3.1.71 \(\int \frac {a+b x}{\sqrt [4]{1-x^2} (2-x^2)} \, dx\) [71]

Optimal. Leaf size=149 \[ \frac {b \tan ^{-1}\left (\frac {1-\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {1}{2} a \tan ^{-1}\left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac {b \tanh ^{-1}\left (\frac {1+\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {1}{2} a \tanh ^{-1}\left (\frac {1+\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right ) \]

[Out]

1/2*a*arctan((1-(-x^2+1)^(1/2))/x/(-x^2+1)^(1/4))+1/2*a*arctanh((1+(-x^2+1)^(1/2))/x/(-x^2+1)^(1/4))+1/2*b*arc
tan(1/2*(1-(-x^2+1)^(1/2))/(-x^2+1)^(1/4)*2^(1/2))*2^(1/2)+1/2*b*arctanh(1/2*(1+(-x^2+1)^(1/2))/(-x^2+1)^(1/4)
*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1024, 406, 450} \begin {gather*} \frac {1}{2} a \text {ArcTan}\left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac {1}{2} a \tanh ^{-1}\left (\frac {\sqrt {1-x^2}+1}{x \sqrt [4]{1-x^2}}\right )+\frac {b \text {ArcTan}\left (\frac {1-\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {1-x^2}+1}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((1 - x^2)^(1/4)*(2 - x^2)),x]

[Out]

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

Rule 406

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

Rule 450

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

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 10.18, size = 144, normalized size = 0.97 \begin {gather*} \frac {1}{4} b x^2 F_1\left (1;\frac {1}{4},1;2;x^2,\frac {x^2}{2}\right )-\frac {6 a x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )}{\sqrt [4]{1-x^2} \left (-2+x^2\right ) \left (6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )+x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};x^2,\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};x^2,\frac {x^2}{2}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x)/((1 - x^2)^(1/4)*(2 - x^2)),x]

[Out]

(b*x^2*AppellF1[1, 1/4, 1, 2, x^2, x^2/2])/4 - (6*a*x*AppellF1[1/2, 1/4, 1, 3/2, x^2, x^2/2])/((1 - x^2)^(1/4)
*(-2 + x^2)*(6*AppellF1[1/2, 1/4, 1, 3/2, x^2, x^2/2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, x^2, x^2/2] + Appell
F1[3/2, 5/4, 1, 5/2, x^2, x^2/2])))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {b x +a}{\left (-x^{2}+1\right )^{\frac {1}{4}} \left (-x^{2}+2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(-x^2+1)^(1/4)/(-x^2+2),x)

[Out]

int((b*x+a)/(-x^2+1)^(1/4)/(-x^2+2),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)^(1/4)/(-x^2+2),x, algorithm="maxima")

[Out]

-integrate((b*x + a)/((x^2 - 2)*(-x^2 + 1)^(1/4)), x)

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Fricas [F(-1)] Timed out
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((b*x+a)/(-x^2+1)^(1/4)/(-x^2+2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a}{x^{2} \sqrt [4]{1 - x^{2}} - 2 \sqrt [4]{1 - x^{2}}}\, dx - \int \frac {b x}{x^{2} \sqrt [4]{1 - x^{2}} - 2 \sqrt [4]{1 - x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(-x**2+1)**(1/4)/(-x**2+2),x)

[Out]

-Integral(a/(x**2*(1 - x**2)**(1/4) - 2*(1 - x**2)**(1/4)), x) - Integral(b*x/(x**2*(1 - x**2)**(1/4) - 2*(1 -
x**2)**(1/4)), 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)^(1/4)/(-x^2+2),x, algorithm="giac")

[Out]

integrate(-(b*x + a)/((x^2 - 2)*(-x^2 + 1)^(1/4)), 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 )}^{1/4}\,\left (x^2-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*x)/((1 - x^2)^(1/4)*(x^2 - 2)),x)

[Out]

int(-(a + b*x)/((1 - x^2)^(1/4)*(x^2 - 2)), x)

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