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

Optimal. Leaf size=169 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{3 x^2+1}+1}\right )}{4 \sqrt {3}}-\frac {i \tanh ^{-1}\left (\frac {2 i \sqrt {3} x-2 i \sqrt {3} x \sqrt [3]{3 x^2+1}}{3 x^2-4 \left (3 x^2+1\right )^{2/3}+2 \sqrt [3]{3 x^2+1}-1}\right )}{8 \sqrt {3}}+\frac {1}{8} \tanh ^{-1}\left (\frac {6 x \sqrt [3]{3 x^2+1}}{3 x^2+4 \left (3 x^2+1\right )^{2/3}-2 \sqrt [3]{3 x^2+1}+1}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 81, normalized size of antiderivative = 0.48, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {394} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\left (1-\sqrt [3]{3 x^2+1}\right )^2}{3 \sqrt {3} x}\right )}{4 \sqrt {3}}-\frac {1}{4} \tanh ^{-1}\left (\frac {1-\sqrt [3]{3 x^2+1}}{x}\right )+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 394

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b/a, 2]}, Simp[(q*ArcTan[
(q*x)/3])/(12*Rt[a, 3]*d), x] + (Simp[(q*ArcTan[(Rt[a, 3] - (a + b*x^2)^(1/3))^2/(3*Rt[a, 3]^2*q*x)])/(12*Rt[a
, 3]*d), x] - Simp[(q*ArcTanh[(Sqrt[3]*(Rt[a, 3] - (a + b*x^2)^(1/3)))/(Rt[a, 3]*q*x)])/(4*Sqrt[3]*Rt[a, 3]*d)
, x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c - 9*a*d, 0] && PosQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (3+x^2\right ) \sqrt [3]{1+3 x^2}} \, dx &=\frac {\tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\left (1-\sqrt [3]{1+3 x^2}\right )^2}{3 \sqrt {3} x}\right )}{4 \sqrt {3}}-\frac {1}{4} \tanh ^{-1}\left (\frac {1-\sqrt [3]{1+3 x^2}}{x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 126, normalized size = 0.75 \begin {gather*} -\frac {9 x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-3 x^2,-\frac {x^2}{3}\right )}{\left (x^2+3\right ) \sqrt [3]{3 x^2+1} \left (2 x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};-3 x^2,-\frac {x^2}{3}\right )+3 F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};-3 x^2,-\frac {x^2}{3}\right )\right )-9 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-3 x^2,-\frac {x^2}{3}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 4.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (3+x^2\right ) \sqrt [3]{1+3 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 2.51, size = 345, normalized size = 2.04 \begin {gather*} \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (3 \, x^{4} - 10 \, x^{3} - 36 \, x^{2} + 18 \, x + 9\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {2}{3}} - 4 \, \sqrt {3} {\left (x^{5} + 15 \, x^{4} - 26 \, x^{3} - 54 \, x^{2} + 9 \, x - 9\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{6} - 2 \, x^{5} - 105 \, x^{4} - 28 \, x^{3} + 63 \, x^{2} + 126 \, x + 9\right )}}{x^{6} + 126 \, x^{5} - 225 \, x^{4} - 828 \, x^{3} - 81 \, x^{2} - 162 \, x + 81}\right ) - \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {2 \, {\left (2 \, \sqrt {3} {\left (23 \, x^{3} + 9 \, x\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {2}{3}} + \sqrt {3} {\left (x^{5} - 80 \, x^{3} - 9 \, x\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (11 \, x^{5} + 10 \, x^{3} - 9 \, x\right )}\right )}}{x^{6} - 657 \, x^{4} - 189 \, x^{2} - 27}\right ) + \frac {1}{24} \, \log \left (\frac {x^{6} + 108 \, x^{5} + 549 \, x^{4} + 360 \, x^{3} + 99 \, x^{2} + 6 \, {\left (3 \, x^{4} + 32 \, x^{3} + 42 \, x^{2} + 3\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {2}{3}} + 6 \, {\left (x^{5} + 27 \, x^{4} + 70 \, x^{3} + 18 \, x^{2} + 9 \, x + 3\right )} {\left (3 \, x^{2} + 1\right )}^{\frac {1}{3}} + 108 \, x - 9}{x^{6} + 9 \, x^{4} + 27 \, x^{2} + 27}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*sqrt(3)*arctan((4*sqrt(3)*(3*x^4 - 10*x^3 - 36*x^2 + 18*x + 9)*(3*x^2 + 1)^(2/3) - 4*sqrt(3)*(x^5 + 15*x^
4 - 26*x^3 - 54*x^2 + 9*x - 9)*(3*x^2 + 1)^(1/3) + sqrt(3)*(x^6 - 2*x^5 - 105*x^4 - 28*x^3 + 63*x^2 + 126*x +
9))/(x^6 + 126*x^5 - 225*x^4 - 828*x^3 - 81*x^2 - 162*x + 81)) - 1/36*sqrt(3)*arctan(2*(2*sqrt(3)*(23*x^3 + 9*
x)*(3*x^2 + 1)^(2/3) + sqrt(3)*(x^5 - 80*x^3 - 9*x)*(3*x^2 + 1)^(1/3) + sqrt(3)*(11*x^5 + 10*x^3 - 9*x))/(x^6
- 657*x^4 - 189*x^2 - 27)) + 1/24*log((x^6 + 108*x^5 + 549*x^4 + 360*x^3 + 99*x^2 + 6*(3*x^4 + 32*x^3 + 42*x^2
+ 3)*(3*x^2 + 1)^(2/3) + 6*(x^5 + 27*x^4 + 70*x^3 + 18*x^2 + 9*x + 3)*(3*x^2 + 1)^(1/3) + 108*x - 9)/(x^6 + 9
*x^4 + 27*x^2 + 27))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.84, size = 445, normalized size = 2.63

method	result	size
trager	\(\frac {\ln \left (-\frac {12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}} x +6 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x^{2}+\left (3 x^{2}+1\right )^{\frac {2}{3}}-12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}}-\left (3 x^{2}+1\right )^{\frac {1}{3}} x -24 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -x^{2}+\left (3 x^{2}+1\right )^{\frac {1}{3}}-6 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+4 x +1}{x^{2}+3}\right )}{4}-\ln \left (-\frac {12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}} x +6 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x^{2}+\left (3 x^{2}+1\right )^{\frac {2}{3}}-12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}}-\left (3 x^{2}+1\right )^{\frac {1}{3}} x -24 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -x^{2}+\left (3 x^{2}+1\right )^{\frac {1}{3}}-6 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+4 x +1}{x^{2}+3}\right ) \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+\RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \ln \left (-\frac {-24 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}} x -12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x^{2}+2 \left (3 x^{2}+1\right )^{\frac {2}{3}}+24 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \left (3 x^{2}+1\right )^{\frac {1}{3}}+4 \left (3 x^{2}+1\right )^{\frac {1}{3}} x +48 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +x^{2}-4 \left (3 x^{2}+1\right )^{\frac {1}{3}}+12 \RootOf \left (48 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )-4 x -1}{x^{2}+3}\right )\)	\(445\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2+3)/(3*x^2+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(-(12*RootOf(48*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)*x+6*RootOf(48*_Z^2-12*_Z+1)*x^2+(3*x^2+1)^(2/3)-12*RootOf(
48*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)-(3*x^2+1)^(1/3)*x-24*RootOf(48*_Z^2-12*_Z+1)*x-x^2+(3*x^2+1)^(1/3)-6*RootOf(4
8*_Z^2-12*_Z+1)+4*x+1)/(x^2+3))-ln(-(12*RootOf(48*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)*x+6*RootOf(48*_Z^2-12*_Z+1)*x^
2+(3*x^2+1)^(2/3)-12*RootOf(48*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)-(3*x^2+1)^(1/3)*x-24*RootOf(48*_Z^2-12*_Z+1)*x-x^
2+(3*x^2+1)^(1/3)-6*RootOf(48*_Z^2-12*_Z+1)+4*x+1)/(x^2+3))*RootOf(48*_Z^2-12*_Z+1)+RootOf(48*_Z^2-12*_Z+1)*ln
(-(-24*RootOf(48*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)*x-12*RootOf(48*_Z^2-12*_Z+1)*x^2+2*(3*x^2+1)^(2/3)+24*RootOf(48
*_Z^2-12*_Z+1)*(3*x^2+1)^(1/3)+4*(3*x^2+1)^(1/3)*x+48*RootOf(48*_Z^2-12*_Z+1)*x+x^2-4*(3*x^2+1)^(1/3)+12*RootO
f(48*_Z^2-12*_Z+1)-4*x-1)/(x^2+3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (x^2+3\right )\,{\left (3\,x^2+1\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((x**2 + 3)*(3*x**2 + 1)**(1/3)), x)

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