Integrand size = 25, antiderivative size = 133 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {e \log \left (d+e x^2\right )}{2 \left (c d^2-b d e+a e^2\right )}-\frac {e \log \left (a+b x^2+c x^4\right )}{4 \left (c d^2-b d e+a e^2\right )} \]
1/2*e*ln(e*x^2+d)/(a*e^2-b*d*e+c*d^2)-1/4*e*ln(c*x^4+b*x^2+a)/(a*e^2-b*d*e +c*d^2)-1/2*(-b*e+2*c*d)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/(a*e^2-b* d*e+c*d^2)/(-4*a*c+b^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\frac {(-4 c d+2 b e) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} e \left (-2 \log \left (d+e x^2\right )+\log \left (a+b x^2+c x^4\right )\right )}{4 \sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )} \]
((-4*c*d + 2*b*e)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4 *a*c]*e*(-2*Log[d + e*x^2] + Log[a + b*x^2 + c*x^4]))/(4*Sqrt[-b^2 + 4*a*c ]*(-(c*d^2) + e*(b*d - a*e)))
Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1576, 1144, 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.
\(\displaystyle \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (e x^2+d\right ) \left (c x^4+b x^2+a\right )}dx^2\) |
\(\Big \downarrow \) 1144 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {-c e x^2+c d-b e}{c x^4+b x^2+a}dx^2}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} (2 c d-b e) \int \frac {1}{c x^4+b x^2+a}dx^2-\frac {1}{2} e \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {-\left ((2 c d-b e) \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )\right )-\frac {1}{2} e \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {1}{2} e \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {1}{2} e \log \left (a+b x^2+c x^4\right )}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2-b d e+c d^2}\right )\) |
((e*Log[d + e*x^2])/(c*d^2 - b*d*e + a*e^2) + (-(((2*c*d - b*e)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c]) - (e*Log[a + b*x^2 + c* x^4])/2)/(c*d^2 - b*d*e + a*e^2))/2
3.3.99.3.1 Defintions of rubi rules used
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_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {\frac {e \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2}+\frac {2 \left (\frac {b e}{2}-c d \right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {e \ln \left (e \,x^{2}+d \right )}{2 a \,e^{2}-2 b d e +2 c \,d^{2}}\) | \(113\) |
risch | \(\frac {e \ln \left (e \,x^{2}+d \right )}{2 a \,e^{2}-2 b d e +2 c \,d^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c \,e^{2}-a \,b^{2} e^{2}-4 a b c d e +4 a \,c^{2} d^{2}+b^{3} d e -b^{2} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 a c e -b^{2} e \right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (4 a^{2} c \,e^{3}-a \,b^{2} e^{3}+3 a b c d \,e^{2}-4 a \,c^{2} d^{2} e -b^{3} d \,e^{2}+2 b^{2} c \,d^{2} e -b \,c^{2} d^{3}\right ) \textit {\_R}^{2}+\left (4 e^{2} a c -b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \textit {\_R} +e c \right ) x^{2}+\left (6 a^{2} c d \,e^{2}-2 a \,b^{2} d \,e^{2}+2 a b c \,d^{2} e -2 a \,c^{2} d^{3}\right ) \textit {\_R}^{2}+3 a c d e \textit {\_R} \right )\right )}{2}\) | \(271\) |
-1/2/(a*e^2-b*d*e+c*d^2)*(1/2*e*ln(c*x^4+b*x^2+a)+2*(1/2*b*e-c*d)/(4*a*c-b ^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))+1/2*e*ln(e*x^2+d)/(a*e^2- b*d*e+c*d^2)
Time = 7.67 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.41 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{4} + b x^{2} + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x^{2} + d\right ) + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{4 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}, -\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{4} + b x^{2} + a\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} e \log \left (e x^{2} + d\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{4 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}\right ] \]
[-1/4*((b^2 - 4*a*c)*e*log(c*x^4 + b*x^2 + a) - 2*(b^2 - 4*a*c)*e*log(e*x^ 2 + d) + sqrt(b^2 - 4*a*c)*(2*c*d - b*e)*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*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)*e^2), -1/4*((b^2 - 4*a*c)*e*log(c*x^4 + b*x^2 + a) - 2*(b^2 - 4*a*c)*e*log(e*x^2 + d) + 2*sq rt(-b^2 + 4*a*c)*(2*c*d - b*e)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b ^2 - 4*a*c)))/((b^2*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^ 2*c)*e^2)]
Timed out. \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Exception raised: ValueError} \]
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 = 0.59 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\frac {e^{2} \log \left ({\left | e x^{2} + d \right |}\right )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac {e \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (c d^{2} - b d e + a e^{2}\right )}} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]
1/2*e^2*log(abs(e*x^2 + d))/(c*d^2*e - b*d*e^2 + a*e^3) - 1/4*e*log(c*x^4 + b*x^2 + a)/(c*d^2 - b*d*e + a*e^2) + 1/2*(2*c*d - b*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((c*d^2 - b*d*e + a*e^2)*sqrt(-b^2 + 4*a*c))
Time = 13.23 (sec) , antiderivative size = 2434, normalized size of antiderivative = 18.30 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
(e*log(d + e*x^2))/(2*a*e^2 + 2*c*d^2 - 2*b*d*e) - (log(36*a^4*c^3*e^5 - 4 *a*b^6*e^5 - 4*b^7*e^5*x^2 + 32*a^2*b^4*c*e^5 + 36*a^2*c^5*d^4*e - 4*a*c^6 *d^5*x^2 - 4*b^6*e^5*x^2*(b^2 - 4*a*c)^(1/2) - 73*a^3*b^2*c^2*e^5 - 184*a^ 3*c^4*d^2*e^3 + b^2*c^5*d^5*x^2 - 4*a*b^5*e^5*(b^2 - 4*a*c)^(1/2) + 2*a*c^ 5*d^5*(b^2 - 4*a*c)^(1/2) + 16*a*b^5*c*d*e^4 - 60*a^2*c^4*d^3*e^2*(b^2 - 4 *a*c)^(1/2) + 18*a^3*c^3*e^5*x^2*(b^2 - 4*a*c)^(1/2) + 146*a^2*b^2*c^3*d^2 *e^3 - 101*a^2*b^3*c^2*e^5*x^2 + 120*a^2*c^5*d^3*e^2*x^2 + 19*b^4*c^3*d^3* e^2*x^2 - 25*b^5*c^2*d^2*e^3*x^2 - 9*a*b^2*c^4*d^4*e + 184*a^3*b*c^3*d*e^4 + 36*a*b^5*c*e^5*x^2 + 16*b^6*c*d*e^4*x^2 + 24*a^2*b^3*c*e^5*(b^2 - 4*a*c )^(1/2) - 33*a^3*b*c^2*e^5*(b^2 - 4*a*c)^(1/2) + 66*a^3*c^3*d*e^4*(b^2 - 4 *a*c)^(1/2) + b*c^5*d^5*x^2*(b^2 - 4*a*c)^(1/2) + 18*a*b^3*c^3*d^3*e^2 - 2 5*a*b^4*c^2*d^2*e^3 - 72*a^2*b*c^4*d^3*e^2 - 110*a^2*b^3*c^2*d*e^4 + 84*a^ 3*b*c^3*e^5*x^2 - 132*a^3*c^4*d*e^4*x^2 - 7*b^3*c^4*d^4*e*x^2 + 28*a*b^4*c *e^5*x^2*(b^2 - 4*a*c)^(1/2) + 18*a*c^5*d^4*e*x^2*(b^2 - 4*a*c)^(1/2) + 16 *b^5*c*d*e^4*x^2*(b^2 - 4*a*c)^(1/2) - 126*a*b^4*c^2*d*e^4*x^2 + 20*a*b^2* c^3*d^3*e^2*(b^2 - 4*a*c)^(1/2) - 25*a*b^3*c^2*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 90*a^2*b*c^3*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 78*a^2*b^2*c^2*d*e^4*(b^2 - 4*a*c)^(1/2) - 7*b^2*c^4*d^4*e*x^2*(b^2 - 4*a*c)^(1/2) - 106*a*b^2*c^4*d^3 *e^2*x^2 + 168*a*b^3*c^3*d^2*e^3*x^2 - 272*a^2*b*c^4*d^2*e^3*x^2 + 281*a^2 *b^2*c^3*d*e^4*x^2 - 5*a*b*c^4*d^4*e*(b^2 - 4*a*c)^(1/2) + 16*a*b^4*c*d...