### 3.43 $$\int \frac{x^4 (d+e x^4)}{a+b x^4+c x^8} \, dx$$

Optimal. Leaf size=433 $-\frac{\left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{e x}{c}$

[Out]

(e*x)/c - ((c*d - b*e + (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e - (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b + Sqrt
[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e + (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e - (b*c*d -
b^2*e + 2*a*c*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^(1/4)*c^
(5/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

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Rubi [A]  time = 1.13232, antiderivative size = 433, 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 = {1502, 1422, 212, 208, 205} $-\frac{\left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-\sqrt{b^2-4 a c}-b\right )^{3/4}}-\frac{\left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt{b^2-4 a c}-b}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (\sqrt{b^2-4 a c}-b\right )^{3/4}}+\frac{e x}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*x)/c - ((c*d - b*e + (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e - (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b + Sqrt
[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e + (b*c*d - b^2*e + 2*a*c*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^(1/4)*c^(5/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)) - ((c*d - b*e - (b*c*d -
b^2*e + 2*a*c*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^(1/4)*c^
(5/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))

Rule 1502

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

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rule 212

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

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]

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]

Rubi steps

\begin{align*} \int \frac{x^4 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx &=\frac{e x}{c}-\frac{\int \frac{a e-(c d-b e) x^4}{a+b x^4+c x^8} \, dx}{c}\\ &=\frac{e x}{c}+\frac{\left (c d-b e-\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx}{2 c}+\frac{\left (c d-b e+\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^4} \, dx}{2 c}\\ &=\frac{e x}{c}-\frac{\left (c d-b e-\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx}{2 c \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (c d-b e-\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{-b+\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx}{2 c \sqrt{-b+\sqrt{b^2-4 a c}}}-\frac{\left (c d-b e+\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}-\sqrt{2} \sqrt{c} x^2} \, dx}{2 c \sqrt{-b-\sqrt{b^2-4 a c}}}-\frac{\left (c d-b e+\frac{b c d-b^2 e+2 a c e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{-b-\sqrt{b^2-4 a c}}+\sqrt{2} \sqrt{c} x^2} \, dx}{2 c \sqrt{-b-\sqrt{b^2-4 a c}}}\\ &=\frac{e x}{c}-\frac{\left (c d-b e+\frac{b c d-b^2 e+2 a c 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 \sqrt [4]{2} c^{5/4} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{\left (c d-b e-\frac{b c d-b^2 e+2 a c 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 \sqrt [4]{2} c^{5/4} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{\left (c d-b e+\frac{b c d-b^2 e+2 a c 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 \sqrt [4]{2} c^{5/4} \left (-b-\sqrt{b^2-4 a c}\right )^{3/4}}-\frac{\left (c d-b e-\frac{b c d-b^2 e+2 a c 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 \sqrt [4]{2} c^{5/4} \left (-b+\sqrt{b^2-4 a c}\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.07527, size = 88, normalized size = 0.2 $\frac{e x}{c}-\frac{\text{RootSum}\left [\text{\#1}^4 b+\text{\#1}^8 c+a\& ,\frac{\text{\#1}^4 b e \log (x-\text{\#1})+\text{\#1}^4 (-c) d \log (x-\text{\#1})+a e \log (x-\text{\#1})}{\text{\#1}^3 b+2 \text{\#1}^7 c}\& \right ]}{4 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*x)/c - RootSum[a + b*#1^4 + c*#1^8 & , (a*e*Log[x - #1] - c*d*Log[x - #1]*#1^4 + b*e*Log[x - #1]*#1^4)/(b*#
1^3 + 2*c*#1^7) & ]/(4*c)

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Maple [C]  time = 0.004, size = 67, normalized size = 0.2 \begin{align*}{\frac{ex}{c}}+{\frac{1}{4\,c}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+{{\it \_Z}}^{4}b+a \right ) }{\frac{ \left ( \left ( -be+cd \right ){{\it \_R}}^{4}-ae \right ) \ln \left ( x-{\it \_R} \right ) }{2\,{{\it \_R}}^{7}c+{{\it \_R}}^{3}b}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{e x}{c} - \frac{-\int \frac{{\left (c d - b e\right )} x^{4} - a e}{c x^{8} + b x^{4} + a}\,{d x}}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x/c - integrate(-((c*d - b*e)*x^4 - a*e)/(c*x^8 + b*x^4 + a), x)/c

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

[Out]

Exception raised: TypeError