Integrand size = 19, antiderivative size = 57 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=-\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))+b c d \log (x)-\frac {b \left (c^2 d+e\right ) \log \left (1+c^2 x^2\right )}{2 c} \]
Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=-\frac {a d}{x}+a e x-\frac {b d \arctan (c x)}{x}+b e x \arctan (c x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1+c^2 x^2\right )-\frac {b e \log \left (1+c^2 x^2\right )}{2 c} \]
-((a*d)/x) + a*e*x - (b*d*ArcTan[c*x])/x + b*e*x*ArcTan[c*x] + b*c*d*Log[x ] - (b*c*d*Log[1 + c^2*x^2])/2 - (b*e*Log[1 + c^2*x^2])/(2*c)
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5511, 25, 354, 86, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx\) |
\(\Big \downarrow \) 5511 |
\(\displaystyle -b c \int -\frac {d-e x^2}{x \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b c \int \frac {d-e x^2}{x \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} b c \int \frac {d-e x^2}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} b c \int \left (\frac {d}{x^2}+\frac {-d c^2-e}{c^2 x^2+1}\right )dx^2-\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d (a+b \arctan (c x))}{x}+e x (a+b \arctan (c x))+\frac {1}{2} b c \left (d \log \left (x^2\right )-\frac {\left (c^2 d+e\right ) \log \left (c^2 x^2+1\right )}{c^2}\right )\) |
-((d*(a + b*ArcTan[c*x]))/x) + e*x*(a + b*ArcTan[c*x]) + (b*c*(d*Log[x^2] - ((c^2*d + e)*Log[1 + c^2*x^2])/c^2))/2
3.12.19.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(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 = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \left (\arctan \left (c x \right ) e c x -\frac {\arctan \left (c x \right ) d c}{x}-\frac {\left (c^{2} d +e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}+d \,c^{2} \ln \left (c x \right )\right )}{c^{2}}\right )\) | \(74\) |
default | \(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \left (\arctan \left (c x \right ) e c x -\frac {\arctan \left (c x \right ) d c}{x}-\frac {\left (c^{2} d +e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}+d \,c^{2} \ln \left (c x \right )\right )}{c^{2}}\right )\) | \(74\) |
parts | \(a \left (e x -\frac {d}{x}\right )+b c \left (\frac {\arctan \left (c x \right ) x e}{c}-\frac {\arctan \left (c x \right ) d}{c x}-\frac {-d \,c^{2} \ln \left (c x \right )+\frac {\left (c^{2} d +e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{c^{2}}\right )\) | \(76\) |
parallelrisch | \(\frac {2 b \,c^{2} d \ln \left (x \right ) x -\ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d x +2 x^{2} \arctan \left (c x \right ) b c e +2 a e \,x^{2} c -\ln \left (c^{2} x^{2}+1\right ) b e x -2 \arctan \left (c x \right ) b c d -2 a d c}{2 c x}\) | \(87\) |
risch | \(\frac {i b \left (-e \,x^{2}+d \right ) \ln \left (i c x +1\right )}{2 x}+\frac {i b c e \,x^{2} \ln \left (-i c x +1\right )+2 b \,c^{2} d \ln \left (x \right ) x -\ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d x -i b c d \ln \left (-i c x +1\right )+2 a e \,x^{2} c -\ln \left (c^{2} x^{2}+1\right ) b e x -2 a d c}{2 c x}\) | \(121\) |
c*(a/c^2*(e*c*x-d*c/x)+b/c^2*(arctan(c*x)*e*c*x-arctan(c*x)*d*c/x-1/2*(c^2 *d+e)*ln(c^2*x^2+1)+d*c^2*ln(c*x)))
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.30 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=\frac {2 \, b c^{2} d x \log \left (x\right ) + 2 \, a c e x^{2} - 2 \, a c d - {\left (b c^{2} d + b e\right )} x \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (b c e x^{2} - b c d\right )} \arctan \left (c x\right )}{2 \, c x} \]
1/2*(2*b*c^2*d*x*log(x) + 2*a*c*e*x^2 - 2*a*c*d - (b*c^2*d + b*e)*x*log(c^ 2*x^2 + 1) + 2*(b*c*e*x^2 - b*c*d)*arctan(c*x))/(c*x)
Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.40 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=\begin {cases} - \frac {a d}{x} + a e x + b c d \log {\left (x \right )} - \frac {b c d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d \operatorname {atan}{\left (c x \right )}}{x} + b e x \operatorname {atan}{\left (c x \right )} - \frac {b e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{x} + e x\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*d/x + a*e*x + b*c*d*log(x) - b*c*d*log(x**2 + c**(-2))/2 - b *d*atan(c*x)/x + b*e*x*atan(c*x) - b*e*log(x**2 + c**(-2))/(2*c), Ne(c, 0) ), (a*(-d/x + e*x), True))
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=-\frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d + a e x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e}{2 \, c} - \frac {a d}{x} \]
-1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d + a*e*x + 1/2 *(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*e/c - a*d/x
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^2} \, dx=a\,e\,x-\frac {a\,d}{x}+b\,e\,x\,\mathrm {atan}\left (c\,x\right )-\frac {b\,c\,d\,\ln \left (c^2\,x^2+1\right )}{2}+b\,c\,d\,\ln \left (x\right )-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {b\,e\,\ln \left (c^2\,x^2+1\right )}{2\,c} \]