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

Optimal. Leaf size=88 \[ \frac{a \tan ^{-1}\left (\frac{x}{\sqrt{2} \sqrt [4]{-x^2-1}}\right )}{2 \sqrt{2}}+\frac{a \tanh ^{-1}\left (\frac{x}{\sqrt{2} \sqrt [4]{-x^2-1}}\right )}{2 \sqrt{2}}+b \tan ^{-1}\left (\sqrt [4]{-x^2-1}\right )-b \tanh ^{-1}\left (\sqrt [4]{-x^2-1}\right ) \]

[Out]

(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)]

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Rubi [A]  time = 0.0475753, antiderivative size = 88, normalized size of antiderivative = 1., 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)/((-1 - x^2)^(1/4)*(2 + x^2)),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 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 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 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 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 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])

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{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}{\sqrt [4]{-1-x} (2+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*}

Mathematica [C]  time = 0.251017, size = 162, normalized size = 1.84 \[ \frac{x \left (b x \sqrt [4]{x^2+1} F_1\left (1;\frac{1}{4},1;2;-x^2,-\frac{x^2}{2}\right )-\frac{24 a F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-x^2,-\frac{x^2}{2}\right )}{\left (x^2+2\right ) \left (x^2 \left (2 F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};-x^2,-\frac{x^2}{2}\right )+F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};-x^2,-\frac{x^2}{2}\right )\right )-6 F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-x^2,-\frac{x^2}{2}\right )\right )}\right )}{4 \sqrt [4]{-x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(b*x*(1 + x^2)^(1/4)*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])
/((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] + A
ppellF1[3/2, 5/4, 1, 5/2, -x^2, -x^2/2])))))/(4*(-1 - x^2)^(1/4))

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Maple [F]  time = 0.029, 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 + b x}{\sqrt [4]{- x^{2} - 1} \left (x^{2} + 2\right )}\, 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 + b*x)/((-x**2 - 1)**(1/4)*(x**2 + 2)), 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)