Integrand size = 18, antiderivative size = 89 \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{4 a x^4}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^4+c x^8\right )}{8 a^2} \] Output:
-1/4/a/x^4-1/4*(-2*a*c+b^2)*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/a^2/(- 4*a*c+b^2)^(1/2)-b*ln(x)/a^2+1/8*b*ln(c*x^8+b*x^4+a)/a^2
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{4 a x^4}-\frac {b \log (x)}{a^2}+\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b^2 \log (x-\text {$\#$1})-a c \log (x-\text {$\#$1})+b c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b+2 c \text {$\#$1}^4}\&\right ]}{4 a^2} \] Input:
Integrate[1/(x^5*(a + b*x^4 + c*x^8)),x]
Output:
-1/4*1/(a*x^4) - (b*Log[x])/a^2 + RootSum[a + b*#1^4 + c*#1^8 & , (b^2*Log [x - #1] - a*c*Log[x - #1] + b*c*Log[x - #1]*#1^4)/(b + 2*c*#1^4) & ]/(4*a ^2)
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1693, 1145, 25, 1200, 2009}
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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (c x^8+b x^4+a\right )}dx^4\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {c x^4+b}{x^4 \left (c x^8+b x^4+a\right )}dx^4}{a}-\frac {1}{a x^4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {c x^4+b}{x^4 \left (c x^8+b x^4+a\right )}dx^4}{a}-\frac {1}{a x^4}\right )\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int \left (\frac {b}{a x^4}+\frac {-b c x^4-b^2+a c}{a \left (c x^8+b x^4+a\right )}\right )dx^4}{a}-\frac {1}{a x^4}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^4+c x^8\right )}{2 a}+\frac {b \log \left (x^4\right )}{a}}{a}-\frac {1}{a x^4}\right )\) |
Input:
Int[1/(x^5*(a + b*x^4 + c*x^8)),x]
Output:
(-(1/(a*x^4)) - (((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/ (a*Sqrt[b^2 - 4*a*c]) + (b*Log[x^4])/a - (b*Log[a + b*x^4 + c*x^8])/(2*a)) /a)/4
Int[((d_.) + (e_.)*(x_))^(m_)/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp [1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x)^(m + 1)*(Simp[c*d - b*e - c*e*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
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 = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94
method | result | size |
default | \(-\frac {-\frac {b \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4}+\frac {\left (a c -\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{2}}-\frac {1}{4 a \,x^{4}}-\frac {b \ln \left (x \right )}{a^{2}}\) | \(84\) |
risch | \(-\frac {1}{4 a \,x^{4}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c -a^{2} b^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c +b^{3}\right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (18 a^{3} c -5 a^{2} b^{2}\right ) \textit {\_R}^{2}-8 \textit {\_R} a b c +4 c^{2}\right ) x^{4}-a^{3} b \,\textit {\_R}^{2}+\left (a^{2} c -4 b^{2} a \right ) \textit {\_R} +4 b c \right )\right )}{4}\) | \(124\) |
Input:
int(1/x^5/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/2/a^2*(-1/4*b*ln(c*x^8+b*x^4+a)+(a*c-1/2*b^2)/(4*a*c-b^2)^(1/2)*arctan( (2*c*x^4+b)/(4*a*c-b^2)^(1/2)))-1/4/a/x^4-b*ln(x)/a^2
Time = 0.18 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.29 \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x^{4} \log \left (\frac {2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c + {\left (2 \, c x^{4} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \, {\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{8 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{4}}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x^{4} \arctan \left (-\frac {{\left (2 \, c x^{4} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \, {\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{8 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{4}}\right ] \] Input:
integrate(1/x^5/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
[-1/8*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*x^4*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^3 - 4*a*b*c)*x^4*log(c*x^8 + b*x^4 + a) + 8*(b^3 - 4*a*b*c)*x^4*log(x) + 2*a* b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^4), -1/8*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*x^4*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - (b^ 3 - 4*a*b*c)*x^4*log(c*x^8 + b*x^4 + a) + 8*(b^3 - 4*a*b*c)*x^4*log(x) + 2 *a*b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^4)]
Timed out. \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**5/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x^5/(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 = 94, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\frac {b \log \left (c x^{8} + b x^{4} + a\right )}{8 \, a^{2}} - \frac {b \log \left (x^{4}\right )}{4 \, a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} + \frac {b x^{4} - a}{4 \, a^{2} x^{4}} \] Input:
integrate(1/x^5/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
1/8*b*log(c*x^8 + b*x^4 + a)/a^2 - 1/4*b*log(x^4)/a^2 + 1/4*(b^2 - 2*a*c)* arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^2) + 1/4*(b *x^4 - a)/(a^2*x^4)
Time = 20.55 (sec) , antiderivative size = 8817, normalized size of antiderivative = 99.07 \[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:
int(1/(x^5*(a + b*x^4 + c*x^8)),x)
Output:
(atan((4*a^5*(4*a*c - b^2)^2*(5*b^7 - 23*a^3*b*c^3 + 66*a^2*b^3*c^2 - 35*a *b^5*c)*(((4*b^3 - 16*a*b*c)*((((((((256*a^4*b^5*c^4 - 256*a^5*b^3*c^5)/a^ 5 - (128*a*b^4*c^4*(4*b^3 - 16*a*b*c))/(64*a^3*c - 16*a^2*b^2))*(2*a*c - b ^2))/(8*a^2*(4*a*c - b^2)^(1/2)) - (16*b^4*c^4*(4*b^3 - 16*a*b*c)*(2*a*c - b^2))/(a*(4*a*c - b^2)^(1/2)*(64*a^3*c - 16*a^2*b^2)))*(2*a*c - b^2))/(8* a^2*(4*a*c - b^2)^(1/2)) - (2*b^4*c^4*(4*b^3 - 16*a*b*c)*(2*a*c - b^2)^2)/ (a^3*(4*a*c - b^2)*(64*a^3*c - 16*a^2*b^2)))*(2*a*c - b^2))/(8*a^2*(4*a*c - b^2)^(1/2)) - (b^4*c^4*(4*b^3 - 16*a*b*c)*(2*a*c - b^2)^3)/(4*a^5*(4*a*c - b^2)^(3/2)*(64*a^3*c - 16*a^2*b^2))))/(2*(64*a^3*c - 16*a^2*b^2)) - ((4 *b^3 - 16*a*b*c)*(((4*b^3 - 16*a*b*c)*(((4*b^3 - 16*a*b*c)*((((256*a^4*b^5 *c^4 - 256*a^5*b^3*c^5)/a^5 - (128*a*b^4*c^4*(4*b^3 - 16*a*b*c))/(64*a^3*c - 16*a^2*b^2))*(2*a*c - b^2))/(8*a^2*(4*a*c - b^2)^(1/2)) - (16*b^4*c^4*( 4*b^3 - 16*a*b*c)*(2*a*c - b^2))/(a*(4*a*c - b^2)^(1/2)*(64*a^3*c - 16*a^2 *b^2))))/(2*(64*a^3*c - 16*a^2*b^2)) + (((256*a^3*b^4*c^5 - 96*a^4*b^2*c^6 )/a^5 + ((4*b^3 - 16*a*b*c)*((256*a^4*b^5*c^4 - 256*a^5*b^3*c^5)/a^5 - (12 8*a*b^4*c^4*(4*b^3 - 16*a*b*c))/(64*a^3*c - 16*a^2*b^2)))/(2*(64*a^3*c - 1 6*a^2*b^2)))*(2*a*c - b^2))/(8*a^2*(4*a*c - b^2)^(1/2))))/(2*(64*a^3*c - 1 6*a^2*b^2)) - (((16*a^3*b*c^7 - 96*a^2*b^3*c^6)/a^5 - ((4*b^3 - 16*a*b*c)* ((256*a^3*b^4*c^5 - 96*a^4*b^2*c^6)/a^5 + ((4*b^3 - 16*a*b*c)*((256*a^4*b^ 5*c^4 - 256*a^5*b^3*c^5)/a^5 - (128*a*b^4*c^4*(4*b^3 - 16*a*b*c))/(64*a...
\[ \int \frac {1}{x^5 \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{5} \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(1/x^5/(c*x^8+b*x^4+a),x)
Output:
int(1/x^5/(c*x^8+b*x^4+a),x)