\(\int x (a+b \arctan (\frac {c}{x}))^2 \, dx\) [142]

Optimal result

Integrand size = 14, antiderivative size = 82 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=b c x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+\frac {1}{2} c^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} x^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} b^2 c^2 \log \left (1+\frac {c^2}{x^2}\right )+b^2 c^2 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4948, 4946, 5038, 272, 36, 29, 31, 5004} \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} c^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} x^2 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+b c x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+\frac {1}{2} b^2 c^2 \log \left (\frac {c^2}{x^2}+1\right )+b^2 c^2 \log (x) \]

[In]

Int[x*(a + b*ArcTan[c/x])^2,x]

[Out]

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

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 4948

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \left (a x (2 b c+a x)+2 b \left (b c x+a \left (c^2+x^2\right )\right ) \arctan \left (\frac {c}{x}\right )+b^2 \left (c^2+x^2\right ) \arctan \left (\frac {c}{x}\right )^2+b^2 c^2 \log \left (c^2+x^2\right )\right ) \]

[In]

Integrate[x*(a + b*ArcTan[c/x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 3.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.28

method	result	size
parts	\(\frac {a^{2} x^{2}}{2}-b^{2} c^{2} \left (-\frac {x^{2} \arctan \left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {\arctan \left (\frac {c}{x}\right )^{2}}{2}-\frac {x \arctan \left (\frac {c}{x}\right )}{c}-\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{2}+\ln \left (\frac {c}{x}\right )\right )-a b \,c^{2} \arctan \left (\frac {x}{c}\right )+x^{2} \arctan \left (\frac {c}{x}\right ) a b +a b c x\)	\(105\)
parallelrisch	\(\frac {x^{2} \arctan \left (\frac {c}{x}\right )^{2} b^{2}}{2}+\frac {\arctan \left (\frac {c}{x}\right )^{2} b^{2} c^{2}}{2}+\frac {b^{2} c^{2} \ln \left (c^{2}+x^{2}\right )}{2}+x^{2} \arctan \left (\frac {c}{x}\right ) a b +x \arctan \left (\frac {c}{x}\right ) b^{2} c +\arctan \left (\frac {c}{x}\right ) a b \,c^{2}+\frac {a^{2} x^{2}}{2}+a b c x -\frac {a^{2} c^{2}}{2}\)	\(107\)
derivativedivides	\(-c^{2} \left (-\frac {a^{2} x^{2}}{2 c^{2}}+b^{2} \left (-\frac {x^{2} \arctan \left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {\arctan \left (\frac {c}{x}\right )^{2}}{2}-\frac {x \arctan \left (\frac {c}{x}\right )}{c}-\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{2}+\ln \left (\frac {c}{x}\right )\right )-\frac {a b \,x^{2} \arctan \left (\frac {c}{x}\right )}{c^{2}}-a b \arctan \left (\frac {c}{x}\right )-\frac {a b x}{c}\right )\)	\(113\)
default	\(-c^{2} \left (-\frac {a^{2} x^{2}}{2 c^{2}}+b^{2} \left (-\frac {x^{2} \arctan \left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {\arctan \left (\frac {c}{x}\right )^{2}}{2}-\frac {x \arctan \left (\frac {c}{x}\right )}{c}-\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{2}+\ln \left (\frac {c}{x}\right )\right )-\frac {a b \,x^{2} \arctan \left (\frac {c}{x}\right )}{c^{2}}-a b \arctan \left (\frac {c}{x}\right )-\frac {a b x}{c}\right )\)	\(113\)
risch	\(\text {Expression too large to display}\)	\(13965\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=-a b c^{2} \arctan \left (\frac {x}{c}\right ) + \frac {1}{2} \, b^{2} c^{2} \log \left (c^{2} + x^{2}\right ) + a b c x + \frac {1}{2} \, a^{2} x^{2} + \frac {1}{2} \, {\left (b^{2} c^{2} + b^{2} x^{2}\right )} \arctan \left (\frac {c}{x}\right )^{2} + {\left (b^{2} c x + a b x^{2}\right )} \arctan \left (\frac {c}{x}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^{2} x^{2}}{2} + a b c^{2} \operatorname {atan}{\left (\frac {c}{x} \right )} + a b c x + a b x^{2} \operatorname {atan}{\left (\frac {c}{x} \right )} + \frac {b^{2} c^{2} \log {\left (c^{2} + x^{2} \right )}}{2} + \frac {b^{2} c^{2} \operatorname {atan}^{2}{\left (\frac {c}{x} \right )}}{2} + b^{2} c x \operatorname {atan}{\left (\frac {c}{x} \right )} + \frac {b^{2} x^{2} \operatorname {atan}^{2}{\left (\frac {c}{x} \right )}}{2} \]

[In]

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

[Out]

a**2*x**2/2 + a*b*c**2*atan(c/x) + a*b*c*x + a*b*x**2*atan(c/x) + b**2*c**2*log(c**2 + x**2)/2 + b**2*c**2*ata
n(c/x)**2/2 + b**2*c*x*atan(c/x) + b**2*x**2*atan(c/x)**2/2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.27 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \, b^{2} x^{2} \arctan \left (\frac {c}{x}\right )^{2} + \frac {1}{2} \, a^{2} x^{2} + {\left (x^{2} \arctan \left (\frac {c}{x}\right ) - {\left (c \arctan \left (\frac {x}{c}\right ) - x\right )} c\right )} a b - \frac {1}{2} \, {\left ({\left (\arctan \left (\frac {x}{c}\right )^{2} - \log \left (c^{2} + x^{2}\right )\right )} c^{2} + 2 \, {\left (c \arctan \left (\frac {x}{c}\right ) - x\right )} c \arctan \left (\frac {c}{x}\right )\right )} b^{2} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((b*arctan(c/x) + a)^2*x, x)

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.20 \[ \int x \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^2\,x^2}{2}+\frac {b^2\,c^2\,{\mathrm {atan}\left (\frac {c}{x}\right )}^2}{2}+\frac {b^2\,c^2\,\ln \left (c^2+x^2\right )}{2}+\frac {b^2\,x^2\,{\mathrm {atan}\left (\frac {c}{x}\right )}^2}{2}+a\,b\,c^2\,\mathrm {atan}\left (\frac {c}{x}\right )+a\,b\,x^2\,\mathrm {atan}\left (\frac {c}{x}\right )+b^2\,c\,x\,\mathrm {atan}\left (\frac {c}{x}\right )+a\,b\,c\,x \]

[In]

int(x*(a + b*atan(c/x))^2,x)

[Out]

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