Integrand size = 30, antiderivative size = 121 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=-\frac {1}{2 a e (d+e x)^2}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c} e}-\frac {b \log (d+e x)}{a^2 e}+\frac {b \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 a^2 e} \]
-1/2/a/e/(e*x+d)^2-b*ln(e*x+d)/a^2/e+1/4*b*ln(a+b*(e*x+d)^2+c*(e*x+d)^4)/a ^2/e-1/2*(-2*a*c+b^2)*arctanh((b+2*c*(e*x+d)^2)/(-4*a*c+b^2)^(1/2))/a^2/e/ (-4*a*c+b^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\frac {-\frac {2 a}{(d+e x)^2}-4 b \log (d+e x)+\frac {\left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c (d+e x)^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c (d+e x)^2\right )}{\sqrt {b^2-4 a c}}}{4 a^2 e} \]
((-2*a)/(d + e*x)^2 - 4*b*Log[d + e*x] + ((b^2 - 2*a*c + b*Sqrt[b^2 - 4*a* c])*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*(d + e*x)^2])/Sqrt[b^2 - 4*a*c] + ((-b ^2 + 2*a*c + b*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*(d + e*x )^2])/Sqrt[b^2 - 4*a*c])/(4*a^2*e)
Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1462, 1434, 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.
\(\displaystyle \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx\) |
\(\Big \downarrow \) 1462 |
\(\displaystyle \frac {\int \frac {1}{(d+e x)^3 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{e}\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {\int \frac {1}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)^2}{2 e}\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {\frac {\int -\frac {c (d+e x)^2+b}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)^2}{a}-\frac {1}{a (d+e x)^2}}{2 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {c (d+e x)^2+b}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)^2}{a}-\frac {1}{a (d+e x)^2}}{2 e}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {-\frac {\int \left (\frac {b}{a (d+e x)^2}+\frac {-b^2-c (d+e x)^2 b+a c}{a \left (c (d+e x)^4+b (d+e x)^2+a\right )}\right )d(d+e x)^2}{a}-\frac {1}{a (d+e x)^2}}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{2 a}+\frac {b \log \left ((d+e x)^2\right )}{a}}{a}-\frac {1}{a (d+e x)^2}}{2 e}\) |
(-(1/(a*(d + e*x)^2)) - (((b^2 - 2*a*c)*ArcTanh[(b + 2*c*(d + e*x)^2)/Sqrt [b^2 - 4*a*c]])/(a*Sqrt[b^2 - 4*a*c]) + (b*Log[(d + e*x)^2])/a - (b*Log[a + b*(d + e*x)^2 + c*(d + e*x)^4])/(2*a))/a)/(2*e)
3.7.19.3.1 Defintions of rubi rules used
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_) + (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]
Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Si mp[u^m/(Coefficient[v, x, 1]*v^m) Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p , x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.66 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.76
method | result | size |
default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 d^{3} e c +2 b d e \right ) \textit {\_Z} +d^{4} c +b \,d^{2}+a \right )}{\sum }\frac {\left (b c \,e^{3} \textit {\_R}^{3}+3 b c d \,e^{2} \textit {\_R}^{2}+e \left (3 b c \,d^{2}-a c +b^{2}\right ) \textit {\_R} +b c \,d^{3}-a c d +b^{2} d \right ) \ln \left (x -\textit {\_R} \right )}{2 e^{3} c \,\textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 d^{3} c +b e \textit {\_R} +b d}}{2 a^{2} e}-\frac {1}{2 a e \left (e x +d \right )^{2}}-\frac {b \ln \left (e x +d \right )}{a^{2} e}\) | \(213\) |
risch | \(-\frac {1}{2 a e \left (e x +d \right )^{2}}-\frac {b \ln \left (e x +d \right )}{a^{2} e}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c \,e^{2}-a^{2} b^{2} e^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c e +b^{3} e \right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 a^{3} c \,e^{4}-3 a^{2} b^{2} e^{4}\right ) \textit {\_R}^{2}-4 a b c \,e^{3} \textit {\_R} +2 e^{2} c^{2}\right ) x^{2}+\left (\left (20 a^{3} c d \,e^{3}-6 a^{2} b^{2} d \,e^{3}\right ) \textit {\_R}^{2}-8 a b c d \,e^{2} \textit {\_R} +4 c^{2} d e \right ) x +\left (10 a^{3} c \,d^{2} e^{2}-3 a^{2} b^{2} d^{2} e^{2}-a^{3} b \,e^{2}\right ) \textit {\_R}^{2}+\left (-4 a b c \,d^{2} e +a^{2} c e -2 a \,b^{2} e \right ) \textit {\_R} +2 c^{2} d^{2}+2 b c \right )\right )}{2}\) | \(256\) |
1/2/a^2/e*sum((b*c*e^3*_R^3+3*b*c*d*e^2*_R^2+e*(3*b*c*d^2-a*c+b^2)*_R+b*c* d^3-a*c*d+b^2*d)/(2*_R^3*c*e^3+6*_R^2*c*d*e^2+6*_R*c*d^2*e+2*c*d^3+_R*b*e+ b*d)*ln(x-_R),_R=RootOf(c*e^4*_Z^4+4*c*d*e^3*_Z^3+(6*c*d^2*e^2+b*e^2)*_Z^2 +(4*c*d^3*e+2*b*d*e)*_Z+d^4*c+b*d^2+a))-1/2/a/e/(e*x+d)^2-b*ln(e*x+d)/a^2/ e
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (111) = 222\).
Time = 0.29 (sec) , antiderivative size = 810, normalized size of antiderivative = 6.69 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\left [-\frac {2 \, a b^{2} - 8 \, a^{2} c + {\left ({\left (b^{2} - 2 \, a c\right )} e^{2} x^{2} + 2 \, {\left (b^{2} - 2 \, a c\right )} d e x + {\left (b^{2} - 2 \, a c\right )} d^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} e^{4} x^{4} + 8 \, c^{2} d e^{3} x^{3} + 2 \, c^{2} d^{4} + 2 \, {\left (6 \, c^{2} d^{2} + b c\right )} e^{2} x^{2} + 2 \, b c d^{2} + 4 \, {\left (2 \, c^{2} d^{3} + b c d\right )} e x + b^{2} - 2 \, a c + {\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \log \left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a\right ) + 4 \, {\left ({\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \log \left (e x + d\right )}{4 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{3} x^{2} + 2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d e^{2} x + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e\right )}}, -\frac {2 \, a b^{2} - 8 \, a^{2} c + 2 \, {\left ({\left (b^{2} - 2 \, a c\right )} e^{2} x^{2} + 2 \, {\left (b^{2} - 2 \, a c\right )} d e x + {\left (b^{2} - 2 \, a c\right )} d^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \log \left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a\right ) + 4 \, {\left ({\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \log \left (e x + d\right )}{4 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{3} x^{2} + 2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d e^{2} x + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e\right )}}\right ] \]
[-1/4*(2*a*b^2 - 8*a^2*c + ((b^2 - 2*a*c)*e^2*x^2 + 2*(b^2 - 2*a*c)*d*e*x + (b^2 - 2*a*c)*d^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*e^4*x^4 + 8*c^2*d*e^3*x^ 3 + 2*c^2*d^4 + 2*(6*c^2*d^2 + b*c)*e^2*x^2 + 2*b*c*d^2 + 4*(2*c^2*d^3 + b *c*d)*e*x + b^2 - 2*a*c + (2*c*e^2*x^2 + 4*c*d*e*x + 2*c*d^2 + b)*sqrt(b^2 - 4*a*c))/(c*e^4*x^4 + 4*c*d*e^3*x^3 + c*d^4 + (6*c*d^2 + b)*e^2*x^2 + b* d^2 + 2*(2*c*d^3 + b*d)*e*x + a)) - ((b^3 - 4*a*b*c)*e^2*x^2 + 2*(b^3 - 4* a*b*c)*d*e*x + (b^3 - 4*a*b*c)*d^2)*log(c*e^4*x^4 + 4*c*d*e^3*x^3 + c*d^4 + (6*c*d^2 + b)*e^2*x^2 + b*d^2 + 2*(2*c*d^3 + b*d)*e*x + a) + 4*((b^3 - 4 *a*b*c)*e^2*x^2 + 2*(b^3 - 4*a*b*c)*d*e*x + (b^3 - 4*a*b*c)*d^2)*log(e*x + d))/((a^2*b^2 - 4*a^3*c)*e^3*x^2 + 2*(a^2*b^2 - 4*a^3*c)*d*e^2*x + (a^2*b ^2 - 4*a^3*c)*d^2*e), -1/4*(2*a*b^2 - 8*a^2*c + 2*((b^2 - 2*a*c)*e^2*x^2 + 2*(b^2 - 2*a*c)*d*e*x + (b^2 - 2*a*c)*d^2)*sqrt(-b^2 + 4*a*c)*arctan(-(2* c*e^2*x^2 + 4*c*d*e*x + 2*c*d^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - ( (b^3 - 4*a*b*c)*e^2*x^2 + 2*(b^3 - 4*a*b*c)*d*e*x + (b^3 - 4*a*b*c)*d^2)*l og(c*e^4*x^4 + 4*c*d*e^3*x^3 + c*d^4 + (6*c*d^2 + b)*e^2*x^2 + b*d^2 + 2*( 2*c*d^3 + b*d)*e*x + a) + 4*((b^3 - 4*a*b*c)*e^2*x^2 + 2*(b^3 - 4*a*b*c)*d *e*x + (b^3 - 4*a*b*c)*d^2)*log(e*x + d))/((a^2*b^2 - 4*a^3*c)*e^3*x^2 + 2 *(a^2*b^2 - 4*a^3*c)*d*e^2*x + (a^2*b^2 - 4*a^3*c)*d^2*e)]
Timed out. \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\int { \frac {1}{{\left ({\left (e x + d\right )}^{4} c + {\left (e x + d\right )}^{2} b + a\right )} {\left (e x + d\right )}^{3}} \,d x } \]
-1/2/(a*e^3*x^2 + 2*a*d*e^2*x + a*d^2*e) + integrate((b*c*e^3*x^3 + 3*b*c* d*e^2*x^2 + b*c*d^3 + (3*b*c*d^2 + b^2 - a*c)*e*x + (b^2 - a*c)*d)/(c*e^4* x^4 + 4*c*d*e^3*x^3 + c*d^4 + (6*c*d^2 + b)*e^2*x^2 + b*d^2 + 2*(2*c*d^3 + b*d)*e*x + a), x)/a^2 - b*log(e*x + d)/(a^2*e)
Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\frac {b \log \left (c + \frac {b}{{\left (e x + d\right )}^{2}} + \frac {a}{{\left (e x + d\right )}^{4}}\right )}{4 \, a^{2} e} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (-\frac {b + \frac {2 \, a}{{\left (e x + d\right )}^{2}}}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2} e} - \frac {1}{2 \, {\left (e x + d\right )}^{2} a e} \]
1/4*b*log(c + b/(e*x + d)^2 + a/(e*x + d)^4)/(a^2*e) + 1/2*(b^2 - 2*a*c)*a rctan(-(b + 2*a/(e*x + d)^2)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^2*e ) - 1/2/((e*x + d)^2*a*e)
Time = 12.24 (sec) , antiderivative size = 4950, normalized size of antiderivative = 40.91 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )} \, dx=\text {Too large to display} \]
(atan((16*a^6*x^2*(4*a*c - b^2)^(3/2)*(((3*b^4 + a^2*c^2 - 9*a*b^2*c)*(((( ((20*a^3*c^4*e^18 + 2*a^2*b^2*c^3*e^18)/a^3 + ((2*b^3*e - 8*a*b*c*e)*(40*a ^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/(2*a^3*(16*a^3*c*e^2 - 4*a^2*b^2*e^2 )))*(2*b^3*e - 8*a*b*c*e))/(2*(16*a^3*c*e^2 - 4*a^2*b^2*e^2)) + (6*b*c^4*e ^17)/a^2)*(2*b^3*e - 8*a*b*c*e))/(2*(16*a^3*c*e^2 - 4*a^2*b^2*e^2)) + (c^5 *e^16)/a^3 - (((((20*a^3*c^4*e^18 + 2*a^2*b^2*c^3*e^18)/a^3 + ((2*b^3*e - 8*a*b*c*e)*(40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/(2*a^3*(16*a^3*c*e^2 - 4*a^2*b^2*e^2)))*(2*a*c - b^2))/(4*a^2*e*(4*a*c - b^2)^(1/2)) + ((2*a*c - b^2)*(2*b^3*e - 8*a*b*c*e)*(40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/( 8*a^5*e*(16*a^3*c*e^2 - 4*a^2*b^2*e^2)*(4*a*c - b^2)^(1/2)))*(2*a*c - b^2) )/(4*a^2*e*(4*a*c - b^2)^(1/2)) - ((2*a*c - b^2)^2*(2*b^3*e - 8*a*b*c*e)*( 40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/(32*a^7*e^2*(16*a^3*c*e^2 - 4*a^ 2*b^2*e^2)*(4*a*c - b^2))))/(8*a^3*c^2*(a^2*c^2 - 6*b^4 + 24*a*b^2*c)) + ( ((((((20*a^3*c^4*e^18 + 2*a^2*b^2*c^3*e^18)/a^3 + ((2*b^3*e - 8*a*b*c*e)*( 40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/(2*a^3*(16*a^3*c*e^2 - 4*a^2*b^2 *e^2)))*(2*a*c - b^2))/(4*a^2*e*(4*a*c - b^2)^(1/2)) + ((2*a*c - b^2)*(2*b ^3*e - 8*a*b*c*e)*(40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^19))/(8*a^5*e*(16* a^3*c*e^2 - 4*a^2*b^2*e^2)*(4*a*c - b^2)^(1/2)))*(2*b^3*e - 8*a*b*c*e))/(2 *(16*a^3*c*e^2 - 4*a^2*b^2*e^2)) + (((((20*a^3*c^4*e^18 + 2*a^2*b^2*c^3*e^ 18)/a^3 + ((2*b^3*e - 8*a*b*c*e)*(40*a^4*b*c^3*e^19 - 12*a^3*b^3*c^2*e^...