3.29.97 \(\int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx\)

Optimal. Leaf size=319 \[ -\frac {i \left (\sqrt {3}-i\right ) \log \left (\sqrt [3]{-1} \sqrt [3]{a^3 x^3-b^2 x^2}+\sqrt [3]{a} x \sqrt [3]{a^2-b}\right )}{2 \sqrt [3]{a} b \sqrt [3]{a^2-b}}+\frac {\sqrt {-3+3 i \sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{a} x \sqrt [3]{a^2-b}}{\sqrt [3]{a} x \sqrt [3]{a^2-b}-2 \sqrt [3]{-1} \sqrt [3]{a^3 x^3-b^2 x^2}}\right )}{\sqrt {2} \sqrt [3]{a} b \sqrt [3]{a^2-b}}+\frac {\left (1+i \sqrt {3}\right ) \log \left ((-1)^{2/3} \left (a^3 x^3-b^2 x^2\right )^{2/3}+a^{2/3} x^2 \left (a^2-b\right )^{2/3}-\sqrt [3]{-1} \sqrt [3]{a} x \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3-b^2 x^2}\right )}{4 \sqrt [3]{a} b \sqrt [3]{a^2-b}} \]

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Rubi [A]  time = 0.14, antiderivative size = 291, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2056, 91} \begin {gather*} -\frac {x^{2/3} \sqrt [3]{a^3 x-b^2} \log (a x-b)}{2 \sqrt [3]{a} b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3-b^2 x^2}}+\frac {3 x^{2/3} \sqrt [3]{a^3 x-b^2} \log \left (\frac {\sqrt [3]{a^3 x-b^2}}{\sqrt [3]{a} \sqrt [3]{a^2-b}}-\sqrt [3]{x}\right )}{2 \sqrt [3]{a} b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3-b^2 x^2}}+\frac {\sqrt {3} x^{2/3} \sqrt [3]{a^3 x-b^2} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3 x-b^2}}{\sqrt {3} \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{a^2-b}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{a} b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3-b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 91

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 56, normalized size = 0.18 \begin {gather*} -\frac {3 x \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {\left (a^3-a b\right ) x}{a^3 x-b^2}\right )}{b \sqrt [3]{x^2 \left (a^3 x-b^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Hypergeometric2F1[1/3, 1, 4/3, ((a^3 - a*b)*x)/(-b^2 + a^3*x)])/(b*(x^2*(-b^2 + a^3*x))^(1/3))

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IntegrateAlgebraic [A]  time = 1.97, size = 365, normalized size = 1.14 \begin {gather*} \frac {\sqrt {-3+3 i \sqrt {3}} \tan ^{-1}\left (\frac {3 \sqrt [3]{a} \sqrt [3]{a^2-b} x}{\sqrt {3} \sqrt [3]{a} \sqrt [3]{a^2-b} x-3 i \sqrt [3]{-b^2 x^2+a^3 x^3}-\sqrt {3} \sqrt [3]{-b^2 x^2+a^3 x^3}}\right )}{\sqrt {2} \sqrt [3]{a} \sqrt [3]{a^2-b} b}-\frac {i \left (-i+\sqrt {3}\right ) \log \left (2 \sqrt [3]{a} \sqrt [3]{a^2-b} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{-b^2 x^2+a^3 x^3}\right )}{2 \sqrt [3]{a} \sqrt [3]{a^2-b} b}+\frac {\left (1+i \sqrt {3}\right ) \log \left (-2 i a^{2/3} \left (a^2-b\right )^{2/3} x^2+\sqrt [3]{a} \sqrt [3]{a^2-b} \left (i x-\sqrt {3} x\right ) \sqrt [3]{-b^2 x^2+a^3 x^3}+\left (i+\sqrt {3}\right ) \left (-b^2 x^2+a^3 x^3\right )^{2/3}\right )}{4 \sqrt [3]{a} \sqrt [3]{a^2-b} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-3 + (3*I)*Sqrt[3]]*ArcTan[(3*a^(1/3)*(a^2 - b)^(1/3)*x)/(Sqrt[3]*a^(1/3)*(a^2 - b)^(1/3)*x - (3*I)*(-(b
^2*x^2) + a^3*x^3)^(1/3) - Sqrt[3]*(-(b^2*x^2) + a^3*x^3)^(1/3))])/(Sqrt[2]*a^(1/3)*(a^2 - b)^(1/3)*b) - ((I/2
)*(-I + Sqrt[3])*Log[2*a^(1/3)*(a^2 - b)^(1/3)*x + (1 + I*Sqrt[3])*(-(b^2*x^2) + a^3*x^3)^(1/3)])/(a^(1/3)*(a^
2 - b)^(1/3)*b) + ((1 + I*Sqrt[3])*Log[(-2*I)*a^(2/3)*(a^2 - b)^(2/3)*x^2 + a^(1/3)*(a^2 - b)^(1/3)*(I*x - Sqr
t[3]*x)*(-(b^2*x^2) + a^3*x^3)^(1/3) + (I + Sqrt[3])*(-(b^2*x^2) + a^3*x^3)^(2/3)])/(4*a^(1/3)*(a^2 - b)^(1/3)
*b)

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fricas [A]  time = 0.60, size = 549, normalized size = 1.72 \begin {gather*} \left [\frac {\sqrt {3} {\left (a^{3} - a b\right )} \sqrt {-\frac {1}{{\left (a^{3} - a b\right )}^{\frac {2}{3}}}} \log \left (-\frac {2 \, b^{2} x - {\left (3 \, a^{3} - a b\right )} x^{2} + 3 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {2}{3}} x + \sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {4}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )} x - 2 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}} {\left (a^{3} - a b\right )}^{\frac {2}{3}}\right )} \sqrt {-\frac {1}{{\left (a^{3} - a b\right )}^{\frac {2}{3}}}}}{a x^{2} - b x}\right ) + 2 \, {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} x - {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (\frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {1}{3}} x + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, {\left (a^{3} b - a b^{2}\right )}}, \frac {2 \, \sqrt {3} {\left (a^{3} - a b\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {1}{3}} x + 2 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - a b\right )}^{\frac {1}{3}} x}\right ) + 2 \, {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} x - {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (\frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {1}{3}} x + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, {\left (a^{3} b - a b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a x -b \right ) \left (a^{3} x^{3}-b^{2} x^{2}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a x - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a^3*x^3 - b^2*x^2)^(1/3)*(a*x - b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {1}{{\left (a^3\,x^3-b^2\,x^2\right )}^{1/3}\,\left (b-a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{x^{2} \left (a^{3} x - b^{2}\right )} \left (a x - b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((x**2*(a**3*x - b**2))**(1/3)*(a*x - b)), x)

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