Integrand size = 18, antiderivative size = 69 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 a \sqrt {b^2-4 a c}}+\frac {\log (x)}{a}-\frac {\log \left (a+b x^4+c x^8\right )}{8 a} \] Output:
1/4*b*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(1/2)+ln(x)/a -1/8*ln(c*x^8+b*x^4+a)/a
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\frac {\log (x)}{a}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log (x-\text {$\#$1})+c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b+2 c \text {$\#$1}^4}\&\right ]}{4 a} \] Input:
Integrate[1/(x*(a + b*x^4 + c*x^8)),x]
Output:
Log[x]/a - RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[x - #1] + c*Log[x - #1]* #1^4)/(b + 2*c*#1^4) & ]/(4*a)
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1693, 1144, 25, 1142, 1083, 219, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^4 \left (c x^8+b x^4+a\right )}dx^4\) |
\(\Big \downarrow \) 1144 |
\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {c x^4+b}{c x^8+b x^4+a}dx^4}{a}+\frac {\log \left (x^4\right )}{a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right )}{a}-\frac {\int \frac {c x^4+b}{c x^8+b x^4+a}dx^4}{a}\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right )}{a}-\frac {\frac {1}{2} b \int \frac {1}{c x^8+b x^4+a}dx^4+\frac {1}{2} \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{a}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4-b \int \frac {1}{-x^8+b^2-4 a c}d\left (2 c x^4+b\right )}{a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4-\frac {b \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right )}{a}-\frac {\frac {1}{2} \log \left (a+b x^4+c x^8\right )-\frac {b \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}\right )\) |
Input:
Int[1/(x*(a + b*x^4 + c*x^8)),x]
Output:
(Log[x^4]/a - (-((b*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4 *a*c]) + Log[a + b*x^4 + c*x^8]/2)/a)/4
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[e*(Log[RemoveContent[d + e*x, x]]/(c*d^2 - b*d*e + a*e^2)), x] + S imp[1/(c*d^2 - b*d*e + a*e^2) Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ [Simplify[(m + 1)/n]]
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {\frac {\ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4}+\frac {b \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}}}{2 a}+\frac {\ln \left (x \right )}{a}\) | \(66\) |
risch | \(\frac {\ln \left (x \right )}{a}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c -b^{2} a \right ) \textit {\_Z}^{2}+\left (4 a c -b^{2}\right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (18 a c -5 b^{2}\right ) \textit {\_R} +9 c \right ) x^{4}-\textit {\_R} a b +4 b \right )\right )}{4}\) | \(77\) |
Input:
int(1/x/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/2/a*(1/4*ln(c*x^8+b*x^4+a)+1/2*b/(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/( 4*a*c-b^2)^(1/2)))+ln(x)/a
Time = 0.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.23 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c + {\left (2 \, c x^{4} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{8 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{4} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{8 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \] Input:
integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
[1/8*(sqrt(b^2 - 4*a*c)*b*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)) - (b^2 - 4*a*c)*log(c*x^8 + b*x^4 + a) + 8*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2*c), 1/8*(2*sqrt(-b^ 2 + 4*a*c)*b*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - (b^ 2 - 4*a*c)*log(c*x^8 + b*x^4 + a) + 8*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2 *c)]
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (60) = 120\).
Time = 9.57 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.67 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\left (- \frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) \log {\left (x^{4} + \frac {- 16 a^{2} c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) + 4 a b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) \log {\left (x^{4} + \frac {- 16 a^{2} c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) + 4 a b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac {1}{8 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \frac {\log {\left (x \right )}}{a} \] Input:
integrate(1/x/(c*x**8+b*x**4+a),x)
Output:
(-b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a))*log(x**4 + (-16*a* *2*c*(-b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) + 4*a*b**2*(- b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) - 2*a*c + b**2)/(b*c )) + (b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a))*log(x**4 + (-1 6*a**2*c*(b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) + 4*a*b**2 *(b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) - 2*a*c + b**2)/(b *c)) + log(x)/a
Exception generated. \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 1.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=-\frac {b \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} a} - \frac {\log \left (c x^{8} + b x^{4} + a\right )}{8 \, a} + \frac {\log \left (x^{4}\right )}{4 \, a} \] Input:
integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
-1/4*b*arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a) - 1 /8*log(c*x^8 + b*x^4 + a)/a + 1/4*log(x^4)/a
Time = 19.91 (sec) , antiderivative size = 1690, normalized size of antiderivative = 24.49 \[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:
int(1/(x*(a + b*x^4 + c*x^8)),x)
Output:
log(x)/a + (log(a + b*x^4 + c*x^8)*(16*a*c - 4*b^2))/(2*(16*a*b^2 - 64*a^2 *c)) - (b*atan((4*(4*a*c - b^2)^2*(5*b^6 - 18*a^3*c^3 + 61*a^2*b^2*c^2 - 3 4*a*b^4*c)*((b^9*c^4)/(128*a^4*(4*a*c - b^2)^(5/2)) + (2*b^5*c^4*(16*a*c - 4*b^2)^4)/((16*a*b^2 - 64*a^2*c)^4*(4*a*c - b^2)^(1/2)) - (b*(16*a*c - 4* b^2)^3*(256*b^4*c^4 - (128*a*b^4*c^4*(16*a*c - 4*b^2))/(16*a*b^2 - 64*a^2* c)))/(16*a*(16*a*b^2 - 64*a^2*c)^3*(4*a*c - b^2)^(1/2)) + (b^3*(16*a*c - 4 *b^2)*(256*b^4*c^4 - (128*a*b^4*c^4*(16*a*c - 4*b^2))/(16*a*b^2 - 64*a^2*c )))/(256*a^3*(16*a*b^2 - 64*a^2*c)*(4*a*c - b^2)^(3/2)) - (3*b^7*c^4*(16*a *c - 4*b^2)^2)/(4*a^2*(16*a*b^2 - 64*a^2*c)^2*(4*a*c - b^2)^(3/2))))/(b^4* c^8*(81*a*c - 20*b^2)) + (128*a^5*x^4*(((5*b^5 + 23*a^2*b*c^2 - 24*a*b^3*c )*(((576*b^3*c^5 - ((1280*b^5*c^4 - 4608*a*b^3*c^5)*(16*a*c - 4*b^2))/(2*( 16*a*b^2 - 64*a^2*c)))*(16*a*c - 4*b^2)^4)/(16*(16*a*b^2 - 64*a^2*c)^4) + (b^4*(576*b^3*c^5 - ((1280*b^5*c^4 - 4608*a*b^3*c^5)*(16*a*c - 4*b^2))/(2* (16*a*b^2 - 64*a^2*c))))/(4096*a^4*(4*a*c - b^2)^2) + (b^2*(1280*b^5*c^4 - 4608*a*b^3*c^5)*(16*a*c - 4*b^2)^3)/(128*a^2*(16*a*b^2 - 64*a^2*c)^3*(4*a *c - b^2)) - (3*b^2*(576*b^3*c^5 - ((1280*b^5*c^4 - 4608*a*b^3*c^5)*(16*a* c - 4*b^2))/(2*(16*a*b^2 - 64*a^2*c)))*(16*a*c - 4*b^2)^2)/(128*a^2*(16*a* b^2 - 64*a^2*c)^2*(4*a*c - b^2)) - (b^4*(1280*b^5*c^4 - 4608*a*b^3*c^5)*(1 6*a*c - 4*b^2))/(2048*a^4*(16*a*b^2 - 64*a^2*c)*(4*a*c - b^2)^2)))/(32*a^5 *c^4*(81*a*c - 20*b^2)) + ((5*b^6 - 18*a^3*c^3 + 61*a^2*b^2*c^2 - 34*a*...
\[ \int \frac {1}{x \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(1/x/(c*x^8+b*x^4+a),x)
Output:
int(1/x/(c*x^8+b*x^4+a),x)