\(\int \frac {(d+e x^2) (a+b \arctan (c x))}{x^4} \, dx\) [1121]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d}{6 x^2}-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {e (a+b \arctan (c x))}{x}-\frac {1}{3} b c \left (c^2 d-3 e\right ) \log (x)+\frac {1}{6} b c \left (c^2 d-3 e\right ) \log \left (1+c^2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, 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} \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {e (a+b \arctan (c x))}{x}+\frac {1}{6} b c \left (c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )-\frac {1}{3} b c \log (x) \left (c^2 d-3 e\right )-\frac {b c d}{6 x^2} \]

[In]

Int[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=-\frac {a d}{3 x^3}-\frac {a e}{x}-\frac {b d \arctan (c x)}{3 x^3}-\frac {b e \arctan (c x)}{x}+b c e \log (x)-\frac {1}{2} b c e \log \left (1+c^2 x^2\right )+\frac {1}{6} b c d \left (-\frac {1}{x^2}-2 c^2 \log (x)+c^2 \log \left (1+c^2 x^2\right )\right ) \]

[In]

Integrate[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17

method	result	size
parts	\(a \left (-\frac {e}{x}-\frac {d}{3 x^{3}}\right )+b \,c^{3} \left (-\frac {\arctan \left (c x \right ) e}{c^{3} x}-\frac {\arctan \left (c x \right ) d}{3 c^{3} x^{3}}-\frac {\left (c^{2} d -3 e \right ) \ln \left (c x \right )+\frac {d}{2 x^{2}}+\frac {\left (-c^{2} d +3 e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{3 c^{2}}\right )\)	\(97\)
derivativedivides	\(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) e}{c x}-\frac {\arctan \left (c x \right ) d}{3 c \,x^{3}}-\frac {\left (-c^{2} d +3 e \right ) \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {\left (c^{2} d -3 e \right ) \ln \left (c x \right )}{3}-\frac {d}{6 x^{2}}\right )}{c^{2}}\right )\)	\(105\)
default	\(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) e}{c x}-\frac {\arctan \left (c x \right ) d}{3 c \,x^{3}}-\frac {\left (-c^{2} d +3 e \right ) \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {\left (c^{2} d -3 e \right ) \ln \left (c x \right )}{3}-\frac {d}{6 x^{2}}\right )}{c^{2}}\right )\)	\(105\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) b \,c^{3} d \,x^{3}-\ln \left (c^{2} x^{2}+1\right ) b \,c^{3} d \,x^{3}-b \,c^{3} d \,x^{3}-6 \ln \left (x \right ) b c e \,x^{3}+3 \ln \left (c^{2} x^{2}+1\right ) b c e \,x^{3}+6 \arctan \left (c x \right ) b e \,x^{2}+6 a e \,x^{2}+b c d x +2 \arctan \left (c x \right ) b d +2 a d}{6 x^{3}}\)	\(112\)
risch	\(\frac {i b \left (3 e \,x^{2}+d \right ) \ln \left (i c x +1\right )}{6 x^{3}}-\frac {2 \ln \left (x \right ) b \,c^{3} d \,x^{3}-\ln \left (c^{2} x^{2}+1\right ) b \,c^{3} d \,x^{3}-6 \ln \left (x \right ) b c e \,x^{3}+3 \ln \left (c^{2} x^{2}+1\right ) b c e \,x^{3}+3 i b e \ln \left (-i c x +1\right ) x^{2}+i b d \ln \left (-i c x +1\right )+6 a e \,x^{2}+b c d x +2 a d}{6 x^{3}}\)	\(136\)

[In]

int((e*x^2+d)*(a+b*arctan(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=\frac {{\left (b c^{3} d - 3 \, b c e\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{3} d - 3 \, b c e\right )} x^{3} \log \left (x\right ) - b c d x - 6 \, a e x^{2} - 2 \, a d - 2 \, {\left (3 \, b e x^{2} + b d\right )} \arctan \left (c x\right )}{6 \, x^{3}} \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.40 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=\begin {cases} - \frac {a d}{3 x^{3}} - \frac {a e}{x} - \frac {b c^{3} d \log {\left (x \right )}}{3} + \frac {b c^{3} d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d}{6 x^{2}} + b c e \log {\left (x \right )} - \frac {b c e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {b e \operatorname {atan}{\left (c x \right )}}{x} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{3 x^{3}} - \frac {e}{x}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x**2+d)*(a+b*atan(c*x))/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^4} \, dx=b\,c\,e\,\ln \left (x\right )-\frac {a\,e}{x}-\frac {b\,c\,e\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c\,d}{6\,x^2}-\frac {a\,d}{3\,x^3}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}-\frac {b\,e\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,c^3\,d\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,c^3\,d\,\ln \left (x\right )}{3} \]

[In]

int(((a + b*atan(c*x))*(d + e*x^2))/x^4,x)

[Out]

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