frac {a}{b}}+B\right )-a^{2/3} B\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt {3} \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{b} \sqrt [3]{\frac {a}{b}} \left (2 C \sqrt [3]{\frac {a}{b}}+B\right )+\sqrt [3]{a} B\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} x-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )+2 a C \log \left (a+b x^3\right )}{6 a b} \]

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[(-a^(1/3) + 2*b^(1/3
)*x)/(Sqrt[3]*a^(1/3))] + 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 = 3.54, size = 429, normalized size = 6.04 \[ \left [\frac {C \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) + \sqrt {\frac {1}{3}} \sqrt {-\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} + B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} + C^{2} a}{a}} \log \left (-\frac {C^{3} a^{2} + B^{3} a b - 2 \, {\left (C^{3} a b + B^{3} b^{2}\right )} x^{3} + 3 \, {\left (C^{3} a b + B^{3} b^{2}\right )} x \left (\frac {a}{b}\right )^{\frac {2}{3}} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, B C a b x^{2} - B^{2} a b x + C^{2} a^{2} - {\left (2 \, B^{2} b^{2} x^{2} + C^{2} a b x + B C a b\right )} \left (\frac {a}{b}\right )^{\frac {2}{3}} - {\left (2 \, C^{2} a b x^{2} - B C a b x - B^{2} a b\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} + B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} + C^{2} a}{a}}}{b x^{3} + a}\right )}{b}, \frac {2 \, \sqrt {\frac {1}{3}} \sqrt {\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} + B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} + C^{2} a}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, B^{2} b x - C^{2} a + {\left (2 \, C^{2} b x + B C b\right )} \left (\frac {a}{b}\right )^{\frac {2}{3}} - {\left (2 \, B C b x + B^{2} b\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \sqrt {\frac {2 \, B C b \left (\frac {a}{b}\right )^{\frac {2}{3}} + B^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}} + C^{2} a}{a}}}{C^{3} a + B^{3} b}\right ) + C \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b}\right ] \]

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 [B]  time = 0.20, size = 242, normalized size = 3.41 \[ \frac {{\left (2 \, C a b + \left (-a^{2} b^{4}\right )^{\frac {1}{3}} B\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 \, a b^{2} - \sqrt {3} \sqrt {a^{2} b^{4}} i} - \frac {{\left (C b^{2} \left (-\frac {a}{b}\right )^{\frac {2}{3}} + B b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (a b^{2}\right )^{\frac {1}{3}} B b + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} C\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{2}} + \frac {\sqrt {3} {\left ({\left (9 \, \left (-a^{2} b^{4}\right )^{\frac {1}{3}} a b^{2} - 27^{\frac {5}{6}} \left (-a^{2} b^{4}\right )^{\frac {5}{6}}\right )} B + 18 \, {\left (a^{2} b^{3} - \sqrt {3} \sqrt {a^{4} b^{6}} i\right )} C\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{54 \, a^{2} b^{4}} \]

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*C*a*b + (-a^2*b^4)^(1/3)*B)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(3*a*b^2 - sqrt(3)*sqrt(a^2*b^4)*i) -
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) + 1/54*sqrt(3)*((9*(-a^2*b^4)^(1/3)*a*b^2 - 27^(5/6)*(-a^2*b^4)^(5/6))*B + 18*(a^2*b^3
- sqrt(3)*sqrt(a^4*b^6)*i)*C)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^2*b^4)

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maple [A]  time = 0.05, size = 121, normalized size = 1.70 \[ \frac {2 \sqrt {3}\, B \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {2 \sqrt {3}\, C \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b}+\frac {2 C \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b}-\frac {C \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{3 b}+\frac {C \ln \left (b \,x^{3}+a \right )}{3 b} \]

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

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maxima [A]  time = 2.95, size = 78, normalized size = 1.10 \[ \frac {C \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b} - \frac {2 \, \sqrt {3} {\left (C a - {\left (3 \, B \left (\frac {a}{b}\right )^{\frac {2}{3}} + \frac {4 \, C a}{b}\right )} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b} \]

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 = 6.08, size = 436, normalized size = 6.14 \[ \sum _{k=1}^3\ln \left (\frac {C^2\,a+B^2\,b\,{\left (\frac {a}{b}\right )}^{1/3}+2\,B\,C\,b\,{\left (\frac {a}{b}\right )}^{2/3}}{b^3}+\frac {\mathrm {root}\left (27\,a^2\,b^3\,z^3-27\,C\,a^2\,b^2\,z^2+18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}-9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}-9\,C^3\,a^2,z,k\right )\,\left (-6\,C\,a+\mathrm {root}\left (27\,a^2\,b^3\,z^3-27\,C\,a^2\,b^2\,z^2+18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}-9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}-9\,C^3\,a^2,z,k\right )\,a\,b\,9+3\,B\,b\,x\,{\left (\frac {a}{b}\right )}^{1/3}+6\,C\,b\,x\,{\left (\frac {a}{b}\right )}^{2/3}\right )}{b^2}-\frac {x\,\left (2\,C^2\,{\left (\frac {a}{b}\right )}^{2/3}-B^2+B\,C\,{\left (\frac {a}{b}\right )}^{1/3}\right )}{b^2}\right )\,\mathrm {root}\left (27\,a^2\,b^3\,z^3-27\,C\,a^2\,b^2\,z^2+18\,B\,C\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{2/3}+9\,B^2\,a\,b^2\,z\,{\left (\frac {a}{b}\right )}^{1/3}+9\,C^2\,a^2\,b\,z-18\,B\,C^2\,a\,b\,{\left (\frac {a}{b}\right )}^{2/3}-9\,B^2\,C\,a\,b\,{\left (\frac {a}{b}\right )}^{1/3}-9\,C^3\,a^2,z,k\right ) \]

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((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 + 1
8*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)*(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 - 6*C*a + 3*B*b*x*(a/b)^(1/3) + 6*C*b*x*(a/b)^(2/3)))/b^2 - (x*(2*C^2*(a/b)^(2/3) - B^2 + 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), 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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