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

Optimal. Leaf size=354 \[ \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )}{2^{2/3} a b^{2/3}}-\frac {\log \left (b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}}-\frac {\log \left (2 \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}+2^{2/3} a x+2^{2/3} b\right )}{2^{2/3} a b^{2/3}}+\frac {\log \left (-2^{2/3} a \sqrt [3]{b} x \sqrt [3]{b^2-a^2 x^2}-2^{2/3} b^{4/3} \sqrt [3]{b^2-a^2 x^2}+2 b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}+\sqrt [3]{2} a^2 x^2+2 \sqrt [3]{2} a b x+\sqrt [3]{2} b^2\right )}{2\ 2^{2/3} a b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}{\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}-2^{2/3} a x-2^{2/3} b}\right )}{2^{2/3} a b^{2/3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 229, normalized size of antiderivative = 0.65, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1009, 1008} \begin {gather*} \frac {\sqrt [3]{1-\frac {a^2 x^2}{b^2}} \log \left (a^2 x^2+3 b^2\right )}{2\ 2^{2/3} a \sqrt [3]{b^2-a^2 x^2}}-\frac {3 \sqrt [3]{1-\frac {a^2 x^2}{b^2}} \log \left (\left (\frac {a x}{b}+1\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{1-\frac {a x}{b}}\right )}{2\ 2^{2/3} a \sqrt [3]{b^2-a^2 x^2}}+\frac {\sqrt {3} \sqrt [3]{1-\frac {a^2 x^2}{b^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (\frac {a x}{b}+1\right )^{2/3}}{\sqrt {3} \sqrt [3]{1-\frac {a x}{b}}}\right )}{2^{2/3} a \sqrt [3]{b^2-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1008

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)^(1/3)*((d_) + (f_.)*(x_)^2)), x_Symbol] :> Simp[(Sqrt[3]*h*ArcT
an[1/Sqrt[3] - (2^(2/3)*(1 - (3*h*x)/g)^(2/3))/(Sqrt[3]*(1 + (3*h*x)/g)^(1/3))])/(2^(2/3)*a^(1/3)*f), x] + (-S
imp[(3*h*Log[(1 - (3*h*x)/g)^(2/3) + 2^(1/3)*(1 + (3*h*x)/g)^(1/3)])/(2^(5/3)*a^(1/3)*f), x] + Simp[(h*Log[d +
f*x^2])/(2^(5/3)*a^(1/3)*f), x]) /; FreeQ[{a, c, d, f, g, h}, x] && EqQ[c*d + 3*a*f, 0] && EqQ[c*g^2 + 9*a*h^
2, 0] && GtQ[a, 0]

Rule 1009

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)^(1/3)*((d_) + (f_.)*(x_)^2)), x_Symbol] :> Dist[(1 + (c*x^2)/a)
^(1/3)/(a + c*x^2)^(1/3), Int[(g + h*x)/((1 + (c*x^2)/a)^(1/3)*(d + f*x^2)), x], x] /; FreeQ[{a, c, d, f, g, h
}, x] && EqQ[c*d + 3*a*f, 0] && EqQ[c*g^2 + 9*a*h^2, 0] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.47, size = 250, normalized size = 0.71 \begin {gather*} \frac {x \left (a x \sqrt [3]{1-\frac {a^2 x^2}{b^2}} F_1\left (1;\frac {1}{3},1;2;\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{3 b^2}\right )-\frac {162 b^5 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{3 b^2}\right )}{\left (a^2 x^2+3 b^2\right ) \left (9 b^2 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{3 b^2}\right )+2 a^2 x^2 \left (F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{3 b^2}\right )-F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{3 b^2}\right )\right )\right )}\right )}{6 b^2 \sqrt [3]{b^2-a^2 x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(a*x*(1 - (a^2*x^2)/b^2)^(1/3)*AppellF1[1, 1/3, 1, 2, (a^2*x^2)/b^2, -1/3*(a^2*x^2)/b^2] - (162*b^5*AppellF
1[1/2, 1/3, 1, 3/2, (a^2*x^2)/b^2, -1/3*(a^2*x^2)/b^2])/((3*b^2 + a^2*x^2)*(9*b^2*AppellF1[1/2, 1/3, 1, 3/2, (
a^2*x^2)/b^2, -1/3*(a^2*x^2)/b^2] + 2*a^2*x^2*(-AppellF1[3/2, 1/3, 2, 5/2, (a^2*x^2)/b^2, -1/3*(a^2*x^2)/b^2]
+ AppellF1[3/2, 4/3, 1, 5/2, (a^2*x^2)/b^2, -1/3*(a^2*x^2)/b^2])))))/(6*b^2*(b^2 - a^2*x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.59, size = 354, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}{-2^{2/3} b-2^{2/3} a x+\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}\right )}{2^{2/3} a b^{2/3}}+\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )}{2^{2/3} a b^{2/3}}-\frac {\log \left (b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}}-\frac {\log \left (2^{2/3} b+2^{2/3} a x+2 \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )}{2^{2/3} a b^{2/3}}+\frac {\log \left (\sqrt [3]{2} b^2+2 \sqrt [3]{2} a b x+\sqrt [3]{2} a^2 x^2-2^{2/3} b^{4/3} \sqrt [3]{b^2-a^2 x^2}-2^{2/3} a \sqrt [3]{b} x \sqrt [3]{b^2-a^2 x^2}+2 b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {a x -3 b}{\left (-a^{2} x^{2}+b^{2}\right )^{\frac {1}{3}} \left (a^{2} x^{2}+3 b^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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