3.22.9 \(\int \frac {1}{(d+e x) \sqrt [3]{c^2 d^2-b c d e+b^2 e^2+3 b c e^2 x+3 c^2 e^2 x^2}} \, dx\)

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

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Rubi [A]  time = 0.14, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {750} \begin {gather*} \frac {\log \left (-3 c e^2 \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}+3 c e^2 (c d-b e)-3 c^2 e^3 x\right )}{2 e (2 c d-b e)^{2/3}}-\frac {\tan ^{-1}\left (\frac {2 (-b e+c d-c e x)}{\sqrt {3} \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} e (2 c d-b e)^{2/3}}-\frac {\log (d+e x)}{2 e (2 c d-b e)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[1/Sqrt[3] + (2*(c*d - b*e - c*e*x))/(Sqrt[3]*(2*c*d - b*e)^(1/3)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c
*e^2*x + 3*c^2*e^2*x^2)^(1/3))]/(Sqrt[3]*e*(2*c*d - b*e)^(2/3))) - Log[d + e*x]/(2*e*(2*c*d - b*e)^(2/3)) + Lo
g[3*c*e^2*(c*d - b*e) - 3*c^2*e^3*x - 3*c*e^2*(2*c*d - b*e)^(1/3)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x +
3*c^2*e^2*x^2)^(1/3)]/(2*e*(2*c*d - b*e)^(2/3))

Rule 750

Int[1/(((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)), x_Symbol] :> With[{q = Rt[3*c*e^2*(2*c*
d - b*e), 3]}, -Simp[(Sqrt[3]*c*e*ArcTan[1/Sqrt[3] + (2*(c*d - b*e - c*e*x))/(Sqrt[3]*q*(a + b*x + c*x^2)^(1/3
))])/q^2, x] + (-Simp[(3*c*e*Log[d + e*x])/(2*q^2), x] + Simp[(3*c*e*Log[c*d - b*e - c*e*x - q*(a + b*x + c*x^
2)^(1/3)])/(2*q^2), x])] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && EqQ[c^2*d^2 - b*c*d*e + b^2*e^
2 - 3*a*c*e^2, 0] && PosQ[c*e^2*(2*c*d - b*e)]

Rubi steps

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

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Mathematica [C]  time = 0.48, size = 317, normalized size = 1.31 \begin {gather*} -\frac {\sqrt [3]{3} \sqrt [3]{\frac {-\sqrt {3} \sqrt {-c^2 e^2 (b e-2 c d)^2}+3 b c e^2+6 c^2 e^2 x}{c^2 e (d+e x)}} \sqrt [3]{\frac {\sqrt {3} \sqrt {-c^2 e^2 (b e-2 c d)^2}+3 b c e^2+6 c^2 e^2 x}{c^2 e (d+e x)}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};-\frac {-6 d e c^2+3 b e^2 c+\sqrt {3} \sqrt {-c^2 e^2 (b e-2 c d)^2}}{6 c^2 e (d+e x)},\frac {6 d e c^2-3 b e^2 c+\sqrt {3} \sqrt {-c^2 e^2 (b e-2 c d)^2}}{6 c^2 e (d+e x)}\right )}{2\ 2^{2/3} e \sqrt [3]{b^2 e^2+b c e (3 e x-d)+c^2 \left (d^2+3 e^2 x^2\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(3^(1/3)*((3*b*c*e^2 - Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)] + 6*c^2*e^2*x)/(c^2*e*(d + e*x)))^(1/3)*
((3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)] + 6*c^2*e^2*x)/(c^2*e*(d + e*x)))^(1/3)*AppellF1[2/3,
1/3, 1/3, 5/3, -1/6*(-6*c^2*d*e + 3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)])/(c^2*e*(d + e*x)), (6
*c^2*d*e - 3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)])/(6*c^2*e*(d + e*x))])/(2^(2/3)*e*(b^2*e^2 +
b*c*e*(-d + 3*e*x) + c^2*(d^2 + 3*e^2*x^2))^(1/3))

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IntegrateAlgebraic [C]  time = 1.68, size = 712, normalized size = 2.94 \begin {gather*} \frac {\left (1-i \sqrt {3}\right ) \log \left (e \left (-b^2 e^2+2 b c d e-2 b c e^2 x+c^2 \left (-d^2\right )+2 c^2 d e x-c^2 e^2 x^2\right )+2 e (2 c d-b e)^{2/3} \left (b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2\right )^{2/3}+e \sqrt [3]{2 c d-b e} (b e-c d+c e x) \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}+\sqrt {3} \left (e \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2} (i b e-i c d+i c e x)+e \left (i b^2 e^2-2 i b c d e+2 i b c e^2 x+i c^2 d^2-2 i c^2 d e x+i c^2 e^2 x^2\right )\right )\right )}{12 e (2 c d-b e)^{2/3}}-\frac {i \left (\sqrt {3}-3 i\right ) \tanh ^{-1}\left (\frac {i \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}+\sqrt {3} (c d-b e)+i b e-i c d-\sqrt {3} c e x+i c e x}{\sqrt {3} \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}}\right )}{6 e (2 c d-b e)^{2/3}}+\frac {i \left (\sqrt {3}+i\right ) \log \left (\sqrt {e} \left (2 \sqrt [3]{2 c d-b e} \sqrt [3]{b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2}+i \sqrt {3} c d+c d\right )+e^{3/2} \left (\sqrt {3} (-i b-i c x)-b-c x\right )\right )}{6 e (2 c d-b e)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/6*I)*(-3*I + Sqrt[3])*ArcTanh[((-I)*c*d + I*b*e + Sqrt[3]*(c*d - b*e) + I*c*e*x - Sqrt[3]*c*e*x + I*(2*c*
d - b*e)^(1/3)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(1/3))/(Sqrt[3]*(2*c*d - b*e)^(1/3)
*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(1/3))])/(e*(2*c*d - b*e)^(2/3)) + ((I/6)*(I + Sq
rt[3])*Log[e^(3/2)*(-b - c*x + Sqrt[3]*((-I)*b - I*c*x)) + Sqrt[e]*(c*d + I*Sqrt[3]*c*d + 2*(2*c*d - b*e)^(1/3
)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(1/3))])/(e*(2*c*d - b*e)^(2/3)) + ((1 - I*Sqrt[
3])*Log[e*(-(c^2*d^2) + 2*b*c*d*e - b^2*e^2 + 2*c^2*d*e*x - 2*b*c*e^2*x - c^2*e^2*x^2) + e*(2*c*d - b*e)^(1/3)
*(-(c*d) + b*e + c*e*x)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(1/3) + 2*e*(2*c*d - b*e)^
(2/3)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(2/3) + Sqrt[3]*(e*(I*c^2*d^2 - (2*I)*b*c*d*
e + I*b^2*e^2 - (2*I)*c^2*d*e*x + (2*I)*b*c*e^2*x + I*c^2*e^2*x^2) + e*(2*c*d - b*e)^(1/3)*((-I)*c*d + I*b*e +
I*c*e*x)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2*x^2)^(1/3))])/(12*e*(2*c*d - b*e)^(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(1/(e*x+d)/(3*c^2*e^2*x^2+3*b*c*e^2*x+b^2*e^2-b*c*d*e+c^2*d^2)^(1/3),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 {1}{{\left (3 \, c^{2} e^{2} x^{2} + 3 \, b c e^{2} x + c^{2} d^{2} - b c d e + b^{2} e^{2}\right )}^{\frac {1}{3}} {\left (e x + d\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((3*c^2*e^2*x^2 + 3*b*c*e^2*x + c^2*d^2 - b*c*d*e + b^2*e^2)^(1/3)*(e*x + d)), x)

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maple [F]  time = 2.56, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (e x +d \right ) \left (3 c^{2} e^{2} x^{2}+3 b c \,e^{2} x +b^{2} e^{2}-b c d e +c^{2} d^{2}\right )^{\frac {1}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(3*c^2*e^2*x^2+3*b*c*e^2*x+b^2*e^2-b*c*d*e+c^2*d^2)^(1/3),x)

[Out]

int(1/(e*x+d)/(3*c^2*e^2*x^2+3*b*c*e^2*x+b^2*e^2-b*c*d*e+c^2*d^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 (3 \, c^{2} e^{2} x^{2} + 3 \, b c e^{2} x + c^{2} d^{2} - b c d e + b^{2} e^{2}\right )}^{\frac {1}{3}} {\left (e x + d\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((3*c^2*e^2*x^2 + 3*b*c*e^2*x + c^2*d^2 - b*c*d*e + b^2*e^2)^(1/3)*(e*x + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)*(b^2*e^2 + c^2*d^2 + 3*c^2*e^2*x^2 + 3*b*c*e^2*x - b*c*d*e)^(1/3)),x)

[Out]

int(1/((d + e*x)*(b^2*e^2 + c^2*d^2 + 3*c^2*e^2*x^2 + 3*b*c*e^2*x - b*c*d*e)^(1/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(3*c**2*e**2*x**2+3*b*c*e**2*x+b**2*e**2-b*c*d*e+c**2*d**2)**(1/3),x)

[Out]

Integral(1/((d + e*x)*(b**2*e**2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 3*c**2*e**2*x**2)**(1/3)), x)

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