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

Optimal. Leaf size=135 \[ -\frac {1}{2} a \tan ^{-1}\left (\frac {\sqrt {x^2+1}+1}{x \sqrt [4]{x^2+1}}\right )-\frac {1}{2} a \tanh ^{-1}\left (\frac {1-\sqrt {x^2+1}}{x \sqrt [4]{x^2+1}}\right )-\frac {b \tan ^{-1}\left (\frac {1-\sqrt {x^2+1}}{\sqrt {2} \sqrt [4]{x^2+1}}\right )}{\sqrt {2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {x^2+1}+1}{\sqrt {2} \sqrt [4]{x^2+1}}\right )}{\sqrt {2}} \]

[Out]

-1/2*a*arctan(((x^2+1)^(1/2)+1)/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*arctan
(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*((x^2+1)^(1/2)+1)/(x^2+1)^(1/4)*2^(1/2
))*2^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1010, 397, 439} \[ -\frac {1}{2} a \tan ^{-1}\left (\frac {\sqrt {x^2+1}+1}{x \sqrt [4]{x^2+1}}\right )-\frac {1}{2} a \tanh ^{-1}\left (\frac {1-\sqrt {x^2+1}}{x \sqrt [4]{x^2+1}}\right )-\frac {b \tan ^{-1}\left (\frac {1-\sqrt {x^2+1}}{\sqrt {2} \sqrt [4]{x^2+1}}\right )}{\sqrt {2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {x^2+1}+1}{\sqrt {2} \sqrt [4]{x^2+1}}\right )}{\sqrt {2}} \]

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 - (a*ArcTanh[(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]

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]

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}{\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 {1}{2} a \tanh ^{-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}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 152, normalized size = 1.13 \[ \frac {1}{4} b x^2 F_1\left (1;\frac {1}{4},1;2;-x^2,-\frac {x^2}{2}\right )-\frac {6 a x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-x^2,-\frac {x^2}{2}\right )}{\sqrt [4]{x^2+1} \left (x^2+2\right ) \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-x^2,-\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-x^2,-\frac {x^2}{2}\right )\right )-6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-x^2,-\frac {x^2}{2}\right )\right )} \]

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

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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+1)^(1/4)/(x^2+2),x, algorithm="fricas")

[Out]

Timed out

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

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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maple [F]  time = 0.12, 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 \[ \int \frac {b x + a}{{\left (x^{2} + 2\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}\,{d x} \]

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

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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