\(\int \frac {1}{a x+b x^3+c x^5} \, dx\) [32]

Optimal result

Integrand size = 16, antiderivative size = 69 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a \sqrt {b^2-4 a c}}+\frac {\log (x)}{a}-\frac {\log \left (a+b x^2+c x^4\right )}{4 a} \] Output:

1/2*b*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(1/2)+ln(x)/a 
-1/4*ln(c*x^4+b*x^2+a)/a
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.64 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\frac {4 \sqrt {b^2-4 a c} \log (x)-\left (b+\sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )+\left (b-\sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{4 a \sqrt {b^2-4 a c}} \] Input:

Integrate[(a*x + b*x^3 + c*x^5)^(-1),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1949, 1434, 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.

\(\displaystyle \int \frac {1}{a x+b x^3+c x^5} \, dx\)

\(\Big \downarrow \) 1949

\(\displaystyle \int \frac {1}{x \left (a+b x^2+c x^4\right )}dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (c x^4+b x^2+a\right )}dx^2\)

\(\Big \downarrow \) 1144

\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {c x^2+b}{c x^4+b x^2+a}dx^2}{a}+\frac {\log \left (x^2\right )}{a}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right )}{a}-\frac {\int \frac {c x^2+b}{c x^4+b x^2+a}dx^2}{a}\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right )}{a}-\frac {\frac {1}{2} b \int \frac {1}{c x^4+b x^2+a}dx^2+\frac {1}{2} \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{a}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2-b \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{a}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2-\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right )}{a}-\frac {\frac {1}{2} \log \left (a+b x^2+c x^4\right )-\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}\right )\)

Input:

Int[(a*x + b*x^3 + c*x^5)^(-1),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

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])
 

rule 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

rule 1144

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]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

rule 1949

Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol 
] :> Int[x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p, x] /; FreeQ[{a, b, 
c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && IntegerQ[p]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\frac {\ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2}+\frac {b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a}+\frac {\ln \left (x \right )}{a}\)	\(65\)
risch	\(\frac {\ln \left (x \right )}{a}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c -b^{2} a \right ) \textit {\_Z}^{2}+\left (4 a c -b^{2}\right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 a c -3 b^{2}\right ) \textit {\_R} +5 c \right ) x^{2}-\textit {\_R} a b +2 b \right )\right )}{2}\)	\(77\)

Input:

int(1/(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.23 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{4 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{4 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \] Input:

integrate(1/(c*x^5+b*x^3+a*x),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (60) = 120\).

Time = 3.18 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.67 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) \log {\left (x^{2} + \frac {- 8 a^{2} c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) + 2 a b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) \log {\left (x^{2} + \frac {- 8 a^{2} c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) + 2 a b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac {1}{4 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \frac {\log {\left (x \right )}}{a} \] Input:

integrate(1/(c*x**5+b*x**3+a*x),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\int { \frac {1}{c x^{5} + b x^{3} + a x} \,d x } \] Input:

integrate(1/(c*x^5+b*x^3+a*x),x, algorithm="maxima")
 

Output:

-integrate((c*x^3 + b*x)/(c*x^4 + b*x^2 + a), x)/a + log(x)/a
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=-\frac {b \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a} - \frac {\log \left (c x^{4} + b x^{2} + a\right )}{4 \, a} + \frac {\log \left (x^{2}\right )}{2 \, a} \] Input:

integrate(1/(c*x^5+b*x^3+a*x),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 1014, normalized size of antiderivative = 14.70 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx =\text {Too large to display} \] Input:

int(1/(a*x + b*x^3 + c*x^5),x)
 

Output:

log(x)/a + (log(a + b*x^2 + c*x^4)*(8*a*c - 2*b^2))/(2*(4*a*b^2 - 16*a^2*c 
)) + (b*atan((16*a^3*x^2*(((3*b^3 - 8*a*b*c)*(((8*a*c - 2*b^2)^2*(10*b*c^3 
 - ((12*b^3*c^2 - 40*a*b*c^3)*(8*a*c - 2*b^2))/(2*(4*a*b^2 - 16*a^2*c))))/ 
(4*(4*a*b^2 - 16*a^2*c)^2) - (b^2*(10*b*c^3 - ((12*b^3*c^2 - 40*a*b*c^3)*( 
8*a*c - 2*b^2))/(2*(4*a*b^2 - 16*a^2*c))))/(16*a^2*(4*a*c - b^2)) + (b^2*( 
12*b^3*c^2 - 40*a*b*c^3)*(8*a*c - 2*b^2))/(16*a^2*(4*a*b^2 - 16*a^2*c)*(4* 
a*c - b^2))))/(8*a^3*c^2*(25*a*c - 6*b^2)) - ((3*b^4 + 10*a^2*c^2 - 14*a*b 
^2*c)*((b^3*(12*b^3*c^2 - 40*a*b*c^3))/(64*a^3*(4*a*c - b^2)^(3/2)) - (b*( 
12*b^3*c^2 - 40*a*b*c^3)*(8*a*c - 2*b^2)^2)/(16*a*(4*a*b^2 - 16*a^2*c)^2*( 
4*a*c - b^2)^(1/2)) + (b*(8*a*c - 2*b^2)*(10*b*c^3 - ((12*b^3*c^2 - 40*a*b 
*c^3)*(8*a*c - 2*b^2))/(2*(4*a*b^2 - 16*a^2*c))))/(4*a*(4*a*b^2 - 16*a^2*c 
)*(4*a*c - b^2)^(1/2))))/(8*a^3*c^2*(4*a*c - b^2)^(1/2)*(25*a*c - 6*b^2))) 
*(4*a*c - b^2)^(3/2))/(b^2*c^2) + (2*(3*b^3 - 8*a*b*c)*(4*a*c - b^2)^(3/2) 
*(((8*a*c - 2*b^2)^2*(4*b^2*c^2 - (2*a*b^2*c^2*(8*a*c - 2*b^2))/(4*a*b^2 - 
 16*a^2*c)))/(4*(4*a*b^2 - 16*a^2*c)^2) - (b^2*(4*b^2*c^2 - (2*a*b^2*c^2*( 
8*a*c - 2*b^2))/(4*a*b^2 - 16*a^2*c)))/(16*a^2*(4*a*c - b^2)) + (b^4*c^2*( 
8*a*c - 2*b^2))/(4*a*(4*a*b^2 - 16*a^2*c)*(4*a*c - b^2))))/(b^2*c^4*(25*a* 
c - 6*b^2)) - (2*(4*a*c - b^2)*(3*b^4 + 10*a^2*c^2 - 14*a*b^2*c)*((b^5*c^2 
)/(16*a^2*(4*a*c - b^2)^(3/2)) - (b^3*c^2*(8*a*c - 2*b^2)^2)/(4*(4*a*b^2 - 
 16*a^2*c)^2*(4*a*c - b^2)^(1/2)) + (b*(8*a*c - 2*b^2)*(4*b^2*c^2 - (2*...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 247, normalized size of antiderivative = 3.58 \[ \int \frac {1}{a x+b x^3+c x^5} \, dx=\frac {2 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b +2 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b -4 \,\mathrm {log}\left (-\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a c +\mathrm {log}\left (-\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) b^{2}-4 \,\mathrm {log}\left (\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a c +\mathrm {log}\left (\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) b^{2}+16 \,\mathrm {log}\left (x \right ) a c -4 \,\mathrm {log}\left (x \right ) b^{2}}{4 a \left (4 a c -b^{2}\right )} \] Input:

int(1/(c*x^5+b*x^3+a*x),x)
 

Output:

(2*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sq 
rt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b + 2*sqrt( 
2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sq 
rt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b - 4*log( - sqrt(2 
*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*a*c + log( - sqrt(2*sqrt 
(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*b**2 - 4*log(sqrt(2*sqrt(c)*s 
qrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*a*c + log(sqrt(2*sqrt(c)*sqrt(a) - 
 b)*x + sqrt(a) + sqrt(c)*x**2)*b**2 + 16*log(x)*a*c - 4*log(x)*b**2)/(4*a 
*(4*a*c - b**2))