Integrand size = 22, antiderivative size = 407 \[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=-\frac {\left (b-\sqrt {8 a^2+b^2-4 a c}\right ) \arctan \left (\frac {b-\sqrt {8 a^2+b^2-4 a c}+4 a x}{\sqrt {2} \sqrt {4 a^2+2 a c-b \left (b-\sqrt {8 a^2+b^2-4 a c}\right )}}\right )}{\sqrt {2} \sqrt {8 a^2+b^2-4 a c} \sqrt {4 a^2+2 a c-b \left (b-\sqrt {8 a^2+b^2-4 a c}\right )}}+\frac {\left (b+\sqrt {8 a^2+b^2-4 a c}\right ) \arctan \left (\frac {b+\sqrt {8 a^2+b^2-4 a c}+4 a x}{\sqrt {2} \sqrt {4 a^2+2 a c-b \left (b+\sqrt {8 a^2+b^2-4 a c}\right )}}\right )}{\sqrt {2} \sqrt {8 a^2+b^2-4 a c} \sqrt {4 a^2+2 a c-b \left (b+\sqrt {8 a^2+b^2-4 a c}\right )}}-\frac {\log \left (2 a+b x-\sqrt {8 a^2+b^2-4 a c} x+2 a x^2\right )}{2 \sqrt {8 a^2+b^2-4 a c}}+\frac {\log \left (2 a+b x+\sqrt {8 a^2+b^2-4 a c} x+2 a x^2\right )}{2 \sqrt {8 a^2+b^2-4 a c}} \] Output:
-1/2*(b-(8*a^2-4*a*c+b^2)^(1/2))*arctan(1/2*(b-(8*a^2-4*a*c+b^2)^(1/2)+4*a *x)*2^(1/2)/(4*a^2+2*a*c-b*(b-(8*a^2-4*a*c+b^2)^(1/2)))^(1/2))*2^(1/2)/(8* a^2-4*a*c+b^2)^(1/2)/(4*a^2+2*a*c-b*(b-(8*a^2-4*a*c+b^2)^(1/2)))^(1/2)+1/2 *(b+(8*a^2-4*a*c+b^2)^(1/2))*arctan(1/2*(b+(8*a^2-4*a*c+b^2)^(1/2)+4*a*x)* 2^(1/2)/(4*a^2+2*a*c-b*(b+(8*a^2-4*a*c+b^2)^(1/2)))^(1/2))*2^(1/2)/(8*a^2- 4*a*c+b^2)^(1/2)/(4*a^2+2*a*c-b*(b+(8*a^2-4*a*c+b^2)^(1/2)))^(1/2)-1/2*ln( 2*a+b*x-(8*a^2-4*a*c+b^2)^(1/2)*x+2*a*x^2)/(8*a^2-4*a*c+b^2)^(1/2)+1/2*ln( 2*a+b*x+(8*a^2-4*a*c+b^2)^(1/2)*x+2*a*x^2)/(8*a^2-4*a*c+b^2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.14 \[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\text {RootSum}\left [a+b \text {$\#$1}+c \text {$\#$1}^2+b \text {$\#$1}^3+a \text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{b+2 c \text {$\#$1}+3 b \text {$\#$1}^2+4 a \text {$\#$1}^3}\&\right ] \] Input:
Integrate[(a + b*x + c*x^2 + b*x^3 + a*x^4)^(-1),x]
Output:
RootSum[a + b*#1 + c*#1^2 + b*#1^3 + a*#1^4 & , Log[x - #1]/(b + 2*c*#1 + 3*b*#1^2 + 4*a*#1^3) & ]
Time = 1.23 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2492, 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.
\(\displaystyle \int \frac {1}{a x^4+a+b x^3+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 2492 |
\(\displaystyle \frac {\int \left (\frac {a \left (b+2 a x+\sqrt {8 a^2-4 c a+b^2}\right )}{\sqrt {8 a^2-4 c a+b^2} \left (2 a x^2+\left (b+\sqrt {8 a^2-4 c a+b^2}\right ) x+2 a\right )}-\frac {a \left (b+2 a x-\sqrt {8 a^2-4 c a+b^2}\right )}{\sqrt {8 a^2-4 c a+b^2} \left (2 a x^2+\left (b-\sqrt {8 a^2-4 c a+b^2}\right ) x+2 a\right )}\right )dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a \left (b-\sqrt {8 a^2-4 a c+b^2}\right ) \arctan \left (\frac {-\sqrt {8 a^2-4 a c+b^2}+4 a x+b}{\sqrt {2} \sqrt {-b \left (b-\sqrt {8 a^2-4 a c+b^2}\right )+4 a^2+2 a c}}\right )}{\sqrt {2} \sqrt {8 a^2-4 a c+b^2} \sqrt {-b \left (b-\sqrt {8 a^2-4 a c+b^2}\right )+4 a^2+2 a c}}+\frac {a \left (\sqrt {8 a^2-4 a c+b^2}+b\right ) \arctan \left (\frac {\sqrt {8 a^2-4 a c+b^2}+4 a x+b}{\sqrt {2} \sqrt {-b \left (\sqrt {8 a^2-4 a c+b^2}+b\right )+4 a^2+2 a c}}\right )}{\sqrt {2} \sqrt {8 a^2-4 a c+b^2} \sqrt {-b \left (\sqrt {8 a^2-4 a c+b^2}+b\right )+4 a^2+2 a c}}-\frac {a \log \left (x \left (b-\sqrt {8 a^2-4 a c+b^2}\right )+2 a x^2+2 a\right )}{2 \sqrt {8 a^2-4 a c+b^2}}+\frac {a \log \left (x \left (\sqrt {8 a^2-4 a c+b^2}+b\right )+2 a x^2+2 a\right )}{2 \sqrt {8 a^2-4 a c+b^2}}}{a}\) |
Input:
Int[(a + b*x + c*x^2 + b*x^3 + a*x^4)^(-1),x]
Output:
(-((a*(b - Sqrt[8*a^2 + b^2 - 4*a*c])*ArcTan[(b - Sqrt[8*a^2 + b^2 - 4*a*c ] + 4*a*x)/(Sqrt[2]*Sqrt[4*a^2 + 2*a*c - b*(b - Sqrt[8*a^2 + b^2 - 4*a*c]) ])])/(Sqrt[2]*Sqrt[8*a^2 + b^2 - 4*a*c]*Sqrt[4*a^2 + 2*a*c - b*(b - Sqrt[8 *a^2 + b^2 - 4*a*c])])) + (a*(b + Sqrt[8*a^2 + b^2 - 4*a*c])*ArcTan[(b + S qrt[8*a^2 + b^2 - 4*a*c] + 4*a*x)/(Sqrt[2]*Sqrt[4*a^2 + 2*a*c - b*(b + Sqr t[8*a^2 + b^2 - 4*a*c])])])/(Sqrt[2]*Sqrt[8*a^2 + b^2 - 4*a*c]*Sqrt[4*a^2 + 2*a*c - b*(b + Sqrt[8*a^2 + b^2 - 4*a*c])]) - (a*Log[2*a + (b - Sqrt[8*a ^2 + b^2 - 4*a*c])*x + 2*a*x^2])/(2*Sqrt[8*a^2 + b^2 - 4*a*c]) + (a*Log[2* a + (b + Sqrt[8*a^2 + b^2 - 4*a*c])*x + 2*a*x^2])/(2*Sqrt[8*a^2 + b^2 - 4* a*c]))/a
Int[(Px_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2 + (d_.)*(x_)^3 + (e_.)*(x_)^4) ^(p_), x_Symbol] :> Simp[e^p Int[ExpandIntegrand[Px*(b/d + ((d + Sqrt[e*( (b^2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p*(b/d + ((d - Sqrt[e*((b^ 2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x] && ILtQ[p, 0] && EqQ[a*d^2 - b^2*e, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.13
method | result | size |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{4}+b \,\textit {\_Z}^{3}+\textit {\_Z}^{2} c +b \textit {\_Z} +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3} a +3 \textit {\_R}^{2} b +2 \textit {\_R} c +b}\) | \(52\) |
default | \(4 a \left (\frac {\frac {\left (16 a^{3}-8 a^{2} c +2 b^{2} a \right ) \ln \left (2 a +b x +\sqrt {8 a^{2}-4 a c +b^{2}}\, x +2 a \,x^{2}\right )}{4 a}+\frac {2 \left (8 \sqrt {8 a^{2}-4 a c +b^{2}}\, a^{2}-4 \sqrt {8 a^{2}-4 a c +b^{2}}\, a c +b^{2} \sqrt {8 a^{2}-4 a c +b^{2}}+8 b \,a^{2}-4 a b c +b^{3}-\frac {\left (16 a^{3}-8 a^{2} c +2 b^{2} a \right ) \left (b +\sqrt {8 a^{2}-4 a c +b^{2}}\right )}{4 a}\right ) \arctan \left (\frac {b +\sqrt {8 a^{2}-4 a c +b^{2}}+4 x a}{\sqrt {8 a^{2}+4 a c -2 b^{2}-2 b \sqrt {8 a^{2}-4 a c +b^{2}}}}\right )}{\sqrt {8 a^{2}+4 a c -2 b^{2}-2 b \sqrt {8 a^{2}-4 a c +b^{2}}}}}{4 \left (8 a^{2}-4 a c +b^{2}\right )^{\frac {3}{2}} a}+\frac {-\frac {\left (16 a^{3}-8 a^{2} c +2 b^{2} a \right ) \ln \left (-2 a \,x^{2}+\sqrt {8 a^{2}-4 a c +b^{2}}\, x -b x -2 a \right )}{4 a}+\frac {2 \left (-8 \sqrt {8 a^{2}-4 a c +b^{2}}\, a^{2}+4 \sqrt {8 a^{2}-4 a c +b^{2}}\, a c -b^{2} \sqrt {8 a^{2}-4 a c +b^{2}}+8 b \,a^{2}-4 a b c +b^{3}+\frac {\left (16 a^{3}-8 a^{2} c +2 b^{2} a \right ) \left (\sqrt {8 a^{2}-4 a c +b^{2}}-b \right )}{4 a}\right ) \arctan \left (\frac {-4 x a +\sqrt {8 a^{2}-4 a c +b^{2}}-b}{\sqrt {8 a^{2}+4 a c -2 b^{2}+2 b \sqrt {8 a^{2}-4 a c +b^{2}}}}\right )}{\sqrt {8 a^{2}+4 a c -2 b^{2}+2 b \sqrt {8 a^{2}-4 a c +b^{2}}}}}{4 \left (8 a^{2}-4 a c +b^{2}\right )^{\frac {3}{2}} a}\right )\) | \(577\) |
Input:
int(1/(a*x^4+b*x^3+c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
sum(1/(4*_R^3*a+3*_R^2*b+2*_R*c+b)*ln(x-_R),_R=RootOf(_Z^4*a+_Z^3*b+_Z^2*c +_Z*b+a))
Result contains complex when optimal does not.
Time = 19.09 (sec) , antiderivative size = 3804183, normalized size of antiderivative = 9346.89 \[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*x^4+b*x^3+c*x^2+b*x+a),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\text {Timed out} \] Input:
integrate(1/(a*x**4+b*x**3+c*x**2+b*x+a),x)
Output:
Timed out
\[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\int { \frac {1}{a x^{4} + b x^{3} + c x^{2} + b x + a} \,d x } \] Input:
integrate(1/(a*x^4+b*x^3+c*x^2+b*x+a),x, algorithm="maxima")
Output:
integrate(1/(a*x^4 + b*x^3 + c*x^2 + b*x + a), x)
\[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\int { \frac {1}{a x^{4} + b x^{3} + c x^{2} + b x + a} \,d x } \] Input:
integrate(1/(a*x^4+b*x^3+c*x^2+b*x+a),x, algorithm="giac")
Output:
integrate(1/(a*x^4 + b*x^3 + c*x^2 + b*x + a), x)
Time = 23.27 (sec) , antiderivative size = 4180, normalized size of antiderivative = 10.27 \[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\text {Too large to display} \] Input:
int(1/(a + b*x + a*x^4 + b*x^3 + c*x^2),x)
Output:
symsum(log(-root(36*a*b^4*c*z^4 - 80*a^2*b^2*c^2*z^4 + 288*a^3*b^2*c*z^4 - 8*a*b^2*c^3*z^4 - 128*a^4*c^2*z^4 + 16*a^2*c^4*z^4 - 192*a^4*b^2*z^4 - 60 *a^2*b^4*z^4 - 4*b^6*z^4 + 256*a^6*z^4 + b^4*c^2*z^4 - 14*a*b^2*c*z^2 - b^ 2*c^2*z^2 + 24*a^2*b^2*z^2 - 16*a^3*c*z^2 + 4*a*c^3*z^2 + 3*b^4*z^2 - 4*a* b*c*z + 8*a^2*b*z + b^3*z + a^2, z, k)*a*(a*b + 16*root(36*a*b^4*c*z^4 - 8 0*a^2*b^2*c^2*z^4 + 288*a^3*b^2*c*z^4 - 8*a*b^2*c^3*z^4 - 128*a^4*c^2*z^4 + 16*a^2*c^4*z^4 - 192*a^4*b^2*z^4 - 60*a^2*b^4*z^4 - 4*b^6*z^4 + 256*a^6* z^4 + b^4*c^2*z^4 - 14*a*b^2*c*z^2 - b^2*c^2*z^2 + 24*a^2*b^2*z^2 - 16*a^3 *c*z^2 + 4*a*c^3*z^2 + 3*b^4*z^2 - 4*a*b*c*z + 8*a^2*b*z + b^3*z + a^2, z, k)*a^3 + 4*a^2*x + 6*root(36*a*b^4*c*z^4 - 80*a^2*b^2*c^2*z^4 + 288*a^3*b ^2*c*z^4 - 8*a*b^2*c^3*z^4 - 128*a^4*c^2*z^4 + 16*a^2*c^4*z^4 - 192*a^4*b^ 2*z^4 - 60*a^2*b^4*z^4 - 4*b^6*z^4 + 256*a^6*z^4 + b^4*c^2*z^4 - 14*a*b^2* c*z^2 - b^2*c^2*z^2 + 24*a^2*b^2*z^2 - 16*a^3*c*z^2 + 4*a*c^3*z^2 + 3*b^4* z^2 - 4*a*b*c*z + 8*a^2*b*z + b^3*z + a^2, z, k)^2*a*b^3 + 48*root(36*a*b^ 4*c*z^4 - 80*a^2*b^2*c^2*z^4 + 288*a^3*b^2*c*z^4 - 8*a*b^2*c^3*z^4 - 128*a ^4*c^2*z^4 + 16*a^2*c^4*z^4 - 192*a^4*b^2*z^4 - 60*a^2*b^4*z^4 - 4*b^6*z^4 + 256*a^6*z^4 + b^4*c^2*z^4 - 14*a*b^2*c*z^2 - b^2*c^2*z^2 + 24*a^2*b^2*z ^2 - 16*a^3*c*z^2 + 4*a*c^3*z^2 + 3*b^4*z^2 - 4*a*b*c*z + 8*a^2*b*z + b^3* z + a^2, z, k)^2*a^3*b - root(36*a*b^4*c*z^4 - 80*a^2*b^2*c^2*z^4 + 288*a^ 3*b^2*c*z^4 - 8*a*b^2*c^3*z^4 - 128*a^4*c^2*z^4 + 16*a^2*c^4*z^4 - 192*...
\[ \int \frac {1}{a+b x+c x^2+b x^3+a x^4} \, dx=\int \frac {1}{a \,x^{4}+b \,x^{3}+c \,x^{2}+b x +a}d x \] Input:
int(1/(a*x^4+b*x^3+c*x^2+b*x+a),x)
Output:
int(1/(a*x**4 + a + b*x**3 + b*x + c*x**2),x)