3.1.53 \(\int \frac {a+b x}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\) [53]

Optimal. Leaf size=198 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {a \tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]

[Out]

-1/12*a*arctanh(x)*2^(1/3)+1/4*a*arctanh(x/(1+2^(1/3)*(-x^2+1)^(1/3)))*2^(1/3)-1/8*b*ln(x^2+3)*2^(1/3)+3/8*b*l
n(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+1/12*a*arctan(1/x*3^(1/2))*2^(1/3)*3^(1/2)+1/12*a*arctan((1-2^(1/3)*(-x^2+1)
^(1/3))*3^(1/2)/x)*2^(1/3)*3^(1/2)+1/4*b*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)

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Rubi [A]
time = 0.06, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1024, 402, 455, 57, 631, 210, 31} \begin {gather*} \frac {a \text {ArcTan}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \text {ArcTan}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \tanh ^{-1}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[Sqrt[3]/x])/(2*2^(2/3)*Sqrt[3]) + (Sqrt[3]*b*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) +
(a*ArcTan[(Sqrt[3]*(1 - 2^(1/3)*(1 - x^2)^(1/3)))/x])/(2*2^(2/3)*Sqrt[3]) - (a*ArcTanh[x])/(6*2^(2/3)) + (a*Ar
cTanh[x/(1 + 2^(1/3)*(1 - x^2)^(1/3))])/(2*2^(2/3)) - (b*Log[3 + x^2])/(4*2^(2/3)) + (3*b*Log[2^(2/3) - (1 - x
^2)^(1/3)])/(4*2^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 402

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-b/a, 2]}, Simp[q*(ArcTan
[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x] + (Simp[q*(ArcTanh[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a + b*
x^2)^(1/3))]/(2*2^(2/3)*a^(1/3)*d)), x] - Simp[q*(ArcTanh[q*x]/(6*2^(2/3)*a^(1/3)*d)), x] + Simp[q*(ArcTan[Sqr
t[3]*((a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3))/(a^(1/3)*q*x))]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x])] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && NegQ[b/a]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1024

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rubi steps

\begin {align*} \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx &=a \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx+b \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {a \tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}+\frac {1}{2} b \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {a \tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {1}{4} (3 b) \text {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {(3 b) \text {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {a \tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{2\ 2^{2/3}}\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {a \tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 10.17, size = 145, normalized size = 0.73 \begin {gather*} \frac {1}{6} b x^2 F_1\left (1;\frac {1}{3},1;2;x^2,-\frac {x^2}{3}\right )-\frac {9 a x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{1-x^2} \left (3+x^2\right ) \left (-9 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};x^2,-\frac {x^2}{3}\right )+2 x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};x^2,-\frac {x^2}{3}\right )-F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};x^2,-\frac {x^2}{3}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*x^2*AppellF1[1, 1/3, 1, 2, x^2, -1/3*x^2])/6 - (9*a*x*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2])/((1 - x^2)
^(1/3)*(3 + x^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, x^2, -1/3*x
^2] - AppellF1[3/2, 4/3, 1, 5/2, x^2, -1/3*x^2])))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {b x +a}{\left (-x^{2}+1\right )^{\frac {1}{3}} \left (x^{2}+3\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(-x**2+1)**(1/3)/(x**2+3),x)

[Out]

Integral((a + b*x)/((-(x - 1)*(x + 1))**(1/3)*(x**2 + 3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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