### 3.49 $$\int \frac{d+e x^4}{x^2 (a+b x^4+c x^8)} \, dx$$

Optimal. Leaf size=392 $-\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b d-2 a e}{\sqrt{b^2-4 a c}}+d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b d-2 a e}{\sqrt{b^2-4 a c}}+d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{d}{a x}$

[Out]

-(d/(a*x)) - (c^(1/4)*(d - (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c]
)^(1/4)])/(2*2^(3/4)*a*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(d + (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTan
[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*a*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*
(d - (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*
a*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(d + (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*x)
/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*a*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

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Rubi [A]  time = 0.682775, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {1504, 1510, 298, 205, 208} $-\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}-\frac{\sqrt [4]{c} \left (\frac{b d-2 a e}{\sqrt{b^2-4 a c}}+d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{-\sqrt{b^2-4 a c}-b}}+\frac{\sqrt [4]{c} \left (\frac{b d-2 a e}{\sqrt{b^2-4 a c}}+d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2\ 2^{3/4} a \sqrt [4]{\sqrt{b^2-4 a c}-b}}-\frac{d}{a x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x^4)/(x^2*(a + b*x^4 + c*x^8)),x]

[Out]

-(d/(a*x)) - (c^(1/4)*(d - (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c]
)^(1/4)])/(2*2^(3/4)*a*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - (c^(1/4)*(d + (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTan
[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*a*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*
(d - (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*
a*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) + (c^(1/4)*(d + (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*x)
/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)*a*(-b + Sqrt[b^2 - 4*a*c])^(1/4))

Rule 1504

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
Simp[(d*(f*x)^(m + 1)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^n*(m + 1)), Int[(f*x)^
(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(p + 1) + 1)*x^n,
x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -
1] && IntegerQ[p]

Rule 1510

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 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 205

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{d+e x^4}{x^2 \left (a+b x^4+c x^8\right )} \, dx &=-\frac{d}{a x}-\frac{\int \frac{x^2 \left (b d-a e+c d x^4\right )}{a+b x^4+c x^8} \, dx}{a}\\ &=-\frac{d}{a x}-\frac{\left (c \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{x^2}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx}{2 a}-\frac{\left (c \left (d+\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{x^2}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx}{2 a}\\ &=-\frac{d}{a x}+\frac{\left (\sqrt{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx}{2 \sqrt{2} a}-\frac{\left (\sqrt{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx}{2 \sqrt{2} a}+\frac{\left (\sqrt{c} \left (d+\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx}{2 \sqrt{2} a}-\frac{\left (\sqrt{c} \left (d+\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx}{2 \sqrt{2} a}\\ &=-\frac{d}{a x}-\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt{b^2-4 a c}}}-\frac{\sqrt [4]{c} \left (d+\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (d-\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt{b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt{b^2-4 a c}}}+\frac{\sqrt [4]{c} \left (d+\frac{b d-2 a e}{\sqrt{b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt{b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.0653445, size = 85, normalized size = 0.22 $-\frac{\text{RootSum}\left [\text{\#1}^4 b+\text{\#1}^8 c+a\& ,\frac{\text{\#1}^4 c d \log (x-\text{\#1})-a e \log (x-\text{\#1})+b d \log (x-\text{\#1})}{2 \text{\#1}^5 c+\text{\#1} b}\& \right ]}{4 a}-\frac{d}{a x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x^4)/(x^2*(a + b*x^4 + c*x^8)),x]

[Out]

-(d/(a*x)) - RootSum[a + b*#1^4 + c*#1^8 & , (b*d*Log[x - #1] - a*e*Log[x - #1] + c*d*Log[x - #1]*#1^4)/(b*#1
+ 2*c*#1^5) & ]/(4*a)

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Maple [C]  time = 0.006, size = 72, normalized size = 0.2 \begin{align*} -{\frac{1}{4\,a}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{ \left ( cd{{\it \_R}}^{6}+ \left ( -ae+bd \right ){{\it \_R}}^{2} \right ) \ln \left ( x-{\it \_R} \right ) }{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}}}-{\frac{d}{ax}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^4+d)/x^2/(c*x^8+b*x^4+a),x)

[Out]

-1/4/a*sum((c*d*_R^6+(-a*e+b*d)*_R^2)/(2*_R^7*c+_R^3*b)*ln(x-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))-d/a/x

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Maxima [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((e*x^4+d)/x^2/(c*x^8+b*x^4+a),x, algorithm="maxima")

[Out]

Timed out

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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((e*x^4+d)/x^2/(c*x^8+b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((e*x**4+d)/x**2/(c*x**8+b*x**4+a),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^4+d)/x^2/(c*x^8+b*x^4+a),x, algorithm="giac")

[Out]

Exception raised: TypeError