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

Optimal. Leaf size=69 \[ \frac {\text {RootSum}\left [\text {$\#$1}^6-2 \text {$\#$1}^3 a^3+a^6-a b^3\& ,\frac {\log \left (\sqrt [3]{a^3 x^3+b^2 x^2}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}}\& \right ]}{2 b} \]

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Rubi [B]  time = 0.39, antiderivative size = 624, normalized size of antiderivative = 9.04, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2056, 912, 91} \begin {gather*} -\frac {x^{2/3} \sqrt [3]{a^3 x+b^2} \log \left (\sqrt {b}-\sqrt {a} x\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{a^3 x^3+b^2 x^2}}-\frac {x^{2/3} \sqrt [3]{a^3 x+b^2} \log \left (\sqrt {a} x+\sqrt {b}\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{a^{5/2}-b^{3/2}}}-\sqrt [3]{x}\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{a^{5/2}+b^{3/2}}}-\sqrt [3]{x}\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \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 [6]{a} \sqrt [3]{x} \sqrt [3]{a^{5/2}-b^{3/2}}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{x} \sqrt [3]{a^{5/2}+b^{3/2}}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{a^3 x^3+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-b + a*x^2)*(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^(1/6)*(a^(5/2) - b^
(3/2))^(1/3)*x^(1/3))])/(2*a^(1/6)*b*(a^(5/2) - b^(3/2))^(1/3)*(b^2*x^2 + a^3*x^3)^(1/3)) + (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^(1/6)*(a^(5/2) + b^(3/2))^(1/3)*x^(1/
3))])/(2*a^(1/6)*b*(a^(5/2) + b^(3/2))^(1/3)*(b^2*x^2 + a^3*x^3)^(1/3)) - (x^(2/3)*(b^2 + a^3*x)^(1/3)*Log[Sqr
t[b] - Sqrt[a]*x])/(4*a^(1/6)*b*(a^(5/2) + b^(3/2))^(1/3)*(b^2*x^2 + a^3*x^3)^(1/3)) - (x^(2/3)*(b^2 + a^3*x)^
(1/3)*Log[Sqrt[b] + Sqrt[a]*x])/(4*a^(1/6)*b*(a^(5/2) - b^(3/2))^(1/3)*(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^(1/6)*(a^(5/2) - b^(3/2))^(1/3))])/(4*a^(1/6)*b*(a^
(5/2) - b^(3/2))^(1/3)*(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^(1/6)*(a^(5/2) + b^(3/2))^(1/3))])/(4*a^(1/6)*b*(a^(5/2) + b^(3/2))^(1/3)*(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 912

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

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}{\left (-b+a x^2\right ) \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} \sqrt [3]{b^2+a^3 x} \left (-b+a x^2\right )} \, dx}{\sqrt [3]{b^2 x^2+a^3 x^3}}\\ &=\frac {\left (x^{2/3} \sqrt [3]{b^2+a^3 x}\right ) \int \left (-\frac {1}{2 \sqrt {b} x^{2/3} \left (\sqrt {b}-\sqrt {a} x\right ) \sqrt [3]{b^2+a^3 x}}-\frac {1}{2 \sqrt {b} x^{2/3} \left (\sqrt {b}+\sqrt {a} x\right ) \sqrt [3]{b^2+a^3 x}}\right ) \, dx}{\sqrt [3]{b^2 x^2+a^3 x^3}}\\ &=-\frac {\left (x^{2/3} \sqrt [3]{b^2+a^3 x}\right ) \int \frac {1}{x^{2/3} \left (\sqrt {b}-\sqrt {a} x\right ) \sqrt [3]{b^2+a^3 x}} \, dx}{2 \sqrt {b} \sqrt [3]{b^2 x^2+a^3 x^3}}-\frac {\left (x^{2/3} \sqrt [3]{b^2+a^3 x}\right ) \int \frac {1}{x^{2/3} \left (\sqrt {b}+\sqrt {a} x\right ) \sqrt [3]{b^2+a^3 x}} \, dx}{2 \sqrt {b} \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 [6]{a} \sqrt [3]{a^{5/2}-b^{3/2}} \sqrt [3]{x}}\right )}{2 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{x}}\right )}{2 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{b^2 x^2+a^3 x^3}}-\frac {x^{2/3} \sqrt [3]{b^2+a^3 x} \log \left (\sqrt {b}-\sqrt {a} x\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{b^2 x^2+a^3 x^3}}-\frac {x^{2/3} \sqrt [3]{b^2+a^3 x} \log \left (\sqrt {b}+\sqrt {a} x\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{a^{5/2}-b^{3/2}}}\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}-b^{3/2}} \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 [6]{a} \sqrt [3]{a^{5/2}+b^{3/2}}}\right )}{4 \sqrt [6]{a} b \sqrt [3]{a^{5/2}+b^{3/2}} \sqrt [3]{b^2 x^2+a^3 x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 102, normalized size = 1.48 \begin {gather*} -\frac {3 x \left (\, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {a^3 x-\sqrt {a} b^{3/2} x}{x a^3+b^2}\right )+\, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {\sqrt {a} \left (a^{5/2}+b^{3/2}\right ) x}{x a^3+b^2}\right )\right )}{2 b \sqrt [3]{x^2 \left (a^3 x+b^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*(Hypergeometric2F1[1/3, 1, 4/3, (Sqrt[a]*(a^(5/2) + b^(3/2))*x)/(b^2 + a^3*x)] + Hypergeometric2F1[1/3,
1, 4/3, (a^3*x - Sqrt[a]*b^(3/2)*x)/(b^2 + a^3*x)]))/(2*b*(x^2*(b^2 + a^3*x))^(1/3))

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IntegrateAlgebraic [A]  time = 0.00, size = 69, normalized size = 1.00 \begin {gather*} \frac {\text {RootSum}\left [a^6-a b^3-2 a^3 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x)+\log \left (\sqrt [3]{b^2 x^2+a^3 x^3}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[a^6 - a*b^3 - 2*a^3*#1^3 + #1^6 & , (-Log[x] + Log[(b^2*x^2 + a^3*x^3)^(1/3) - x*#1])/#1 & ]/(2*b)

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fricas [B]  time = 0.71, size = 2012, normalized size = 29.16

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*arctan(1/3*(2*sqrt
(3)*b*x*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*sqrt(((a^3*b^2*x -
(a^6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*(a^3*x^3 + b^2*x^2)^(1/3)*((a^2 + (a^5*b^3 - b^6)/sqrt
(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(2/3) + (a^3*b*x^2 - (a^6*b^4 - a*b^7)*x^2/sqrt(a^11*b^3 - 2*
a^6*b^6 + a*b^9))*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3) + (a^3*x^
3 + b^2*x^2)^(2/3))/x^2) - 2*sqrt(3)*(a^3*x^3 + b^2*x^2)^(1/3)*b*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6
*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3) - sqrt(3)*x)/x) - sqrt(3)*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*
b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b
^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*sqrt(((a^3*b^2*x + (a^6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))
*(a^3*x^3 + b^2*x^2)^(1/3)*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(2/3)
+ (a^3*b*x^2 + (a^6*b^4 - a*b^7)*x^2/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3
- 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3) + (a^3*x^3 + b^2*x^2)^(2/3))/x^2) - 2*sqrt(3)*(a^3*x^3 + b^2*x^2
)^(1/3)*b*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3) - sqrt(3)*x)/x) +
1/2*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*log(-((a^3*b^2*x - (a^
6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9
))/(a^5*b^3 - b^6))^(2/3) - (a^3*x^3 + b^2*x^2)^(1/3))/x) + 1/2*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*
b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*log(-((a^3*b^2*x + (a^6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9)
)*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(2/3) - (a^3*x^3 + b^2*x^2)^(1/
3))/x) - 1/4*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3)*log(((a^3*b^2*
x - (a^6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*(a^3*x^3 + b^2*x^2)^(1/3)*((a^2 + (a^5*b^3 - b^6)/
sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(2/3) + (a^3*b*x^2 - (a^6*b^4 - a*b^7)*x^2/sqrt(a^11*b^3
- 2*a^6*b^6 + a*b^9))*((a^2 + (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(1/3) + (a^
3*x^3 + b^2*x^2)^(2/3))/x^2) - 1/4*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6)
)^(1/3)*log(((a^3*b^2*x + (a^6*b^5 - a*b^8)*x/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*(a^3*x^3 + b^2*x^2)^(1/3)*((
a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b^3 - b^6))^(2/3) + (a^3*b*x^2 + (a^6*b^4 - a*b
^7)*x^2/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))*((a^2 - (a^5*b^3 - b^6)/sqrt(a^11*b^3 - 2*a^6*b^6 + a*b^9))/(a^5*b
^3 - b^6))^(1/3) + (a^3*x^3 + b^2*x^2)^(2/3))/x^2)

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giac [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^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*x^2-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^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x^2-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^2 - b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {1}{{\left (a^3\,x^3+b^2\,x^2\right )}^{1/3}\,\left (b-a\,x^2\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^2)),x)

[Out]

-int(1/((a^3*x^3 + b^2*x^2)^(1/3)*(b - a*x^2)), 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^{2} - b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x**2-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**2 - b)), x)

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