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

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

[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

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Rubi [A]  time = 0.0450532, antiderivative size = 149, normalized size of antiderivative = 1., 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 = {1010, 397, 439} \[ \frac{1}{2} a \tan ^{-1}\left (\frac{1-\sqrt{1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac{1}{2} a \tanh ^{-1}\left (\frac{\sqrt{1-x^2}+1}{x \sqrt [4]{1-x^2}}\right )+\frac{b \tan ^{-1}\left (\frac{1-\sqrt{1-x^2}}{\sqrt{2} \sqrt [4]{1-x^2}}\right )}{\sqrt{2}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{1-x^2}+1}{\sqrt{2} \sqrt [4]{1-x^2}}\right )}{\sqrt{2}} \]

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 1010

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]

Rule 397

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

Rule 439

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

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*}

Mathematica [C]  time = 0.203389, size = 144, normalized size = 0.97 \[ \frac{1}{4} b x^2 F_1\left (1;\frac{1}{4},1;2;x^2,\frac{x^2}{2}\right )-\frac{6 a x F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};x^2,\frac{x^2}{2}\right )}{\sqrt [4]{1-x^2} \left (x^2-2\right ) \left (x^2 \left (2 F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};x^2,\frac{x^2}{2}\right )+F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};x^2,\frac{x^2}{2}\right )\right )+6 F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};x^2,\frac{x^2}{2}\right )\right )} \]

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 = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{bx+a}{-{x}^{2}+2}{\frac{1}{\sqrt [4]{-{x}^{2}+1}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\int \frac{b x + a}{{\left (x^{2} - 2\right )}{\left (-x^{2} + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} - \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int -\frac{b x + a}{{\left (x^{2} - 2\right )}{\left (-x^{2} + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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)