\(\int \frac {(d+e x^2) (a+b \arctan (c x))}{x^6} \, dx\) [1123]

Optimal result

Integrand size = 19, antiderivative size = 110 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d}{20 x^4}+\frac {b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}+\frac {1}{15} b c^3 \left (3 c^2 d-5 e\right ) \log (x)-\frac {1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (1+c^2 x^2\right ) \]

[Out]

-1/20*b*c*d/x^4+1/30*b*c*(3*c^2*d-5*e)/x^2-1/5*d*(a+b*arctan(c*x))/x^5-1/3*e*(a+b*arctan(c*x))/x^3+1/15*b*c^3*
(3*c^2*d-5*e)*ln(x)-1/30*b*c^3*(3*c^2*d-5*e)*ln(c^2*x^2+1)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 5096, 12, 457, 78} \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}+\frac {b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac {1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (c^2 x^2+1\right )+\frac {1}{15} b c^3 \log (x) \left (3 c^2 d-5 e\right )-\frac {b c d}{20 x^4} \]

[In]

Int[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-1/20*(b*c*d)/x^4 + (b*c*(3*c^2*d - 5*e))/(30*x^2) - (d*(a + b*ArcTan[c*x]))/(5*x^5) - (e*(a + b*ArcTan[c*x]))
/(3*x^3) + (b*c^3*(3*c^2*d - 5*e)*Log[x])/15 - (b*c^3*(3*c^2*d - 5*e)*Log[1 + c^2*x^2])/30

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}-(b c) \int \frac {-3 d-5 e x^2}{15 x^5 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}-\frac {1}{15} (b c) \int \frac {-3 d-5 e x^2}{x^5 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}-\frac {1}{30} (b c) \text {Subst}\left (\int \frac {-3 d-5 e x}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}-\frac {1}{30} (b c) \text {Subst}\left (\int \left (-\frac {3 d}{x^3}+\frac {3 c^2 d-5 e}{x^2}+\frac {-3 c^4 d+5 c^2 e}{x}+\frac {3 c^6 d-5 c^4 e}{1+c^2 x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b c d}{20 x^4}+\frac {b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}+\frac {1}{15} b c^3 \left (3 c^2 d-5 e\right ) \log (x)-\frac {1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (1+c^2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {a d}{5 x^5}-\frac {b c d}{20 x^4}-\frac {a e}{3 x^3}+\frac {b c^3 d}{10 x^2}-\frac {b d \arctan (c x)}{5 x^5}-\frac {b e \arctan (c x)}{3 x^3}+\frac {1}{5} b c^5 d \log (x)-\frac {1}{10} b c^5 d \log \left (1+c^2 x^2\right )+\frac {1}{6} b c e \left (-\frac {1}{x^2}-2 c^2 \log (x)+c^2 \log \left (1+c^2 x^2\right )\right ) \]

[In]

Integrate[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-1/5*(a*d)/x^5 - (b*c*d)/(20*x^4) - (a*e)/(3*x^3) + (b*c^3*d)/(10*x^2) - (b*d*ArcTan[c*x])/(5*x^5) - (b*e*ArcT
an[c*x])/(3*x^3) + (b*c^5*d*Log[x])/5 - (b*c^5*d*Log[1 + c^2*x^2])/10 + (b*c*e*(-x^(-2) - 2*c^2*Log[x] + c^2*L
og[1 + c^2*x^2]))/6

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08

method	result	size
parts	\(a \left (-\frac {e}{3 x^{3}}-\frac {d}{5 x^{5}}\right )+b \,c^{5} \left (-\frac {\arctan \left (c x \right ) e}{3 c^{5} x^{3}}-\frac {\arctan \left (c x \right ) d}{5 c^{5} x^{5}}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )-\frac {3 c^{2} d -5 e}{2 c^{2} x^{2}}+\frac {3 d}{4 c^{2} x^{4}}+\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{15 c^{2}}\right )\)	\(119\)
derivativedivides	\(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )}{15}+\frac {3 c^{2} d -5 e}{30 c^{2} x^{2}}-\frac {d}{20 c^{2} x^{4}}\right )}{c^{2}}\right )\)	\(127\)
default	\(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )}{15}+\frac {3 c^{2} d -5 e}{30 c^{2} x^{2}}-\frac {d}{20 c^{2} x^{4}}\right )}{c^{2}}\right )\)	\(127\)
parallelrisch	\(\frac {12 \ln \left (x \right ) b \,c^{5} d \,x^{5}-6 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{5} d -6 b \,c^{5} d \,x^{5}-20 \ln \left (x \right ) b \,c^{3} e \,x^{5}+10 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{3} e +10 b \,c^{3} e \,x^{5}+6 b \,c^{3} d \,x^{3}-10 b c e \,x^{3}-20 \arctan \left (c x \right ) b e \,x^{2}-20 a e \,x^{2}-3 b c d x -12 \arctan \left (c x \right ) b d -12 a d}{60 x^{5}}\)	\(145\)
risch	\(\frac {i b \left (5 e \,x^{2}+3 d \right ) \ln \left (i c x +1\right )}{30 x^{5}}-\frac {-12 \ln \left (x \right ) b \,c^{5} d \,x^{5}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{5} d \,x^{5}+20 \ln \left (x \right ) b \,c^{3} e \,x^{5}-10 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{3} e \,x^{5}-6 b \,c^{3} d \,x^{3}+10 i b e \ln \left (-i c x +1\right ) x^{2}+10 b c e \,x^{3}+6 i b d \ln \left (-i c x +1\right )+20 a e \,x^{2}+3 b c d x +12 a d}{60 x^{5}}\)	\(163\)

[In]

int((e*x^2+d)*(a+b*arctan(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

a*(-1/3*e/x^3-1/5*d/x^5)+b*c^5*(-1/3*arctan(c*x)/c^5*e/x^3-1/5*arctan(c*x)*d/c^5/x^5-1/15/c^2*((-3*c^2*d+5*e)*
ln(c*x)-1/2*(3*c^2*d-5*e)/c^2/x^2+3/4*d/c^2/x^4+1/2*(3*c^2*d-5*e)*ln(c^2*x^2+1)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {2 \, {\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (x\right ) + 3 \, b c d x + 20 \, a e x^{2} - 2 \, {\left (3 \, b c^{3} d - 5 \, b c e\right )} x^{3} + 12 \, a d + 4 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \arctan \left (c x\right )}{60 \, x^{5}} \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/60*(2*(3*b*c^5*d - 5*b*c^3*e)*x^5*log(c^2*x^2 + 1) - 4*(3*b*c^5*d - 5*b*c^3*e)*x^5*log(x) + 3*b*c*d*x + 20*
a*e*x^2 - 2*(3*b*c^3*d - 5*b*c*e)*x^3 + 12*a*d + 4*(5*b*e*x^2 + 3*b*d)*arctan(c*x))/x^5

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.39 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\begin {cases} - \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} + \frac {b c^{5} d \log {\left (x \right )}}{5} - \frac {b c^{5} d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d}{10 x^{2}} - \frac {b c^{3} e \log {\left (x \right )}}{3} + \frac {b c^{3} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d}{20 x^{4}} - \frac {b c e}{6 x^{2}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b e \operatorname {atan}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{5 x^{5}} - \frac {e}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x**2+d)*(a+b*atan(c*x))/x**6,x)

[Out]

Piecewise((-a*d/(5*x**5) - a*e/(3*x**3) + b*c**5*d*log(x)/5 - b*c**5*d*log(x**2 + c**(-2))/10 + b*c**3*d/(10*x
**2) - b*c**3*e*log(x)/3 + b*c**3*e*log(x**2 + c**(-2))/6 - b*c*d/(20*x**4) - b*c*e/(6*x**2) - b*d*atan(c*x)/(
5*x**5) - b*e*atan(c*x)/(3*x**3), Ne(c, 0)), (a*(-d/(5*x**5) - e/(3*x**3)), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e - \frac {a e}{3 \, x^{3}} - \frac {a d}{5 \, x^{5}} \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d + 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*e - 1/3*a*e/x^3 - 1/5*a*d/x^5

Giac [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{6}} \,d x } \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\frac {b\,c^3\,e\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,c^5\,d\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {x^3\,\left (\frac {b\,c\,e}{6}-\frac {b\,c^3\,d}{10}\right )+\frac {a\,d}{5}+x^2\,\left (\frac {a\,e}{3}+\frac {b\,e\,\mathrm {atan}\left (c\,x\right )}{3}\right )+\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{5}+\frac {b\,c\,d\,x}{20}}{x^5}+\frac {b\,c^5\,d\,\ln \left (x\right )}{5}-\frac {b\,c^3\,e\,\ln \left (x\right )}{3} \]

[In]

int(((a + b*atan(c*x))*(d + e*x^2))/x^6,x)

[Out]

(b*c^3*e*log(c^2*x^2 + 1))/6 - (b*c^5*d*log(c^2*x^2 + 1))/10 - (x^3*((b*c*e)/6 - (b*c^3*d)/10) + (a*d)/5 + x^2
*((a*e)/3 + (b*e*atan(c*x))/3) + (b*d*atan(c*x))/5 + (b*c*d*x)/20)/x^5 + (b*c^5*d*log(x))/5 - (b*c^3*e*log(x))
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