Integrand size = 28, antiderivative size = 103 \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\frac {f x^2}{2 c}-\frac {\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \sqrt {b^2-4 a c}}+\frac {(c e-b f) \log \left (a+b x^2+c x^4\right )}{4 c^2} \]
1/2*f*x^2/c+1/4*(-b*f+c*e)*ln(c*x^4+b*x^2+a)/c^2-1/2*(-2*a*c*f+b^2*f-b*c*e +2*c^2*d)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\frac {2 c f x^2+\frac {2 \left (2 c^2 d+b^2 f-c (b e+2 a f)\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+(c e-b f) \log \left (a+b x^2+c x^4\right )}{4 c^2} \]
(2*c*f*x^2 + (2*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*ArcTan[(b + 2*c*x^2)/S qrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (c*e - b*f)*Log[a + b*x^2 + c*x^4 ])/(4*c^2)
Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2194, 2188, 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.
\(\displaystyle \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int \frac {f x^4+e x^2+d}{c x^4+b x^2+a}dx^2\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle \frac {1}{2} \int \left (\frac {f}{c}+\frac {(c e-b f) x^2+c d-a f}{c \left (c x^4+b x^2+a\right )}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {(c e-b f) \log \left (a+b x^2+c x^4\right )}{2 c^2}+\frac {f x^2}{c}\right )\) |
((f*x^2)/c - ((2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*ArcTanh[(b + 2*c*x^2)/Sq rt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]) + ((c*e - b*f)*Log[a + b*x^2 + c *x^4])/(2*c^2))/2
3.1.50.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] : > Simp[1/2 Subst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2) ^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && IntegerQ [(m - 1)/2]
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {f \,x^{2}}{2 c}+\frac {\frac {\left (-b f +e c \right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (-a f +c d -\frac {\left (-b f +e c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 c}\) | \(101\) |
risch | \(\text {Expression too large to display}\) | \(1690\) |
1/2*f*x^2/c+1/2/c*(1/2*(-b*f+c*e)/c*ln(c*x^4+b*x^2+a)+2*(-a*f+c*d-1/2*(-b* f+c*e)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))
Time = 0.34 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.09 \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\left [\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} f x^{2} - {\left (2 \, c^{2} d - b c e + {\left (b^{2} - 2 \, a c\right )} f\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e - {\left (b^{3} - 4 \, a b c\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} f x^{2} - 2 \, {\left (2 \, c^{2} d - b c e + {\left (b^{2} - 2 \, a c\right )} f\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e - {\left (b^{3} - 4 \, a b c\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \]
[1/4*(2*(b^2*c - 4*a*c^2)*f*x^2 - (2*c^2*d - b*c*e + (b^2 - 2*a*c)*f)*sqrt (b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c + (2*c*x^2 + b)*sqr t(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + ((b^2*c - 4*a*c^2)*e - (b^3 - 4*a*b *c)*f)*log(c*x^4 + b*x^2 + a))/(b^2*c^2 - 4*a*c^3), 1/4*(2*(b^2*c - 4*a*c^ 2)*f*x^2 - 2*(2*c^2*d - b*c*e + (b^2 - 2*a*c)*f)*sqrt(-b^2 + 4*a*c)*arctan (-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + ((b^2*c - 4*a*c^2)*e - (b^3 - 4*a*b*c)*f)*log(c*x^4 + b*x^2 + a))/(b^2*c^2 - 4*a*c^3)]
Timed out. \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\text {Exception raised: ValueError} \]
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 = 0.64 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\frac {f x^{2}}{2 \, c} + \frac {{\left (c e - b f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} + \frac {{\left (2 \, c^{2} d - b c e + b^{2} f - 2 \, a c f\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{2}} \]
1/2*f*x^2/c + 1/4*(c*e - b*f)*log(c*x^4 + b*x^2 + a)/c^2 + 1/2*(2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b ^2 + 4*a*c)*c^2)
Time = 8.83 (sec) , antiderivative size = 1081, normalized size of antiderivative = 10.50 \[ \int \frac {x \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx=\frac {f\,x^2}{2\,c}+\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}+\frac {\mathrm {atan}\left (\frac {2\,c^2\,\left (4\,a\,c-b^2\right )\,\left (x^2\,\left (\frac {\frac {\left (\frac {6\,f\,b^2\,c^2-6\,e\,b\,c^3+4\,d\,c^4-4\,a\,f\,c^3}{c^2}+\frac {4\,b\,c^2\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )\,\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )}{8\,c^2\,\sqrt {4\,a\,c-b^2}}+\frac {b\,\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{2\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}}{a}-\frac {b\,\left (\frac {b^3\,f^2-2\,b^2\,c\,e\,f+b\,c^2\,e^2+d\,b\,c^2\,f-a\,b\,c\,f^2-d\,c^3\,e+a\,c^2\,e\,f}{c^2}+\frac {\left (\frac {6\,f\,b^2\,c^2-6\,e\,b\,c^3+4\,d\,c^4-4\,a\,f\,c^3}{c^2}+\frac {4\,b\,c^2\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}-\frac {b\,{\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )}^2}{2\,c^2\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )-\frac {\frac {\left (\frac {8\,a\,c^3\,e-8\,a\,b\,c^2\,f}{c^2}-\frac {8\,a\,c^2\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )\,\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )}{8\,c^2\,\sqrt {4\,a\,c-b^2}}-\frac {a\,\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}}{a}+\frac {b\,\left (\frac {\left (\frac {8\,a\,c^3\,e-8\,a\,b\,c^2\,f}{c^2}-\frac {8\,a\,c^2\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )\,\left (2\,f\,b^3-2\,e\,b^2\,c-8\,a\,f\,b\,c+8\,a\,e\,c^2\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}-\frac {a\,b^2\,f^2-2\,a\,b\,c\,e\,f+a\,c^2\,e^2}{c^2}+\frac {a\,{\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )}^2}{c^2\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{4\,a^2\,c^2\,f^2-4\,a\,b^2\,c\,f^2+4\,a\,b\,c^2\,e\,f-8\,a\,c^3\,d\,f+b^4\,f^2-2\,b^3\,c\,e\,f+4\,b^2\,c^2\,d\,f+b^2\,c^2\,e^2-4\,b\,c^3\,d\,e+4\,c^4\,d^2}\right )\,\left (f\,b^2-e\,b\,c+2\,d\,c^2-2\,a\,f\,c\right )}{2\,c^2\,\sqrt {4\,a\,c-b^2}} \]
(f*x^2)/(2*c) + (log(a + b*x^2 + c*x^4)*(2*b^3*f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(2*(16*a*c^3 - 4*b^2*c^2)) + (atan((2*c^2*(4*a*c - b^2)*(x^2* (((((4*c^4*d + 6*b^2*c^2*f - 4*a*c^3*f - 6*b*c^3*e)/c^2 + (4*b*c^2*(2*b^3* f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(16*a*c^3 - 4*b^2*c^2))*(2*c^2*d + b^2*f - 2*a*c*f - b*c*e))/(8*c^2*(4*a*c - b^2)^(1/2)) + (b*(2*c^2*d + b^2 *f - 2*a*c*f - b*c*e)*(2*b^3*f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(2*(4 *a*c - b^2)^(1/2)*(16*a*c^3 - 4*b^2*c^2)))/a - (b*((b^3*f^2 + b*c^2*e^2 - c^3*d*e - a*b*c*f^2 + a*c^2*e*f + b*c^2*d*f - 2*b^2*c*e*f)/c^2 + (((4*c^4* d + 6*b^2*c^2*f - 4*a*c^3*f - 6*b*c^3*e)/c^2 + (4*b*c^2*(2*b^3*f + 8*a*c^2 *e - 2*b^2*c*e - 8*a*b*c*f))/(16*a*c^3 - 4*b^2*c^2))*(2*b^3*f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(2*(16*a*c^3 - 4*b^2*c^2)) - (b*(2*c^2*d + b^2*f - 2*a*c*f - b*c*e)^2)/(2*c^2*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))) - ((((8*a*c^3*e - 8*a*b*c^2*f)/c^2 - (8*a*c^2*(2*b^3*f + 8*a*c^2*e - 2*b^2 *c*e - 8*a*b*c*f))/(16*a*c^3 - 4*b^2*c^2))*(2*c^2*d + b^2*f - 2*a*c*f - b* c*e))/(8*c^2*(4*a*c - b^2)^(1/2)) - (a*(2*c^2*d + b^2*f - 2*a*c*f - b*c*e) *(2*b^3*f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/((4*a*c - b^2)^(1/2)*(16*a *c^3 - 4*b^2*c^2)))/a + (b*((((8*a*c^3*e - 8*a*b*c^2*f)/c^2 - (8*a*c^2*(2* b^3*f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(16*a*c^3 - 4*b^2*c^2))*(2*b^3 *f + 8*a*c^2*e - 2*b^2*c*e - 8*a*b*c*f))/(2*(16*a*c^3 - 4*b^2*c^2)) - (a*b ^2*f^2 + a*c^2*e^2 - 2*a*b*c*e*f)/c^2 + (a*(2*c^2*d + b^2*f - 2*a*c*f -...