3.124 \(\int \frac{c+d x}{1+x^4} \, dx\)

Optimal. Leaf size=98 \[ -\frac{c \log \left (x^2-\sqrt{2} x+1\right )}{4 \sqrt{2}}+\frac{c \log \left (x^2+\sqrt{2} x+1\right )}{4 \sqrt{2}}-\frac{c \tan ^{-1}\left (1-\sqrt{2} x\right )}{2 \sqrt{2}}+\frac{c \tan ^{-1}\left (\sqrt{2} x+1\right )}{2 \sqrt{2}}+\frac{1}{2} d \tan ^{-1}\left (x^2\right ) \]

[Out]

(d*ArcTan[x^2])/2 - (c*ArcTan[1 - Sqrt[2]*x])/(2*Sqrt[2]) + (c*ArcTan[1 + Sqrt[2]*x])/(2*Sqrt[2]) - (c*Log[1 -
Sqrt[2]*x + x^2])/(4*Sqrt[2]) + (c*Log[1 + Sqrt[2]*x + x^2])/(4*Sqrt[2])

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Rubi [A]  time = 0.0672962, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {1876, 211, 1165, 628, 1162, 617, 204, 275, 203} \[ -\frac{c \log \left (x^2-\sqrt{2} x+1\right )}{4 \sqrt{2}}+\frac{c \log \left (x^2+\sqrt{2} x+1\right )}{4 \sqrt{2}}-\frac{c \tan ^{-1}\left (1-\sqrt{2} x\right )}{2 \sqrt{2}}+\frac{c \tan ^{-1}\left (\sqrt{2} x+1\right )}{2 \sqrt{2}}+\frac{1}{2} d \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(1 + x^4),x]

[Out]

(d*ArcTan[x^2])/2 - (c*ArcTan[1 - Sqrt[2]*x])/(2*Sqrt[2]) + (c*ArcTan[1 + Sqrt[2]*x])/(2*Sqrt[2]) - (c*Log[1 -
Sqrt[2]*x + x^2])/(4*Sqrt[2]) + (c*Log[1 + Sqrt[2]*x + x^2])/(4*Sqrt[2])

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 211

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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])

Rubi steps

\begin{align*} \int \frac{c+d x}{1+x^4} \, dx &=\int \left (\frac{c}{1+x^4}+\frac{d x}{1+x^4}\right ) \, dx\\ &=c \int \frac{1}{1+x^4} \, dx+d \int \frac{x}{1+x^4} \, dx\\ &=\frac{1}{2} c \int \frac{1-x^2}{1+x^4} \, dx+\frac{1}{2} c \int \frac{1+x^2}{1+x^4} \, dx+\frac{1}{2} d \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} d \tan ^{-1}\left (x^2\right )+\frac{1}{4} c \int \frac{1}{1-\sqrt{2} x+x^2} \, dx+\frac{1}{4} c \int \frac{1}{1+\sqrt{2} x+x^2} \, dx-\frac{c \int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx}{4 \sqrt{2}}-\frac{c \int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx}{4 \sqrt{2}}\\ &=\frac{1}{2} d \tan ^{-1}\left (x^2\right )-\frac{c \log \left (1-\sqrt{2} x+x^2\right )}{4 \sqrt{2}}+\frac{c \log \left (1+\sqrt{2} x+x^2\right )}{4 \sqrt{2}}+\frac{c \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} x\right )}{2 \sqrt{2}}-\frac{c \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} x\right )}{2 \sqrt{2}}\\ &=\frac{1}{2} d \tan ^{-1}\left (x^2\right )-\frac{c \tan ^{-1}\left (1-\sqrt{2} x\right )}{2 \sqrt{2}}+\frac{c \tan ^{-1}\left (1+\sqrt{2} x\right )}{2 \sqrt{2}}-\frac{c \log \left (1-\sqrt{2} x+x^2\right )}{4 \sqrt{2}}+\frac{c \log \left (1+\sqrt{2} x+x^2\right )}{4 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.067398, size = 99, normalized size = 1.01 \[ \frac{1}{4} \left (-\left (\sqrt [4]{-1} c+i d\right ) \log \left (\sqrt [4]{-1}-x\right )+\left (-(-1)^{3/4} c+i d\right ) \log \left ((-1)^{3/4}-x\right )+\left (\sqrt [4]{-1} c-i d\right ) \log \left (x+\sqrt [4]{-1}\right )+\left ((-1)^{3/4} c+i d\right ) \log \left (x+(-1)^{3/4}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(1 + x^4),x]

[Out]

(-(((-1)^(1/4)*c + I*d)*Log[(-1)^(1/4) - x]) + (-((-1)^(3/4)*c) + I*d)*Log[(-1)^(3/4) - x] + ((-1)^(1/4)*c - I
*d)*Log[(-1)^(1/4) + x] + ((-1)^(3/4)*c + I*d)*Log[(-1)^(3/4) + x])/4

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Maple [A]  time = 0.004, size = 68, normalized size = 0.7 \begin{align*}{\frac{c\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{4}}+{\frac{c\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{4}}+{\frac{c\sqrt{2}}{8}\ln \left ({\frac{1+{x}^{2}+x\sqrt{2}}{1+{x}^{2}-x\sqrt{2}}} \right ) }+{\frac{d\arctan \left ({x}^{2} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(x^4+1),x)

[Out]

1/4*c*arctan(1+x*2^(1/2))*2^(1/2)+1/4*c*arctan(-1+x*2^(1/2))*2^(1/2)+1/8*c*2^(1/2)*ln((1+x^2+x*2^(1/2))/(1+x^2
-x*2^(1/2)))+1/2*d*arctan(x^2)

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Maxima [A]  time = 1.53606, size = 116, normalized size = 1.18 \begin{align*} \frac{1}{8} \, \sqrt{2} c \log \left (x^{2} + \sqrt{2} x + 1\right ) - \frac{1}{8} \, \sqrt{2} c \log \left (x^{2} - \sqrt{2} x + 1\right ) + \frac{1}{4} \,{\left (\sqrt{2} c - 2 \, d\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{4} \,{\left (\sqrt{2} c + 2 \, d\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(x^4+1),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*c*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt(2)*c*log(x^2 - sqrt(2)*x + 1) + 1/4*(sqrt(2)*c - 2*d)*arctan
(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*(sqrt(2)*c + 2*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)))

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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((d*x+c)/(x^4+1),x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]  time = 0.411487, size = 83, normalized size = 0.85 \begin{align*} \operatorname{RootSum}{\left (256 t^{4} + 32 t^{2} d^{2} - 16 t c^{2} d + c^{4} + d^{4}, \left ( t \mapsto t \log{\left (x + \frac{128 t^{3} d^{2} + 16 t^{2} c^{2} d + 4 t c^{4} + 8 t d^{4} - 5 c^{2} d^{3}}{c^{5} - 4 c d^{4}} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(x**4+1),x)

[Out]

RootSum(256*_t**4 + 32*_t**2*d**2 - 16*_t*c**2*d + c**4 + d**4, Lambda(_t, _t*log(x + (128*_t**3*d**2 + 16*_t*
*2*c**2*d + 4*_t*c**4 + 8*_t*d**4 - 5*c**2*d**3)/(c**5 - 4*c*d**4))))

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Giac [A]  time = 1.06388, size = 116, normalized size = 1.18 \begin{align*} \frac{1}{8} \, \sqrt{2} c \log \left (x^{2} + \sqrt{2} x + 1\right ) - \frac{1}{8} \, \sqrt{2} c \log \left (x^{2} - \sqrt{2} x + 1\right ) + \frac{1}{4} \,{\left (\sqrt{2} c - 2 \, d\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{4} \,{\left (\sqrt{2} c + 2 \, d\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(x^4+1),x, algorithm="giac")

[Out]

1/8*sqrt(2)*c*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt(2)*c*log(x^2 - sqrt(2)*x + 1) + 1/4*(sqrt(2)*c - 2*d)*arctan
(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*(sqrt(2)*c + 2*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)))