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

Optimal. Leaf size=80 \[ \frac {a \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}+\frac {a \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}-b \tan ^{-1}\left (\sqrt [4]{x^2-1}\right )+b \tanh ^{-1}\left (\sqrt [4]{x^2-1}\right ) \]

[Out]

-b*arctan((x^2-1)^(1/4))+b*arctanh((x^2-1)^(1/4))+1/4*a*arctan(1/2*x/(x^2-1)^(1/4)*2^(1/2))*2^(1/2)+1/4*a*arct
anh(1/2*x/(x^2-1)^(1/4)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 80, 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 = {1010, 398, 444, 63, 298, 203, 206} \[ \frac {a \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}+\frac {a \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}-b \tan ^{-1}\left (\sqrt [4]{x^2-1}\right )+b \tanh ^{-1}\left (\sqrt [4]{x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rule 206

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 398

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

Rule 444

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.26, size = 157, normalized size = 1.96 \[ \frac {x \left (b x \sqrt [4]{1-x^2} \left (x^2-2\right ) F_1\left (1;\frac {1}{4},1;2;x^2,\frac {x^2}{2}\right )-\frac {24 a F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )}{x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};x^2,\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};x^2,\frac {x^2}{2}\right )\right )+6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )}\right )}{4 \left (x^2-2\right ) \sqrt [4]{x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(b*x*(1 - x^2)^(1/4)*(-2 + x^2)*AppellF1[1, 1/4, 1, 2, x^2, x^2/2] - (24*a*AppellF1[1/2, 1/4, 1, 3/2, x^2,
x^2/2])/(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] + AppellF1[3
/2, 5/4, 1, 5/2, x^2, x^2/2]))))/(4*(-2 + x^2)*(-1 + x^2)^(1/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {b x +a}{\left (-x^{2}+2\right ) \left (x^{2}-1\right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {a+b\,x}{{\left (x^2-1\right )}^{1/4}\,\left (x^2-2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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