Integrand size = 22, antiderivative size = 108 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {b \left (b^2-6 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x+c x^2\right )}{2 a^2} \]
(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/(c*x^2+b*x+a)+b*(-6*a*c+b^2)*arctanh((2*c *x+b)/(-4*a*c+b^2)^(1/2))/a^2/(-4*a*c+b^2)^(3/2)+ln(x)/a^2-1/2*ln(c*x^2+b* x+a)/a^2
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {\frac {2 a \left (b^2-2 a c+b c x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {2 b \left (b^2-6 a c\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+2 \log (x)-\log (a+x (b+c x))}{2 a^2} \]
((2*a*(b^2 - 2*a*c + b*c*x))/((b^2 - 4*a*c)*(a + x*(b + c*x))) + (2*b*(b^2 - 6*a*c)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2) + 2 *Log[x] - Log[a + x*(b + c*x)])/(2*a^2)
Time = 0.34 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {9, 1165, 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.
| \(\displaystyle \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x \left (a+b x+c x^2\right )^2}dx\) |
| \(\Big \downarrow \) 1165 |
| \(\displaystyle \frac {-2 a c+b^2+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int -\frac {b^2+c x b-4 a c}{x \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {b^2+c x b-4 a c}{x \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {\int \left (\frac {b^2-4 a c}{a x}+\frac {-b \left (b^2-5 a c\right )-c \left (b^2-4 a c\right ) x}{a \left (c x^2+b x+a\right )}\right )dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b \left (b^2-6 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {\left (b^2-4 a c\right ) \log \left (a+b x+c x^2\right )}{2 a}+\frac {\log (x) \left (b^2-4 a c\right )}{a}}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
(b^2 - 2*a*c + b*c*x)/(a*(b^2 - 4*a*c)*(a + b*x + c*x^2)) + ((b*(b^2 - 6*a *c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(a*Sqrt[b^2 - 4*a*c]) + ((b^2 - 4*a*c)*Log[x])/a - ((b^2 - 4*a*c)*Log[a + b*x + c*x^2])/(2*a))/(a*(b^2 - 4*a*c))
3.1.24.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) *x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
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]
Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.64
| method | result | size |
| default | \(\frac {\ln \left (x \right )}{a^{2}}-\frac {\frac {\frac {a b c x}{4 a c -b^{2}}-\frac {a \left (2 a c -b^{2}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (4 a \,c^{2}-b^{2} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (5 a b c -b^{3}-\frac {\left (4 a \,c^{2}-b^{2} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{2}}\) | \(177\) |
| risch | \(\frac {-\frac {b x c}{a \left (4 a c -b^{2}\right )}+\frac {2 a c -b^{2}}{a \left (4 a c -b^{2}\right )}}{c \,x^{2}+b x +a}+\frac {\ln \left (x \right )}{a^{2}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (64 a^{5} c^{3}-48 a^{4} b^{2} c^{2}+12 a^{3} b^{4} c -a^{2} b^{6}\right ) \textit {\_Z}^{2}+\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \textit {\_Z} +16 a \,c^{3}-3 b^{2} c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (96 a^{5} c^{3}-80 a^{4} b^{2} c^{2}+22 a^{3} b^{4} c -2 a^{2} b^{6}\right ) \textit {\_R}^{2}+\left (48 c^{3} a^{3}-20 a^{2} b^{2} c^{2}+2 a \,b^{4} c \right ) \textit {\_R} +b^{2} c^{2}\right ) x +\left (-16 a^{5} b \,c^{2}+8 a^{4} b^{3} c -a^{3} b^{5}\right ) \textit {\_R}^{2}+\left (20 a^{3} b \,c^{2}-9 a^{2} b^{3} c +b^{5} a \right ) \textit {\_R} -4 a b \,c^{2}+b^{3} c \right )\right )\) | \(315\) |
ln(x)/a^2-1/a^2*((a*b*c/(4*a*c-b^2)*x-a*(2*a*c-b^2)/(4*a*c-b^2))/(c*x^2+b* x+a)+1/(4*a*c-b^2)*(1/2*(4*a*c^2-b^2*c)/c*ln(c*x^2+b*x+a)+2*(5*a*b*c-b^3-1 /2*(4*a*c^2-b^2*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/ 2))))
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (102) = 204\).
Time = 0.33 (sec) , antiderivative size = 781, normalized size of antiderivative = 7.23 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\left [\frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (a b^{3} - 6 \, a^{2} b c + {\left (b^{3} c - 6 \, a b c^{2}\right )} x^{2} + {\left (b^{4} - 6 \, a b^{2} c\right )} x\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} x - {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}, \frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \, {\left (a b^{3} - 6 \, a^{2} b c + {\left (b^{3} c - 6 \, a b c^{2}\right )} x^{2} + {\left (b^{4} - 6 \, a b^{2} c\right )} x\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} x - {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}\right ] \]
[1/2*(2*a*b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + (a*b^3 - 6*a^2*b*c + (b^3*c - 6*a*b*c^2)*x^2 + (b^4 - 6*a*b^2*c)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2 *b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 2*(a*b^3*c - 4*a^2*b*c^2)*x - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*log( c*x^2 + b*x + a) + 2*(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2* c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*log(x))/(a^3*b ^4 - 8*a^4*b^2*c + 16*a^5*c^2 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x ^2 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x), 1/2*(2*a*b^4 - 12*a^2*b^2* c + 16*a^3*c^2 + 2*(a*b^3 - 6*a^2*b*c + (b^3*c - 6*a*b*c^2)*x^2 + (b^4 - 6 *a*b^2*c)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^ 2 - 4*a*c)) + 2*(a*b^3*c - 4*a^2*b*c^2)*x - (a*b^4 - 8*a^2*b^2*c + 16*a^3* c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b *c^2)*x)*log(c*x^2 + b*x + a) + 2*(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4 *c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*l og(x))/(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^2 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x)]
Timed out. \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.17 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {{\left (b^{3} - 6 \, a b c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {\log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2}} + \frac {a b c x + a b^{2} - 2 \, a^{2} c}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{2}} \]
-(b^3 - 6*a*b*c)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((a^2*b^2 - 4*a^3* c)*sqrt(-b^2 + 4*a*c)) - 1/2*log(c*x^2 + b*x + a)/a^2 + log(abs(x))/a^2 + (a*b*c*x + a*b^2 - 2*a^2*c)/((c*x^2 + b*x + a)*(b^2 - 4*a*c)*a^2)
Time = 9.03 (sec) , antiderivative size = 620, normalized size of antiderivative = 5.74 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {\ln \left (x\right )}{a^2}+\frac {\frac {2\,a\,c-b^2}{a\,\left (4\,a\,c-b^2\right )}-\frac {b\,c\,x}{a\,\left (4\,a\,c-b^2\right )}}{c\,x^2+b\,x+a}+\frac {\ln \left (2\,a\,b^6+2\,b^7\,x-96\,a^4\,c^3+2\,a\,b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-23\,a^2\,b^4\,c+2\,b^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+84\,a^3\,b^2\,c^2+94\,a^2\,b^3\,c^2\,x+12\,a^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-24\,a\,b^5\,c\,x-9\,a^2\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-120\,a^3\,b\,c^3\,x-12\,a\,b^2\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (b^6-64\,a^3\,c^3+b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c-6\,a\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^2\,{\left (4\,a\,c-b^2\right )}^3}+\frac {\ln \left (96\,a^4\,c^3-2\,b^7\,x-2\,a\,b^6+2\,a\,b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+23\,a^2\,b^4\,c+2\,b^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-84\,a^3\,b^2\,c^2-94\,a^2\,b^3\,c^2\,x+12\,a^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+24\,a\,b^5\,c\,x-9\,a^2\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+120\,a^3\,b\,c^3\,x-12\,a\,b^2\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (b^6-64\,a^3\,c^3-b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+6\,a\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^2\,{\left (4\,a\,c-b^2\right )}^3} \]
log(x)/a^2 + ((2*a*c - b^2)/(a*(4*a*c - b^2)) - (b*c*x)/(a*(4*a*c - b^2))) /(a + b*x + c*x^2) + (log(2*a*b^6 + 2*b^7*x - 96*a^4*c^3 + 2*a*b^3*(-(4*a* c - b^2)^3)^(1/2) - 23*a^2*b^4*c + 2*b^4*x*(-(4*a*c - b^2)^3)^(1/2) + 84*a ^3*b^2*c^2 + 94*a^2*b^3*c^2*x + 12*a^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2) - 24 *a*b^5*c*x - 9*a^2*b*c*(-(4*a*c - b^2)^3)^(1/2) - 120*a^3*b*c^3*x - 12*a*b ^2*c*x*(-(4*a*c - b^2)^3)^(1/2))*(b^6 - 64*a^3*c^3 + b^3*(-(4*a*c - b^2)^3 )^(1/2) + 48*a^2*b^2*c^2 - 12*a*b^4*c - 6*a*b*c*(-(4*a*c - b^2)^3)^(1/2))) /(2*a^2*(4*a*c - b^2)^3) + (log(96*a^4*c^3 - 2*b^7*x - 2*a*b^6 + 2*a*b^3*( -(4*a*c - b^2)^3)^(1/2) + 23*a^2*b^4*c + 2*b^4*x*(-(4*a*c - b^2)^3)^(1/2) - 84*a^3*b^2*c^2 - 94*a^2*b^3*c^2*x + 12*a^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2 ) + 24*a*b^5*c*x - 9*a^2*b*c*(-(4*a*c - b^2)^3)^(1/2) + 120*a^3*b*c^3*x - 12*a*b^2*c*x*(-(4*a*c - b^2)^3)^(1/2))*(b^6 - 64*a^3*c^3 - b^3*(-(4*a*c - b^2)^3)^(1/2) + 48*a^2*b^2*c^2 - 12*a*b^4*c + 6*a*b*c*(-(4*a*c - b^2)^3)^( 1/2)))/(2*a^2*(4*a*c - b^2)^3)