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

Optimal. Leaf size=78 \[ \frac{(b d-2 a e) \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a \sqrt{b^2-4 a c}}-\frac{d \log \left (a+b x^4+c x^8\right )}{8 a}+\frac{d \log (x)}{a} \]

[Out]

((b*d - 2*a*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a*Sqrt[b^2 - 4*a*c]) + (d*Log[x])/a - (d*Log[a + b
*x^4 + c*x^8])/(8*a)

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Rubi [A]  time = 0.125543, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1474, 800, 634, 618, 206, 628} \[ \frac{(b d-2 a e) \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a \sqrt{b^2-4 a c}}-\frac{d \log \left (a+b x^4+c x^8\right )}{8 a}+\frac{d \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - 2*a*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a*Sqrt[b^2 - 4*a*c]) + (d*Log[x])/a - (d*Log[a + b
*x^4 + c*x^8])/(8*a)

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

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]

Rubi steps

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

Mathematica [C]  time = 0.0339713, size = 80, normalized size = 1.03 \[ \frac{d \log (x)}{a}-\frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{\text{$\#$1}^4 c d \log (x-\text{$\#$1})-a e \log (x-\text{$\#$1})+b d \log (x-\text{$\#$1})}{2 \text{$\#$1}^4 c+b}\& \right ]}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Log[x])/a - 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 +
2*c*#1^4) & ]/(4*a)

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Maple [A]  time = 0.007, size = 106, normalized size = 1.4 \begin{align*}{\frac{d\ln \left ( x \right ) }{a}}-{\frac{d\ln \left ( c{x}^{8}+b{x}^{4}+a \right ) }{8\,a}}+{\frac{e}{2}\arctan \left ({(2\,c{x}^{4}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{bd}{4\,a}\arctan \left ({(2\,c{x}^{4}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.89877, size = 556, normalized size = 7.13 \begin{align*} \left [-\frac{{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{8} + b x^{4} + a\right ) - 8 \,{\left (b^{2} - 4 \, a c\right )} d \log \left (x\right ) + \sqrt{b^{2} - 4 \, a c}{\left (b d - 2 \, a e\right )} \log \left (\frac{2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c -{\left (2 \, c x^{4} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right )}{8 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}, -\frac{{\left (b^{2} - 4 \, a c\right )} d \log \left (c x^{8} + b x^{4} + a\right ) - 8 \,{\left (b^{2} - 4 \, a c\right )} d \log \left (x\right ) - 2 \, \sqrt{-b^{2} + 4 \, a c}{\left (b d - 2 \, a e\right )} \arctan \left (-\frac{{\left (2 \, c x^{4} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{8 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b^2 - 4*a*c)*d*log(c*x^8 + b*x^4 + a) - 8*(b^2 - 4*a*c)*d*log(x) + sqrt(b^2 - 4*a*c)*(b*d - 2*a*e)*log
((2*c^2*x^8 + 2*b*c*x^4 + b^2 - 2*a*c - (2*c*x^4 + b)*sqrt(b^2 - 4*a*c))/(c*x^8 + b*x^4 + a)))/(a*b^2 - 4*a^2*
c), -1/8*((b^2 - 4*a*c)*d*log(c*x^8 + b*x^4 + a) - 8*(b^2 - 4*a*c)*d*log(x) - 2*sqrt(-b^2 + 4*a*c)*(b*d - 2*a*
e)*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)))/(a*b^2 - 4*a^2*c)]

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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/(c*x**8+b*x**4+a),x)

[Out]

Timed out

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Giac [A]  time = 6.59181, size = 105, normalized size = 1.35 \begin{align*} -\frac{d \log \left (c x^{8} + b x^{4} + a\right )}{8 \, a} + \frac{d \log \left (x^{4}\right )}{4 \, a} - \frac{{\left (b d - 2 \, a e\right )} \arctan \left (\frac{2 \, c x^{4} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt{-b^{2} + 4 \, a c} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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