Integrand size = 21, antiderivative size = 115 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d^2}{6 x^2}-\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))-\frac {1}{3} b c d \left (c^2 d-6 e\right ) \log (x)+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (1+c^2 x^2\right )}{6 c} \]
-1/6*b*c*d^2/x^2-1/3*d^2*(a+b*arctan(c*x))/x^3-2*d*e*(a+b*arctan(c*x))/x+e ^2*x*(a+b*arctan(c*x))-1/3*b*c*d*(c^2*d-6*e)*ln(x)+1/6*b*(c^4*d^2-6*c^2*d* e-3*e^2)*ln(c^2*x^2+1)/c
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=\frac {1}{6} \left (-\frac {2 a d^2}{x^3}-\frac {b c d^2}{x^2}-\frac {12 a d e}{x}+6 a e^2 x-\frac {2 b \left (d^2+6 d e x^2-3 e^2 x^4\right ) \arctan (c x)}{x^3}-2 b c d \left (c^2 d-6 e\right ) \log (x)+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (1+c^2 x^2\right )}{c}\right ) \]
((-2*a*d^2)/x^3 - (b*c*d^2)/x^2 - (12*a*d*e)/x + 6*a*e^2*x - (2*b*(d^2 + 6 *d*e*x^2 - 3*e^2*x^4)*ArcTan[c*x])/x^3 - 2*b*c*d*(c^2*d - 6*e)*Log[x] + (b *(c^4*d^2 - 6*c^2*d*e - 3*e^2)*Log[1 + c^2*x^2])/c)/6
Time = 0.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5511, 27, 1578, 1195, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {-3 e^2 x^4+6 d e x^2+d^2}{3 x^3 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} b c \int \frac {-3 e^2 x^4+6 d e x^2+d^2}{x^3 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{6} b c \int \frac {-3 e^2 x^4+6 d e x^2+d^2}{x^4 \left (c^2 x^2+1\right )}dx^2-\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \frac {1}{6} b c \int \left (\frac {d^2}{x^4}-\frac {\left (c^2 d-6 e\right ) d}{x^2}+\frac {d^2 c^4-6 d e c^2-3 e^2}{c^2 x^2+1}\right )dx^2-\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 (a+b \arctan (c x))}{3 x^3}-\frac {2 d e (a+b \arctan (c x))}{x}+e^2 x (a+b \arctan (c x))+\frac {1}{6} b c \left (-d \log \left (x^2\right ) \left (c^2 d-6 e\right )+\frac {\left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (c^2 x^2+1\right )}{c^2}-\frac {d^2}{x^2}\right )\) |
-1/3*(d^2*(a + b*ArcTan[c*x]))/x^3 - (2*d*e*(a + b*ArcTan[c*x]))/x + e^2*x *(a + b*ArcTan[c*x]) + (b*c*(-(d^2/x^2) - d*(c^2*d - 6*e)*Log[x^2] + ((c^4 *d^2 - 6*c^2*d*e - 3*e^2)*Log[1 + c^2*x^2])/c^2))/6
3.12.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && Int egerQ[(m - 1)/2]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Sim p[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 + c^2 *x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] && !ILt Q[(m - 1)/2, 0]))
Time = 0.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(c^{3} \left (\frac {a \left (c x \,e^{2}-\frac {2 c d e}{x}-\frac {c \,d^{2}}{3 x^{3}}\right )}{c^{4}}+\frac {b \left (\arctan \left (c x \right ) c x \,e^{2}-\frac {2 \arctan \left (c x \right ) c d e}{x}-\frac {\arctan \left (c x \right ) c \,d^{2}}{3 x^{3}}-\frac {\left (-c^{4} d^{2}+6 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {d \,c^{2} \left (c^{2} d -6 e \right ) \ln \left (c x \right )}{3}-\frac {c^{2} d^{2}}{6 x^{2}}\right )}{c^{4}}\right )\) | \(139\) |
| default | \(c^{3} \left (\frac {a \left (c x \,e^{2}-\frac {2 c d e}{x}-\frac {c \,d^{2}}{3 x^{3}}\right )}{c^{4}}+\frac {b \left (\arctan \left (c x \right ) c x \,e^{2}-\frac {2 \arctan \left (c x \right ) c d e}{x}-\frac {\arctan \left (c x \right ) c \,d^{2}}{3 x^{3}}-\frac {\left (-c^{4} d^{2}+6 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {d \,c^{2} \left (c^{2} d -6 e \right ) \ln \left (c x \right )}{3}-\frac {c^{2} d^{2}}{6 x^{2}}\right )}{c^{4}}\right )\) | \(139\) |
| parts | \(a \left (x \,e^{2}-\frac {2 e d}{x}-\frac {d^{2}}{3 x^{3}}\right )+b \,c^{3} \left (\frac {\arctan \left (c x \right ) x \,e^{2}}{c^{3}}-\frac {2 \arctan \left (c x \right ) d e}{c^{3} x}-\frac {\arctan \left (c x \right ) d^{2}}{3 c^{3} x^{3}}-\frac {d \,c^{2} \left (c^{2} d -6 e \right ) \ln \left (c x \right )+\frac {c^{2} d^{2}}{2 x^{2}}+\frac {\left (-c^{4} d^{2}+6 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{3 c^{4}}\right )\) | \(140\) |
| parallelrisch | \(-\frac {2 \ln \left (x \right ) b \,c^{4} d^{2} x^{3}-\ln \left (c^{2} x^{2}+1\right ) x^{3} b \,c^{4} d^{2}-b \,c^{4} d^{2} x^{3}-12 \ln \left (x \right ) b \,c^{2} d e \,x^{3}+6 \ln \left (c^{2} x^{2}+1\right ) x^{3} b \,c^{2} d e -6 x^{4} \arctan \left (c x \right ) b c \,e^{2}-6 x^{4} e^{2} a c +3 \ln \left (c^{2} x^{2}+1\right ) x^{3} b \,e^{2}+12 x^{2} \arctan \left (c x \right ) b c d e +12 a c d e \,x^{2}+b \,c^{2} d^{2} x +2 \arctan \left (c x \right ) b c \,d^{2}+2 a c \,d^{2}}{6 c \,x^{3}}\) | \(184\) |
| risch | \(\frac {i b \left (-3 x^{4} e^{2}+6 x^{2} e d +d^{2}\right ) \ln \left (i c x +1\right )}{6 x^{3}}-\frac {2 \ln \left (x \right ) b \,c^{4} d^{2} x^{3}-\ln \left (-c^{2} x^{2}-1\right ) b \,c^{4} d^{2} x^{3}-3 i b c \,e^{2} x^{4} \ln \left (-i c x +1\right )-12 \ln \left (x \right ) b \,c^{2} d e \,x^{3}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{2} d e \,x^{3}+6 i b c d e \,x^{2} \ln \left (-i c x +1\right )-6 x^{4} e^{2} a c +3 \ln \left (-c^{2} x^{2}-1\right ) b \,e^{2} x^{3}+i b c \,d^{2} \ln \left (-i c x +1\right )+12 a c d e \,x^{2}+b \,c^{2} d^{2} x +2 a c \,d^{2}}{6 c \,x^{3}}\) | \(225\) |
c^3*(a/c^4*(c*x*e^2-2*c*d*e/x-1/3*c*d^2/x^3)+b/c^4*(arctan(c*x)*c*x*e^2-2* arctan(c*x)*c*d*e/x-1/3*arctan(c*x)*c*d^2/x^3-1/6*(-c^4*d^2+6*c^2*d*e+3*e^ 2)*ln(c^2*x^2+1)-1/3*d*c^2*(c^2*d-6*e)*ln(c*x)-1/6*c^2*d^2/x^2))
Time = 0.25 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=\frac {6 \, a c e^{2} x^{4} - b c^{2} d^{2} x - 12 \, a c d e x^{2} + {\left (b c^{4} d^{2} - 6 \, b c^{2} d e - 3 \, b e^{2}\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{4} d^{2} - 6 \, b c^{2} d e\right )} x^{3} \log \left (x\right ) - 2 \, a c d^{2} + 2 \, {\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2}\right )} \arctan \left (c x\right )}{6 \, c x^{3}} \]
1/6*(6*a*c*e^2*x^4 - b*c^2*d^2*x - 12*a*c*d*e*x^2 + (b*c^4*d^2 - 6*b*c^2*d *e - 3*b*e^2)*x^3*log(c^2*x^2 + 1) - 2*(b*c^4*d^2 - 6*b*c^2*d*e)*x^3*log(x ) - 2*a*c*d^2 + 2*(3*b*c*e^2*x^4 - 6*b*c*d*e*x^2 - b*c*d^2)*arctan(c*x))/( c*x^3)
Time = 0.44 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.57 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=\begin {cases} - \frac {a d^{2}}{3 x^{3}} - \frac {2 a d e}{x} + a e^{2} x - \frac {b c^{3} d^{2} \log {\left (x \right )}}{3} + \frac {b c^{3} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{2}}{6 x^{2}} + 2 b c d e \log {\left (x \right )} - b c d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{x} + b e^{2} x \operatorname {atan}{\left (c x \right )} - \frac {b e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{3 x^{3}} - \frac {2 d e}{x} + e^{2} x\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*d**2/(3*x**3) - 2*a*d*e/x + a*e**2*x - b*c**3*d**2*log(x)/3 + b*c**3*d**2*log(x**2 + c**(-2))/6 - b*c*d**2/(6*x**2) + 2*b*c*d*e*log(x) - b*c*d*e*log(x**2 + c**(-2)) - b*d**2*atan(c*x)/(3*x**3) - 2*b*d*e*atan( c*x)/x + b*e**2*x*atan(c*x) - b*e**2*log(x**2 + c**(-2))/(2*c), Ne(c, 0)), (a*(-d**2/(3*x**3) - 2*d*e/x + e**2*x), True))
Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{2} - {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e^{2}}{2 \, c} - \frac {2 \, a d e}{x} - \frac {a d^{2}}{3 \, x^{3}} \]
1/6*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)* b*d^2 - (c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d*e + a*e^2* x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*e^2/c - 2*a*d*e/x - 1/3*a *d^2/x^3
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
Time = 0.78 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^4} \, dx=a\,e^2\,x-\frac {a\,d^2}{3\,x^3}+\frac {b\,c^3\,d^2\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,e^2\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,c^3\,d^2\,\ln \left (x\right )}{3}-\frac {2\,a\,d\,e}{x}+b\,e^2\,x\,\mathrm {atan}\left (c\,x\right )-\frac {b\,c\,d^2}{6\,x^2}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}-b\,c\,d\,e\,\ln \left (c^2\,x^2+1\right )+2\,b\,c\,d\,e\,\ln \left (x\right )-\frac {2\,b\,d\,e\,\mathrm {atan}\left (c\,x\right )}{x} \]