3.365 \(\int \frac {1}{-1+a+b x^3} \, dx\)

Optimal. Leaf size=139 \[ -\frac {\log \left (\sqrt [3]{1-a} \sqrt [3]{b} x+(1-a)^{2/3}+b^{2/3} x^2\right )}{6 (1-a)^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{1-a}-\sqrt [3]{b} x\right )}{3 (1-a)^{2/3} \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{1-a}}+1}{\sqrt {3}}\right )}{\sqrt {3} (1-a)^{2/3} \sqrt [3]{b}} \]

[Out]

1/3*ln((1-a)^(1/3)-b^(1/3)*x)/(1-a)^(2/3)/b^(1/3)-1/6*ln((1-a)^(2/3)+(1-a)^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(1-a)^
(2/3)/b^(1/3)-1/3*arctan(1/3*(1+2*b^(1/3)*x/(1-a)^(1/3))*3^(1/2))/(1-a)^(2/3)/b^(1/3)*3^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {200, 31, 634, 617, 204, 628} \[ -\frac {\log \left (\sqrt [3]{1-a} \sqrt [3]{b} x+(1-a)^{2/3}+b^{2/3} x^2\right )}{6 (1-a)^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{1-a}-\sqrt [3]{b} x\right )}{3 (1-a)^{2/3} \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{1-a}}+1}{\sqrt {3}}\right )}{\sqrt {3} (1-a)^{2/3} \sqrt [3]{b}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + a + b*x^3)^(-1),x]

[Out]

-(ArcTan[(1 + (2*b^(1/3)*x)/(1 - a)^(1/3))/Sqrt[3]]/(Sqrt[3]*(1 - a)^(2/3)*b^(1/3))) + Log[(1 - a)^(1/3) - b^(
1/3)*x]/(3*(1 - a)^(2/3)*b^(1/3)) - Log[(1 - a)^(2/3) + (1 - a)^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(6*(1 - a)^(2/3
)*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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 101, normalized size = 0.73 \[ \frac {-\log \left (-\sqrt [3]{a-1} \sqrt [3]{b} x+(a-1)^{2/3}+b^{2/3} x^2\right )+2 \log \left (\sqrt [3]{a-1}+\sqrt [3]{b} x\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a-1}}-1}{\sqrt {3}}\right )}{6 (a-1)^{2/3} \sqrt [3]{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + a + b*x^3)^(-1),x]

[Out]

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

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fricas [A]  time = 0.94, size = 446, normalized size = 3.21 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (a - 1\right )} b \sqrt {-\frac {\left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, {\left (a - 1\right )} b x^{3} - 3 \, \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}} {\left (a - 1\right )} x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (a - 1\right )} b x^{2} + \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} x - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}} {\left (a - 1\right )}\right )} \sqrt {-\frac {\left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}}}{b}} + 2 \, a - 1}{b x^{3} + a - 1}\right ) - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} \log \left ({\left (a - 1\right )} b x^{2} - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} x + \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}} {\left (a - 1\right )}\right ) + 2 \, \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} \log \left ({\left (a - 1\right )} b x + \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} - 2 \, a + 1\right )} b}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (a - 1\right )} b \sqrt {\frac {\left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} x - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}} {\left (a - 1\right )}\right )} \sqrt {\frac {\left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}}}{b}}}{a^{2} - 2 \, a + 1}\right ) - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} \log \left ({\left (a - 1\right )} b x^{2} - \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} x + \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {1}{3}} {\left (a - 1\right )}\right ) + 2 \, \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}} \log \left ({\left (a - 1\right )} b x + \left ({\left (a^{2} - 2 \, a + 1\right )} b\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} - 2 \, a + 1\right )} b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 142, normalized size = 1.02 \[ \frac {{\left (-a b^{2} + b^{2}\right )}^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a - 1}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a - 1}{b}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} a b - \sqrt {3} b} + \frac {{\left (-a b^{2} + b^{2}\right )}^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a - 1}{b}\right )^{\frac {1}{3}} + \left (-\frac {a - 1}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a b - b\right )}} - \frac {\left (-\frac {a - 1}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a - 1}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (a - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 105, normalized size = 0.76 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a -1}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a -1}{b}\right )^{\frac {2}{3}} b}+\frac {\ln \left (x +\left (\frac {a -1}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a -1}{b}\right )^{\frac {2}{3}} b}-\frac {\ln \left (x^{2}-\left (\frac {a -1}{b}\right )^{\frac {1}{3}} x +\left (\frac {a -1}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a -1}{b}\right )^{\frac {2}{3}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b/((-1+a)/b)^(2/3)*ln(x+((-1+a)/b)^(1/3))-1/6/b/((-1+a)/b)^(2/3)*ln(x^2-((-1+a)/b)^(1/3)*x+((-1+a)/b)^(2/3
))+1/3/b/((-1+a)/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/((-1+a)/b)^(1/3)*x-1))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1.0>0)', see `assume?` for m
ore details)Is a-1.0 positive or negative?

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mupad [B]  time = 1.23, size = 137, normalized size = 0.99 \[ \frac {\ln \left (a+b^{1/3}\,x\,{\left (a-1\right )}^{2/3}-1\right )}{3\,b^{1/3}\,{\left (a-1\right )}^{2/3}}+\frac {\ln \left (3\,b^2\,x+\frac {\left (9\,a\,b^2-9\,b^2\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}\,{\left (a-1\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}\,{\left (a-1\right )}^{2/3}}-\frac {\ln \left (3\,b^2\,x-\frac {\left (9\,a\,b^2-9\,b^2\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}\,{\left (a-1\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}\,{\left (a-1\right )}^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^3 - 1),x)

[Out]

log(a + b^(1/3)*x*(a - 1)^(2/3) - 1)/(3*b^(1/3)*(a - 1)^(2/3)) + (log(3*b^2*x + ((9*a*b^2 - 9*b^2)*(3^(1/2)*1i
- 1))/(6*b^(1/3)*(a - 1)^(2/3)))*(3^(1/2)*1i - 1))/(6*b^(1/3)*(a - 1)^(2/3)) - (log(3*b^2*x - ((9*a*b^2 - 9*b
^2)*(3^(1/2)*1i + 1))/(6*b^(1/3)*(a - 1)^(2/3)))*(3^(1/2)*1i + 1))/(6*b^(1/3)*(a - 1)^(2/3))

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sympy [A]  time = 0.37, size = 32, normalized size = 0.23 \[ \operatorname {RootSum} {\left (t^{3} \left (27 a^{2} b - 54 a b + 27 b\right ) - 1, \left (t \mapsto t \log {\left (3 t a - 3 t + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_t**3*(27*a**2*b - 54*a*b + 27*b) - 1, Lambda(_t, _t*log(3*_t*a - 3*_t + x)))

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