3.22.12 \(\int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx\)

Optimal. Leaf size=564 \[ -\frac {\log (d+e x) \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}+\frac {3 \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \log \left (-\frac {\sqrt [3]{\frac {3}{2}} \left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{\sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e}}-\sqrt [3]{6} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}\right )}{4 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}-\frac {\sqrt {3} \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}}\right )}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}} \]

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Rubi [A]  time = 0.41, antiderivative size = 564, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {755, 123} \begin {gather*} -\frac {\log (d+e x) \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}+\frac {3 \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \log \left (-\frac {\sqrt [3]{\frac {3}{2}} \left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{\sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e}}-\sqrt [3]{6} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}\right )}{4 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}-\frac {\sqrt {3} \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}}\right )}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 + 9*b*c*e^2*x + 9*c^2*e^2*x^2)^(1/3)),x]

[Out]

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

Rule 123

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[
(b*(b*e - a*f))/(b*c - a*d)^2, 3]}, -Simp[Log[a + b*x]/(2*q*(b*c - a*d)), x] + (-Simp[(Sqrt[3]*ArcTan[1/Sqrt[3
] + (2*q*(c + d*x)^(2/3))/(Sqrt[3]*(e + f*x)^(1/3))])/(2*q*(b*c - a*d)), x] + Simp[(3*Log[q*(c + d*x)^(2/3) -
(e + f*x)^(1/3)])/(4*q*(b*c - a*d)), x])] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0]

Rule 755

Int[1/(((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[((b + q + 2*c*x)^(1/3)*(b - q + 2*c*x)^(1/3))/(a + b*x + c*x^2)^(1/3), Int[1/((d + e*x)*(b + q + 2*c
*x)^(1/3)*(b - q + 2*c*x)^(1/3)), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c^2*d^2 -
b*c*d*e - 2*b^2*e^2 + 9*a*c*e^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx &=\frac {\left (\sqrt [3]{9 b c e^2-3 c e (2 c d-b e)+18 c^2 e^2 x} \sqrt [3]{9 b c e^2+3 c e (2 c d-b e)+18 c^2 e^2 x}\right ) \int \frac {1}{(d+e x) \sqrt [3]{9 b c e^2-3 c e (2 c d-b e)+18 c^2 e^2 x} \sqrt [3]{9 b c e^2+3 c e (2 c d-b e)+18 c^2 e^2 x}} \, dx}{\sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}}\\ &=-\frac {\sqrt {3} \sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \left (-c e (c d-2 b e)+3 c^2 e^2 x\right )^{2/3}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x}}\right )}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}}-\frac {\sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \log (d+e x)}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}}+\frac {3 \sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \log \left (-\frac {\sqrt [3]{\frac {3}{2}} \left (-c e (c d-2 b e)+3 c^2 e^2 x\right )^{2/3}}{\sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e}}-\sqrt [3]{6} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x}\right )}{4 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.41, size = 290, normalized size = 0.51 \begin {gather*} -\frac {\sqrt [3]{3} \sqrt [3]{\frac {-\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 {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 {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 {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]{2 b^2 e^2+b c e (d+9 e x)-\left (c^2 \left (d^2-9 e^2 x^2\right )\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((d + e*x)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 + 9*b*c*e^2*x + 9*c^2*e^2*x^2)^(1/3)),x]

[Out]

-1/2*(3^(1/3)*((3*b*c*e^2 - 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[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[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[c^2*e
^2*(-2*c*d + b*e)^2])/(6*c^2*e*(d + e*x))])/(2^(2/3)*e*(2*b^2*e^2 + b*c*e*(d + 9*e*x) - c^2*(d^2 - 9*e^2*x^2))
^(1/3))

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IntegrateAlgebraic [A]  time = 1.84, size = 600, normalized size = 1.06 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2}}{\sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2}-2 \sqrt [3]{2} b e+\sqrt [3]{2} c d-3 \sqrt [3]{2} c e x}\right )}{2 \sqrt [3]{2} e (2 c d-b e)^{2/3}}+\frac {\log \left (2 \sqrt {e} \sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2}+\sqrt [3]{2} e^{3/2} (2 b+3 c x)-\sqrt [3]{2} c d \sqrt {e}\right )}{2 \sqrt [3]{2} e (2 c d-b e)^{2/3}}-\frac {\log \left (4 e (2 c d-b e)^{2/3} \left (2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2\right )^{2/3}+\left (-6 \sqrt [3]{2} c e^2 x \sqrt [3]{2 c d-b e}-4 \sqrt [3]{2} b e^2 \sqrt [3]{2 c d-b e}+2 \sqrt [3]{2} c d e \sqrt [3]{2 c d-b e}\right ) \sqrt [3]{2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2}+4\ 2^{2/3} b^2 e^3-4\ 2^{2/3} b c d e^2+12\ 2^{2/3} b c e^3 x+2^{2/3} c^2 d^2 e-6\ 2^{2/3} c^2 d e^2 x+9\ 2^{2/3} c^2 e^3 x^2\right )}{4 \sqrt [3]{2} 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 + 2*b^2*e^2 + 9*b*c*e^2*x + 9*c^2*e^2*x^2)^(1/3)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*(2*c*d - b*e)^(1/3)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 + 9*b*c*e^2*x + 9*c^2*e^2*x^2)^
(1/3))/(2^(1/3)*c*d - 2*2^(1/3)*b*e - 3*2^(1/3)*c*e*x + (2*c*d - b*e)^(1/3)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2
+ 9*b*c*e^2*x + 9*c^2*e^2*x^2)^(1/3))])/(2*2^(1/3)*e*(2*c*d - b*e)^(2/3)) + Log[-(2^(1/3)*c*d*Sqrt[e]) + 2^(1/
3)*e^(3/2)*(2*b + 3*c*x) + 2*Sqrt[e]*(2*c*d - b*e)^(1/3)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 + 9*b*c*e^2*x + 9*c
^2*e^2*x^2)^(1/3)]/(2*2^(1/3)*e*(2*c*d - b*e)^(2/3)) - Log[2^(2/3)*c^2*d^2*e - 4*2^(2/3)*b*c*d*e^2 + 4*2^(2/3)
*b^2*e^3 - 6*2^(2/3)*c^2*d*e^2*x + 12*2^(2/3)*b*c*e^3*x + 9*2^(2/3)*c^2*e^3*x^2 + (2*2^(1/3)*c*d*e*(2*c*d - b*
e)^(1/3) - 4*2^(1/3)*b*e^2*(2*c*d - b*e)^(1/3) - 6*2^(1/3)*c*e^2*(2*c*d - b*e)^(1/3)*x)*(-(c^2*d^2) + b*c*d*e
+ 2*b^2*e^2 + 9*b*c*e^2*x + 9*c^2*e^2*x^2)^(1/3) + 4*e*(2*c*d - b*e)^(2/3)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 +
9*b*c*e^2*x + 9*c^2*e^2*x^2)^(2/3)]/(4*2^(1/3)*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)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*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 (9 \, c^{2} e^{2} x^{2} + 9 \, b c e^{2} x - c^{2} d^{2} + b c d e + 2 \, 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)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((9*c^2*e^2*x^2 + 9*b*c*e^2*x - c^2*d^2 + b*c*d*e + 2*b^2*e^2)^(1/3)*(e*x + d)), x)

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maple [F]  time = 1.71, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (e x +d \right ) \left (9 c^{2} e^{2} x^{2}+9 b c \,e^{2} x +2 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)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^(1/3),x)

[Out]

int(1/(e*x+d)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*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 (9 \, c^{2} e^{2} x^{2} + 9 \, b c e^{2} x - c^{2} d^{2} + b c d e + 2 \, 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)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((9*c^2*e^2*x^2 + 9*b*c*e^2*x - c^2*d^2 + b*c*d*e + 2*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 (2\,b^2\,e^2+b\,c\,d\,e+9\,b\,c\,e^2\,x-c^2\,d^2+9\,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)*(2*b^2*e^2 - c^2*d^2 + 9*c^2*e^2*x^2 + 9*b*c*e^2*x + b*c*d*e)^(1/3)),x)

[Out]

int(1/((d + e*x)*(2*b^2*e^2 - c^2*d^2 + 9*c^2*e^2*x^2 + 9*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}{\sqrt [3]{\left (b e + c d + 3 c e x\right ) \left (2 b e - c d + 3 c e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(9*c**2*e**2*x**2+9*b*c*e**2*x+2*b**2*e**2+b*c*d*e-c**2*d**2)**(1/3),x)

[Out]

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

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