3.26.79 \(\int \frac {1}{x \sqrt [3]{(1+x) (q+2 q x+x^2)}} \, dx\)

Optimal. Leaf size=223 \[ -\frac {\log \left (q^{2/3} x^2+2 q^{2/3} x+q^{2/3}+\left ((2 q+1) x^2+3 q x+q+x^3\right )^{2/3}+\left (\sqrt [3]{q} x+\sqrt [3]{q}\right ) \sqrt [3]{(2 q+1) x^2+3 q x+q+x^3}\right )}{4 \sqrt [3]{q}}+\frac {\log \left (\sqrt [3]{(2 q+1) x^2+3 q x+q+x^3}-\sqrt [3]{q} x-\sqrt [3]{q}\right )}{2 \sqrt [3]{q}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{(2 q+1) x^2+3 q x+q+x^3}}{\sqrt [3]{(2 q+1) x^2+3 q x+q+x^3}+2 \sqrt [3]{q} x+2 \sqrt [3]{q}}\right )}{2 \sqrt [3]{q}} \]

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Rubi [F]  time = 22.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{x \sqrt [3]{(1+x) \left (q+2 q x+x^2\right )}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/(x*((1 + x)*(q + 2*q*x + x^2))^(1/3)),x]

[Out]

((1 + 2*q + (1 - 5*q + 4*q^2 + (1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[-((-1 + q)^3*q)])^(2/3))/(1 + 6*q -
15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[-((-1 + q)^3*q)])^(1/3) + 3*x)^(1/3)*(-1 + 5*q - 4*q^2 + ((1 - 4*q)^2*(1 - q)^
2)/(1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(2/3) + (1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt
[(1 - q)^3*q])^(2/3) - (3*(1 - 5*q + 4*q^2 + (1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(2/3))*(
(1 + 2*q)/3 + x))/(1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(1/3) + 9*((1 + 2*q)/3 + x)^2)^(1/3
)*Defer[Subst][Defer[Int][1/(((-1 - 2*q)/3 + x)*((1 - 5*q + 4*q^2 + (1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt
[(1 - q)^3*q])^(2/3))/(3*(1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(1/3)) + x)^(1/3)*((-1 + 5*q
- 4*q^2 + ((1 - 4*q)^2*(1 - q)^2)/(1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(2/3) + (1 + 6*q -
15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(2/3))/9 - ((1 - 5*q + 4*q^2 + (1 + 6*q - 15*q^2 + 8*q^3 - 3*Sq
rt[3]*Sqrt[(1 - q)^3*q])^(2/3))*x)/(3*(1 + 6*q - 15*q^2 + 8*q^3 - 3*Sqrt[3]*Sqrt[(1 - q)^3*q])^(1/3)) + x^2)^(
1/3)), x], x, (1 + 2*q)/3 + x])/(3*(q + 3*q*x + (1 + 2*q)*x^2 + x^3)^(1/3))

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt [3]{(1+x) \left (q+2 q x+x^2\right )}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (\frac {1}{3} (-1-2 q)+x\right ) \sqrt [3]{\frac {2}{27} (1-q)^2 (1+8 q)-\frac {1}{3} (1-4 q) (1-q) x+x^3}} \, dx,x,\frac {1}{3} (1+2 q)+x\right )\\ &=\frac {\left (\sqrt [3]{1+2 q+\frac {1-5 q+4 q^2+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {-(-1+q)^3 q}\right )^{2/3}}{\sqrt [3]{1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {-(-1+q)^3 q}}}+3 x} \sqrt [3]{-1+5 q-4 q^2+\frac {(1-4 q)^2 (1-q)^2}{\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}}+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}+9 \left (\frac {1}{3} (1+2 q)+x\right )^2-\frac {\left (1-5 q+4 q^2+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}\right ) (1+2 q+3 x)}{\sqrt [3]{1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}}}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\frac {1}{3} (-1-2 q)+x\right ) \sqrt [3]{\frac {1-5 q+4 q^2+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}}{3 \sqrt [3]{1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}}}+x} \sqrt [3]{\frac {1}{9} \left (-1+5 q-4 q^2+\frac {(1-4 q)^2 (1-q)^2}{\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}}+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}\right )-\frac {\left (1-5 q+4 q^2+\left (1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}\right )^{2/3}\right ) x}{3 \sqrt [3]{1+6 q-15 q^2+8 q^3-3 \sqrt {3} \sqrt {(1-q)^3 q}}}+x^2}} \, dx,x,\frac {1}{3} (1+2 q)+x\right )}{3 \sqrt [3]{q+3 q x+(1+2 q) x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 55, normalized size = 0.25 \begin {gather*} -\frac {3 \left ((x+1) \left (2 q x+q+x^2\right )\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x^2+2 q x+q}{q (x+1)^2}\right )}{4 q (x+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*((1 + x)*(q + 2*q*x + x^2))^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.51, size = 223, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{q+3 q x+(1+2 q) x^2+x^3}}{2 \sqrt [3]{q}+2 \sqrt [3]{q} x+\sqrt [3]{q+3 q x+(1+2 q) x^2+x^3}}\right )}{2 \sqrt [3]{q}}+\frac {\log \left (-\sqrt [3]{q}-\sqrt [3]{q} x+\sqrt [3]{q+3 q x+(1+2 q) x^2+x^3}\right )}{2 \sqrt [3]{q}}-\frac {\log \left (q^{2/3}+2 q^{2/3} x+q^{2/3} x^2+\left (\sqrt [3]{q}+\sqrt [3]{q} x\right ) \sqrt [3]{q+3 q x+(1+2 q) x^2+x^3}+\left (q+3 q x+(1+2 q) x^2+x^3\right )^{2/3}\right )}{4 \sqrt [3]{q}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*((1 + x)*(q + 2*q*x + x^2))^(1/3)),x]

[Out]

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

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fricas [B]  time = 24.00, size = 1383, normalized size = 6.20

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(sqrt(3)*q*sqrt(-1/q^(2/3))*log(-((q^3 - 30*q^2 - 51*q - 1)*x^6 - 54*(q^3 + 6*q^2 + 2*q)*x^5 - 27*(17*q^
3 + 26*q^2 + 2*q)*x^4 - 486*q^3*x - 540*(2*q^3 + q^2)*x^3 - 81*q^3 - 135*(8*q^3 + q^2)*x^2 - 9*((2*q^2 - q - 1
)*x^4 + 6*(q^2 - q)*x^3 + 3*(q^2 - q)*x^2)*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(2/3)*q^(1/3) + 9*((q^2 + 7*q + 1
)*x^5 + (19*q^2 + 25*q + 1)*x^4 + 9*(7*q^2 + 3*q)*x^3 + 45*q^2*x + 9*(9*q^2 + q)*x^2 + 9*q^2)*((2*q + 1)*x^2 +
x^3 + 3*q*x + q)^(1/3)*q^(2/3) - sqrt(3)*(3*((4*q^2 + 13*q + 1)*x^4 + 6*(7*q^2 + 5*q)*x^3 + 72*q^2*x + 3*(31*
q^2 + 5*q)*x^2 + 18*q^2)*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(2/3)*q^(2/3) + 3*((q^3 - 5*q^2 - 5*q)*x^5 - 5*(q^3
+ 7*q^2 + q)*x^4 - 45*q^3*x - 45*(q^3 + q^2)*x^3 - 9*q^3 - 15*(5*q^3 + q^2)*x^2)*((2*q + 1)*x^2 + x^3 + 3*q*x
+ q)^(1/3) - ((q^3 + 24*q^2 + 3*q - 1)*x^6 + 54*(q^3 + 2*q^2)*x^5 + 81*(3*q^3 + 2*q^2)*x^4 + 162*q^3*x + 108*
(4*q^3 + q^2)*x^3 + 27*q^3 + 27*(14*q^3 + q^2)*x^2)*q^(1/3))*sqrt(-1/q^(2/3)))/x^6) + 2*q^(2/3)*log(((q - 1)*q
^(2/3)*x^2 - 3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(1/3)*(q*x + q)*q^(1/3) + 3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)
^(2/3)*q)/x^2) - q^(2/3)*log((3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(2/3)*((2*q + 1)*x^2 + 6*q*x + 3*q)*q^(2/3)
+ 3*((q^2 + 2*q)*x^3 + 9*q^2*x + (7*q^2 + 2*q)*x^2 + 3*q^2)*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(1/3) + ((q^2 +
7*q + 1)*x^4 + 18*(q^2 + q)*x^3 + 36*q^2*x + 9*(5*q^2 + q)*x^2 + 9*q^2)*q^(1/3))/x^4))/q, -1/12*(2*sqrt(3)*q^(
2/3)*arctan(1/3*sqrt(3)*(6*((2*q^2 - q - 1)*x^4 + 6*(q^2 - q)*x^3 + 3*(q^2 - q)*x^2)*((2*q + 1)*x^2 + x^3 + 3*
q*x + q)^(2/3)*q^(2/3) - 6*((q^3 + 7*q^2 + q)*x^5 + (19*q^3 + 25*q^2 + q)*x^4 + 45*q^3*x + 9*(7*q^3 + 3*q^2)*x
^3 + 9*q^3 + 9*(9*q^3 + q^2)*x^2)*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(1/3) + ((q^3 - 12*q^2 - 15*q - 1)*x^6 - 1
8*(q^3 + 6*q^2 + 2*q)*x^5 - 9*(17*q^3 + 26*q^2 + 2*q)*x^4 - 162*q^3*x - 180*(2*q^3 + q^2)*x^3 - 27*q^3 - 45*(8
*q^3 + q^2)*x^2)*q^(1/3))/(((q^3 + 24*q^2 + 3*q - 1)*x^6 + 54*(q^3 + 2*q^2)*x^5 + 81*(3*q^3 + 2*q^2)*x^4 + 162
*q^3*x + 108*(4*q^3 + q^2)*x^3 + 27*q^3 + 27*(14*q^3 + q^2)*x^2)*q^(1/3))) - 2*q^(2/3)*log(((q - 1)*q^(2/3)*x^
2 - 3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(1/3)*(q*x + q)*q^(1/3) + 3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(2/3)*q)
/x^2) + q^(2/3)*log((3*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(2/3)*((2*q + 1)*x^2 + 6*q*x + 3*q)*q^(2/3) + 3*((q^2
+ 2*q)*x^3 + 9*q^2*x + (7*q^2 + 2*q)*x^2 + 3*q^2)*((2*q + 1)*x^2 + x^3 + 3*q*x + q)^(1/3) + ((q^2 + 7*q + 1)*
x^4 + 18*(q^2 + q)*x^3 + 36*q^2*x + 9*(5*q^2 + q)*x^2 + 9*q^2)*q^(1/3))/x^4))/q]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(((2*q*x + x^2 + q)*(x + 1))^(1/3)*x), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (\left (1+x \right ) \left (2 q x +x^{2}+q \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/((1+x)*(2*q*x+x^2+q))^(1/3),x)

[Out]

int(1/x/((1+x)*(2*q*x+x^2+q))^(1/3),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(((2*q*x + x^2 + q)*(x + 1))^(1/3)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*((x + 1)*(q + 2*q*x + x^2))^(1/3)),x)

[Out]

int(1/(x*((x + 1)*(q + 2*q*x + x^2))^(1/3)), x)

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sympy [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/x/((1+x)*(2*q*x+x**2+q))**(1/3),x)

[Out]

Timed out

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