3.2.42 \(\int \frac {g+h x}{\sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2} (g^2+3 h^2 x^2)} \, dx\)

Optimal. Leaf size=242 \[ \frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (g^2+3 h^2 x^2\right )}{6\ 2^{2/3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}}-\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (\left (1-\frac {3 h x}{g}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{\frac {3 h x}{g}+1}\right )}{2\ 2^{2/3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}}+\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h x}{g}\right )^{2/3}}{\sqrt {3} \sqrt [3]{\frac {3 h x}{g}+1}}\right )}{2^{2/3} \sqrt {3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}} \]

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Rubi [A]  time = 0.09, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1009, 1008} \begin {gather*} \frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (g^2+3 h^2 x^2\right )}{6\ 2^{2/3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}}-\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (\left (1-\frac {3 h x}{g}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{\frac {3 h x}{g}+1}\right )}{2\ 2^{2/3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}}+\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h x}{g}\right )^{2/3}}{\sqrt {3} \sqrt [3]{\frac {3 h x}{g}+1}}\right )}{2^{2/3} \sqrt {3} h \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(g + h*x)/((-((c*g^2)/h^2) + 9*c*x^2)^(1/3)*(g^2 + 3*h^2*x^2)),x]

[Out]

((1 - (9*h^2*x^2)/g^2)^(1/3)*ArcTan[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)*Sqrt[3]*h*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3)) + ((1 - (9*h^2*x^2)/g^2)^(1/3)*Log[g^2 + 3*h^2*x^2])/
(6*2^(2/3)*h*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3)) - ((1 - (9*h^2*x^2)/g^2)^(1/3)*Log[(1 - (3*h*x)/g)^(2/3) + 2^(1
/3)*(1 + (3*h*x)/g)^(1/3)])/(2*2^(2/3)*h*(-((c*g^2)/h^2) + 9*c*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 {g+h x}{\sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2} \left (g^2+3 h^2 x^2\right )} \, dx &=\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \int \frac {g+h x}{\left (g^2+3 h^2 x^2\right ) \sqrt [3]{1-\frac {9 h^2 x^2}{g^2}}} \, dx}{\sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2}}\\ &=\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h x}{g}\right )^{2/3}}{\sqrt {3} \sqrt [3]{1+\frac {3 h x}{g}}}\right )}{2^{2/3} \sqrt {3} h \sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2}}+\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (g^2+3 h^2 x^2\right )}{6\ 2^{2/3} h \sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2}}-\frac {\sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} \log \left (\left (1-\frac {3 h x}{g}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\frac {3 h x}{g}}\right )}{2\ 2^{2/3} h \sqrt [3]{-\frac {c g^2}{h^2}+9 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.56, size = 268, normalized size = 1.11 \begin {gather*} \frac {h^2 x \left (-h x \sqrt [3]{1-\frac {9 h^2 x^2}{g^2}} F_1\left (1;\frac {1}{3},1;2;\frac {9 h^2 x^2}{g^2},-\frac {3 h^2 x^2}{g^2}\right )-\frac {2 g^5 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {9 h^2 x^2}{g^2},-\frac {3 h^2 x^2}{g^2}\right )}{\left (g^2+3 h^2 x^2\right ) \left (g^2 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {9 h^2 x^2}{g^2},-\frac {3 h^2 x^2}{g^2}\right )+2 h^2 x^2 \left (F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};\frac {9 h^2 x^2}{g^2},-\frac {3 h^2 x^2}{g^2}\right )-F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};\frac {9 h^2 x^2}{g^2},-\frac {3 h^2 x^2}{g^2}\right )\right )\right )}\right ) \left (c \left (9 x^2-\frac {g^2}{h^2}\right )\right )^{2/3}}{2 c g^2 \left (g^2-9 h^2 x^2\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(g + h*x)/((-((c*g^2)/h^2) + 9*c*x^2)^(1/3)*(g^2 + 3*h^2*x^2)),x]

[Out]

(h^2*x*(c*(-(g^2/h^2) + 9*x^2))^(2/3)*(-(h*x*(1 - (9*h^2*x^2)/g^2)^(1/3)*AppellF1[1, 1/3, 1, 2, (9*h^2*x^2)/g^
2, (-3*h^2*x^2)/g^2]) - (2*g^5*AppellF1[1/2, 1/3, 1, 3/2, (9*h^2*x^2)/g^2, (-3*h^2*x^2)/g^2])/((g^2 + 3*h^2*x^
2)*(g^2*AppellF1[1/2, 1/3, 1, 3/2, (9*h^2*x^2)/g^2, (-3*h^2*x^2)/g^2] + 2*h^2*x^2*(-AppellF1[3/2, 1/3, 2, 5/2,
 (9*h^2*x^2)/g^2, (-3*h^2*x^2)/g^2] + AppellF1[3/2, 4/3, 1, 5/2, (9*h^2*x^2)/g^2, (-3*h^2*x^2)/g^2])))))/(2*c*
g^2*(g^2 - 9*h^2*x^2))

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IntegrateAlgebraic [B]  time = 1.08, size = 499, normalized size = 2.06 \begin {gather*} -\frac {\log \left (\sqrt [3]{2} c^{2/3} g^2-6 \sqrt [3]{2} c^{2/3} g h x+9 \sqrt [3]{2} c^{2/3} h^2 x^2-3\ 2^{2/3} \sqrt [3]{c} \sqrt [3]{g} h^{5/3} x \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}+2^{2/3} \sqrt [3]{c} g^{4/3} h^{2/3} \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}+2 g^{2/3} h^{4/3} \left (9 c x^2-\frac {c g^2}{h^2}\right )^{2/3}\right )}{6\ 2^{2/3} \sqrt [3]{c} g^{2/3} \sqrt [3]{h}}-\frac {\log \left (\sqrt [3]{g} \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}\right )}{3\ 2^{2/3} \sqrt [3]{c} g^{2/3} \sqrt [3]{h}}+\frac {\log \left (g^{2/3} \left (9 c x^2-\frac {c g^2}{h^2}\right )^{2/3}\right )}{6\ 2^{2/3} \sqrt [3]{c} g^{2/3} \sqrt [3]{h}}+\frac {\log \left (-2 \sqrt [3]{g} h^{2/3} \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}+2^{2/3} \sqrt [3]{c} g-3\ 2^{2/3} \sqrt [3]{c} h x\right )}{3\ 2^{2/3} \sqrt [3]{c} g^{2/3} \sqrt [3]{h}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{g} h^{2/3} \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}}{\sqrt [3]{g} h^{2/3} \sqrt [3]{9 c x^2-\frac {c g^2}{h^2}}+2^{2/3} \sqrt [3]{c} g-3\ 2^{2/3} \sqrt [3]{c} h x}\right )}{2^{2/3} \sqrt {3} \sqrt [3]{c} g^{2/3} \sqrt [3]{h}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(g + h*x)/((-((c*g^2)/h^2) + 9*c*x^2)^(1/3)*(g^2 + 3*h^2*x^2)),x]

[Out]

ArcTan[(Sqrt[3]*g^(1/3)*h^(2/3)*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*g - 3*2^(2/3)*c^(1/3)*h*x +
 g^(1/3)*h^(2/3)*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3))]/(2^(2/3)*Sqrt[3]*c^(1/3)*g^(2/3)*h^(1/3)) - Log[g^(1/3)*(-
((c*g^2)/h^2) + 9*c*x^2)^(1/3)]/(3*2^(2/3)*c^(1/3)*g^(2/3)*h^(1/3)) + Log[g^(2/3)*(-((c*g^2)/h^2) + 9*c*x^2)^(
2/3)]/(6*2^(2/3)*c^(1/3)*g^(2/3)*h^(1/3)) + Log[2^(2/3)*c^(1/3)*g - 3*2^(2/3)*c^(1/3)*h*x - 2*g^(1/3)*h^(2/3)*
(-((c*g^2)/h^2) + 9*c*x^2)^(1/3)]/(3*2^(2/3)*c^(1/3)*g^(2/3)*h^(1/3)) - Log[2^(1/3)*c^(2/3)*g^2 - 6*2^(1/3)*c^
(2/3)*g*h*x + 9*2^(1/3)*c^(2/3)*h^2*x^2 + 2^(2/3)*c^(1/3)*g^(4/3)*h^(2/3)*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3) - 3
*2^(2/3)*c^(1/3)*g^(1/3)*h^(5/3)*x*(-((c*g^2)/h^2) + 9*c*x^2)^(1/3) + 2*g^(2/3)*h^(4/3)*(-((c*g^2)/h^2) + 9*c*
x^2)^(2/3)]/(6*2^(2/3)*c^(1/3)*g^(2/3)*h^(1/3))

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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((h*x+g)/(-c*g^2/h^2+9*c*x^2)^(1/3)/(3*h^2*x^2+g^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 {h x + g}{{\left (3 \, h^{2} x^{2} + g^{2}\right )} {\left (9 \, c x^{2} - \frac {c g^{2}}{h^{2}}\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(-c*g^2/h^2+9*c*x^2)^(1/3)/(3*h^2*x^2+g^2),x, algorithm="giac")

[Out]

integrate((h*x + g)/((3*h^2*x^2 + g^2)*(9*c*x^2 - c*g^2/h^2)^(1/3)), x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {h x +g}{\left (9 c \,x^{2}-\frac {c \,g^{2}}{h^{2}}\right )^{\frac {1}{3}} \left (3 h^{2} x^{2}+g^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)/(-c*g^2/h^2+9*c*x^2)^(1/3)/(3*h^2*x^2+g^2),x)

[Out]

int((h*x+g)/(-c*g^2/h^2+9*c*x^2)^(1/3)/(3*h^2*x^2+g^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {h x + g}{{\left (3 \, h^{2} x^{2} + g^{2}\right )} {\left (9 \, c x^{2} - \frac {c g^{2}}{h^{2}}\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(-c*g^2/h^2+9*c*x^2)^(1/3)/(3*h^2*x^2+g^2),x, algorithm="maxima")

[Out]

integrate((h*x + g)/((3*h^2*x^2 + g^2)*(9*c*x^2 - c*g^2/h^2)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {g+h\,x}{\left (g^2+3\,h^2\,x^2\right )\,{\left (9\,c\,x^2-\frac {c\,g^2}{h^2}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g + h*x)/((g^2 + 3*h^2*x^2)*(9*c*x^2 - (c*g^2)/h^2)^(1/3)),x)

[Out]

int((g + h*x)/((g^2 + 3*h^2*x^2)*(9*c*x^2 - (c*g^2)/h^2)^(1/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(-c*g**2/h**2+9*c*x**2)**(1/3)/(3*h**2*x**2+g**2),x)

[Out]

Integral((g + h*x)/((c*(-g/h + 3*x)*(g/h + 3*x))**(1/3)*(g**2 + 3*h**2*x**2)), x)

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