\(\int \frac {(d+e x^2)^2 (a+b \arctan (c x))}{x^2} \, dx\) [1130]

Optimal result

Integrand size = 21, antiderivative size = 109 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=-\frac {b e^2 x^2}{6 c}-\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))+b c d^2 \log (x)-\frac {b \left (3 c^4 d^2+6 c^2 d e-e^2\right ) \log \left (1+c^2 x^2\right )}{6 c^3} \]

[Out]

-1/6*b*e^2*x^2/c-d^2*(a+b*arctan(c*x))/x+2*d*e*x*(a+b*arctan(c*x))+1/3*e^2*x^3*(a+b*arctan(c*x))+b*c*d^2*ln(x)
-1/6*b*(3*c^4*d^2+6*c^2*d*e-e^2)*ln(c^2*x^2+1)/c^3

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {276, 5096, 1265, 907} \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=-\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))-\frac {b \left (3 c^4 d^2+6 c^2 d e-e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}+b c d^2 \log (x)-\frac {b e^2 x^2}{6 c} \]

[In]

Int[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^2,x]

[Out]

-1/6*(b*e^2*x^2)/c - (d^2*(a + b*ArcTan[c*x]))/x + 2*d*e*x*(a + b*ArcTan[c*x]) + (e^2*x^3*(a + b*ArcTan[c*x]))
/3 + b*c*d^2*Log[x] - (b*(3*c^4*d^2 + 6*c^2*d*e - e^2)*Log[1 + c^2*x^2])/(6*c^3)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 907

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}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[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] &&
IntegerQ[(m - 1)/2]

Rule 5096

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]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[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] &&
!ILtQ[(m - 1)/2, 0]))

Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))-(b c) \int \frac {-d^2+2 d e x^2+\frac {e^2 x^4}{3}}{x \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))-\frac {1}{2} (b c) \text {Subst}\left (\int \frac {-d^2+2 d e x+\frac {e^2 x^2}{3}}{x \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))-\frac {1}{2} (b c) \text {Subst}\left (\int \left (\frac {e^2}{3 c^2}-\frac {d^2}{x}+\frac {3 c^4 d^2+6 c^2 d e-e^2}{3 c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b e^2 x^2}{6 c}-\frac {d^2 (a+b \arctan (c x))}{x}+2 d e x (a+b \arctan (c x))+\frac {1}{3} e^2 x^3 (a+b \arctan (c x))+b c d^2 \log (x)-\frac {b \left (3 c^4 d^2+6 c^2 d e-e^2\right ) \log \left (1+c^2 x^2\right )}{6 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\frac {1}{6} \left (-\frac {6 a d^2}{x}+12 a d e x-\frac {b e^2 x^2}{c}+2 a e^2 x^3+\frac {2 b \left (-3 d^2+6 d e x^2+e^2 x^4\right ) \arctan (c x)}{x}+6 b c d^2 \log (x)+\frac {b \left (-3 c^4 d^2-6 c^2 d e+e^2\right ) \log \left (1+c^2 x^2\right )}{c^3}\right ) \]

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^2,x]

[Out]

((-6*a*d^2)/x + 12*a*d*e*x - (b*e^2*x^2)/c + 2*a*e^2*x^3 + (2*b*(-3*d^2 + 6*d*e*x^2 + e^2*x^4)*ArcTan[c*x])/x
+ 6*b*c*d^2*Log[x] + (b*(-3*c^4*d^2 - 6*c^2*d*e + e^2)*Log[1 + c^2*x^2])/c^3)/6

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.23

method	result	size
parts	\(a \left (\frac {e^{2} x^{3}}{3}+2 d e x -\frac {d^{2}}{x}\right )+b c \left (\frac {\arctan \left (c x \right ) e^{2} x^{3}}{3 c}+\frac {2 \arctan \left (c x \right ) x d e}{c}-\frac {\arctan \left (c x \right ) d^{2}}{c x}-\frac {\frac {e^{2} c^{2} x^{2}}{2}-3 c^{4} d^{2} \ln \left (c x \right )+\frac {\left (3 c^{4} d^{2}+6 c^{2} d e -e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{3 c^{4}}\right )\)	\(134\)
derivativedivides	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arctan \left (c x \right ) c^{3} d e x +\frac {\arctan \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arctan \left (c x \right ) c^{3} d^{2}}{x}-\frac {e^{2} c^{2} x^{2}}{6}-\frac {\left (3 c^{4} d^{2}+6 c^{2} d e -e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{6}+c^{4} d^{2} \ln \left (c x \right )\right )}{c^{4}}\right )\)	\(143\)
default	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arctan \left (c x \right ) c^{3} d e x +\frac {\arctan \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arctan \left (c x \right ) c^{3} d^{2}}{x}-\frac {e^{2} c^{2} x^{2}}{6}-\frac {\left (3 c^{4} d^{2}+6 c^{2} d e -e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{6}+c^{4} d^{2} \ln \left (c x \right )\right )}{c^{4}}\right )\)	\(143\)
parallelrisch	\(\frac {2 x^{4} \arctan \left (c x \right ) b \,c^{3} e^{2}+2 a \,c^{3} e^{2} x^{4}+6 b \,c^{4} d^{2} \ln \left (x \right ) x -3 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2} x +12 x^{2} \arctan \left (c x \right ) b \,c^{3} d e -b \,c^{2} e^{2} x^{3}+12 a \,c^{3} d e \,x^{2}-6 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d e x -6 \arctan \left (c x \right ) b \,c^{3} d^{2}-6 a \,c^{3} d^{2}+\ln \left (c^{2} x^{2}+1\right ) b \,e^{2} x}{6 x \,c^{3}}\)	\(165\)
risch	\(\frac {i b \left (-x^{4} e^{2}-6 x^{2} e d +3 d^{2}\right ) \ln \left (i c x +1\right )}{6 x}+\frac {i b \,c^{3} e^{2} x^{4} \ln \left (-i c x +1\right )+6 i b \,c^{3} d e \,x^{2} \ln \left (-i c x +1\right )+2 a \,c^{3} e^{2} x^{4}+6 b \,c^{4} d^{2} \ln \left (x \right ) x -3 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2} x -3 i b \,c^{3} d^{2} \ln \left (-i c x +1\right )+12 a \,c^{3} d e \,x^{2}-b \,c^{2} e^{2} x^{3}-6 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d e x -6 a \,c^{3} d^{2}+\ln \left (c^{2} x^{2}+1\right ) b \,e^{2} x}{6 c^{3} x}\)	\(217\)

[In]

int((e*x^2+d)^2*(a+b*arctan(c*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

a*(1/3*e^2*x^3+2*d*e*x-d^2/x)+b*c*(1/3*arctan(c*x)/c*e^2*x^3+2*arctan(c*x)/c*x*d*e-arctan(c*x)*d^2/c/x-1/3/c^4
*(1/2*e^2*c^2*x^2-3*c^4*d^2*ln(c*x)+1/2*(3*c^4*d^2+6*c^2*d*e-e^2)*ln(c^2*x^2+1)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\frac {2 \, a c^{3} e^{2} x^{4} + 6 \, b c^{4} d^{2} x \log \left (x\right ) + 12 \, a c^{3} d e x^{2} - b c^{2} e^{2} x^{3} - 6 \, a c^{3} d^{2} - {\left (3 \, b c^{4} d^{2} + 6 \, b c^{2} d e - b e^{2}\right )} x \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2}\right )} \arctan \left (c x\right )}{6 \, c^{3} x} \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^2,x, algorithm="fricas")

[Out]

1/6*(2*a*c^3*e^2*x^4 + 6*b*c^4*d^2*x*log(x) + 12*a*c^3*d*e*x^2 - b*c^2*e^2*x^3 - 6*a*c^3*d^2 - (3*b*c^4*d^2 +
6*b*c^2*d*e - b*e^2)*x*log(c^2*x^2 + 1) + 2*(b*c^3*e^2*x^4 + 6*b*c^3*d*e*x^2 - 3*b*c^3*d^2)*arctan(c*x))/(c^3*
x)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.51 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\begin {cases} - \frac {a d^{2}}{x} + 2 a d e x + \frac {a e^{2} x^{3}}{3} + b c d^{2} \log {\left (x \right )} - \frac {b c d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{x} + 2 b d e x \operatorname {atan}{\left (c x \right )} + \frac {b e^{2} x^{3} \operatorname {atan}{\left (c x \right )}}{3} - \frac {b d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{c} - \frac {b e^{2} x^{2}}{6 c} + \frac {b e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6 c^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{x} + 2 d e x + \frac {e^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x**2+d)**2*(a+b*atan(c*x))/x**2,x)

[Out]

Piecewise((-a*d**2/x + 2*a*d*e*x + a*e**2*x**3/3 + b*c*d**2*log(x) - b*c*d**2*log(x**2 + c**(-2))/2 - b*d**2*a
tan(c*x)/x + 2*b*d*e*x*atan(c*x) + b*e**2*x**3*atan(c*x)/3 - b*d*e*log(x**2 + c**(-2))/c - b*e**2*x**2/(6*c) +
b*e**2*log(x**2 + c**(-2))/(6*c**3), Ne(c, 0)), (a*(-d**2/x + 2*d*e*x + e**2*x**3/3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\frac {1}{3} \, a e^{2} x^{3} - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b e^{2} + 2 \, a d e x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d e}{c} - \frac {a d^{2}}{x} \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^2,x, algorithm="maxima")

[Out]

1/3*a*e^2*x^3 - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d^2 + 1/6*(2*x^3*arctan(c*x) - c*(x^
2/c^2 - log(c^2*x^2 + 1)/c^4))*b*e^2 + 2*a*d*e*x + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d*e/c - a*d^2/x

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^2} \, dx=\frac {a\,e^2\,x^3}{3}-\frac {a\,d^2}{x}+2\,a\,d\,e\,x+\frac {b\,e^2\,\ln \left (c^2\,x^2+1\right )}{6\,c^3}-\frac {b\,e^2\,x^2}{6\,c}-\frac {b\,c\,d^2\,\ln \left (c^2\,x^2+1\right )}{2}+b\,c\,d^2\,\ln \left (x\right )-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,e^2\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}-\frac {b\,d\,e\,\ln \left (c^2\,x^2+1\right )}{c}+2\,b\,d\,e\,x\,\mathrm {atan}\left (c\,x\right ) \]

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^2)/x^2,x)

[Out]

(a*e^2*x^3)/3 - (a*d^2)/x + 2*a*d*e*x + (b*e^2*log(c^2*x^2 + 1))/(6*c^3) - (b*e^2*x^2)/(6*c) - (b*c*d^2*log(c^
2*x^2 + 1))/2 + b*c*d^2*log(x) - (b*d^2*atan(c*x))/x + (b*e^2*x^3*atan(c*x))/3 - (b*d*e*log(c^2*x^2 + 1))/c +
2*b*d*e*x*atan(c*x)