\(\int \frac {(a+b \arctan (c x))^2}{x^3} \, dx\) [21]

Optimal result

Integrand size = 14, antiderivative size = 79 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=-\frac {b c (a+b \arctan (c x))}{x}-\frac {1}{2} c^2 (a+b \arctan (c x))^2-\frac {(a+b \arctan (c x))^2}{2 x^2}+b^2 c^2 \log (x)-\frac {1}{2} b^2 c^2 \log \left (1+c^2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4946, 5038, 272, 36, 29, 31, 5004} \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=-\frac {1}{2} c^2 (a+b \arctan (c x))^2-\frac {(a+b \arctan (c x))^2}{2 x^2}-\frac {b c (a+b \arctan (c x))}{x}-\frac {1}{2} b^2 c^2 \log \left (c^2 x^2+1\right )+b^2 c^2 \log (x) \]

[In]

Int[(a + b*ArcTan[c*x])^2/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

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

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5038

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x
^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=-\frac {a^2+2 a b c x+2 b \left (a+b c x+a c^2 x^2\right ) \arctan (c x)+b^2 \left (1+c^2 x^2\right ) \arctan (c x)^2-2 b^2 c^2 x^2 \log (x)+b^2 c^2 x^2 \log \left (1+c^2 x^2\right )}{2 x^2} \]

[In]

Integrate[(a + b*ArcTan[c*x])^2/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.24

method	result	size
parts	\(-\frac {a^{2}}{2 x^{2}}+b^{2} c^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {\arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )-\frac {a b \arctan \left (c x \right )}{x^{2}}-a b \,c^{2} \arctan \left (c x \right )-\frac {a b c}{x}\)	\(98\)
derivativedivides	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {\arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )-\frac {a b \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {a b}{c x}-a b \arctan \left (c x \right )\right )\)	\(104\)
default	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {\arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )-\frac {a b \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {a b}{c x}-a b \arctan \left (c x \right )\right )\)	\(104\)
parallelrisch	\(\frac {-b^{2} \arctan \left (c x \right )^{2} x^{2} c^{2}+2 b^{2} c^{2} \ln \left (x \right ) x^{2}-b^{2} c^{2} \ln \left (c^{2} x^{2}+1\right ) x^{2}-2 a b \arctan \left (c x \right ) x^{2} c^{2}+c^{2} x^{2} a^{2}-2 b^{2} \arctan \left (c x \right ) x c -2 a b c x -b^{2} \arctan \left (c x \right )^{2}-2 a b \arctan \left (c x \right )-a^{2}}{2 x^{2}}\)	\(123\)
risch	\(\frac {b^{2} \left (c^{2} x^{2}+1\right ) \ln \left (i c x +1\right )^{2}}{8 x^{2}}+\frac {i b \left (i b \,c^{2} x^{2} \ln \left (-i c x +1\right )+2 x b c +2 a +i b \ln \left (-i c x +1\right )\right ) \ln \left (i c x +1\right )}{4 x^{2}}-\frac {-4 i \ln \left (\left (-3 i b c -a c \right ) x -3 b +i a \right ) a b \,c^{2} x^{2}+4 i \ln \left (\left (3 i b c -a c \right ) x -3 b -i a \right ) a b \,c^{2} x^{2}-b^{2} c^{2} x^{2} \ln \left (-i c x +1\right )^{2}+4 \ln \left (\left (-3 i b c -a c \right ) x -3 b +i a \right ) b^{2} c^{2} x^{2}+4 \ln \left (\left (3 i b c -a c \right ) x -3 b -i a \right ) b^{2} c^{2} x^{2}-8 b^{2} c^{2} \ln \left (-x \right ) x^{2}+4 i b^{2} c x \ln \left (-i c x +1\right )+4 i \ln \left (-i c x +1\right ) a b +8 a b c x -b^{2} \ln \left (-i c x +1\right )^{2}+4 a^{2}}{8 x^{2}}\)	\(308\)

[In]

int((a+b*arctan(c*x))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=-\frac {b^{2} c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b^{2} c^{2} x^{2} \log \left (x\right ) + 2 \, a b c x + {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x\right )^{2} + a^{2} + 2 \, {\left (a b c^{2} x^{2} + b^{2} c x + a b\right )} \arctan \left (c x\right )}{2 \, x^{2}} \]

[In]

integrate((a+b*arctan(c*x))^2/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=\begin {cases} - \frac {a^{2}}{2 x^{2}} - a b c^{2} \operatorname {atan}{\left (c x \right )} - \frac {a b c}{x} - \frac {a b \operatorname {atan}{\left (c x \right )}}{x^{2}} + b^{2} c^{2} \log {\left (x \right )} - \frac {b^{2} c^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b^{2} c^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2} - \frac {b^{2} c \operatorname {atan}{\left (c x \right )}}{x} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2 x^{2}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{2 x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*atan(c*x))**2/x**3,x)

[Out]

Piecewise((-a**2/(2*x**2) - a*b*c**2*atan(c*x) - a*b*c/x - a*b*atan(c*x)/x**2 + b**2*c**2*log(x) - b**2*c**2*l
og(x**2 + c**(-2))/2 - b**2*c**2*atan(c*x)**2/2 - b**2*c*atan(c*x)/x - b**2*atan(c*x)**2/(2*x**2), Ne(c, 0)),
(-a**2/(2*x**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=-{\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} a b + \frac {1}{2} \, {\left ({\left (\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right ) + 2 \, \log \left (x\right )\right )} c^{2} - 2 \, {\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c \arctan \left (c x\right )\right )} b^{2} - \frac {b^{2} \arctan \left (c x\right )^{2}}{2 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \]

[In]

integrate((a+b*arctan(c*x))^2/x^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \]

[In]

integrate((a+b*arctan(c*x))^2/x^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 2.58 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.77 \[ \int \frac {(a+b \arctan (c x))^2}{x^3} \, dx=b^2\,c^2\,\ln \left (x\right )-\frac {a^2}{2\,x^2}-\frac {b^2\,c^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2}-\frac {b^2\,c^2\,\ln \left (c\,x+1{}\mathrm {i}\right )}{2}-\frac {b^2\,c^2\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )}{2}-\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2\,x^2}-\frac {a\,b\,c}{x}-\frac {a\,b\,\mathrm {atan}\left (c\,x\right )}{x^2}-\frac {b^2\,c\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {a\,b\,c^2\,\ln \left (c\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a\,b\,c^2\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

[In]

int((a + b*atan(c*x))^2/x^3,x)

[Out]

b^2*c^2*log(x) - a^2/(2*x^2) - (b^2*c^2*atan(c*x)^2)/2 - (b^2*c^2*log(c*x + 1i))/2 - (b^2*c^2*log(c*x*1i + 1))
/2 - (b^2*atan(c*x)^2)/(2*x^2) - (a*b*c)/x - (a*b*atan(c*x))/x^2 - (a*b*c^2*log(c*x + 1i)*1i)/2 + (a*b*c^2*log
(c*x*1i + 1)*1i)/2 - (b^2*c*atan(c*x))/x