\(\int \frac {\arctan (a x)}{x^2 (c+a^2 c x^2)^{5/2}} \, dx\) [247]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{c^{5/2}} \]

[Out]

-1/9*a/c/(a^2*c*x^2+c)^(3/2)-1/3*a^2*x*arctan(a*x)/c/(a^2*c*x^2+c)^(3/2)-a*arctanh((a^2*c*x^2+c)^(1/2)/c^(1/2)
)/c^(5/2)-5/3*a/c^2/(a^2*c*x^2+c)^(1/2)-5/3*a^2*x*arctan(a*x)/c^2/(a^2*c*x^2+c)^(1/2)-arctan(a*x)*(a^2*c*x^2+c
)^(1/2)/c^3/x

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5086, 5064, 272, 65, 214, 5014, 5016} \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c^3 x}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {a^2 x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a \text {arctanh}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{c^{5/2}}-\frac {5 a}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {a}{9 c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

Int[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)^(5/2)),x]

[Out]

-1/9*a/(c*(c + a^2*c*x^2)^(3/2)) - (5*a)/(3*c^2*Sqrt[c + a^2*c*x^2]) - (a^2*x*ArcTan[a*x])/(3*c*(c + a^2*c*x^2
)^(3/2)) - (5*a^2*x*ArcTan[a*x])/(3*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(c^3*x) - (a*
ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/c^(5/2)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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 5014

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[b/(c*d*Sqrt[d + e*x^2]),
x] + Simp[x*((a + b*ArcTan[c*x])/(d*Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rule 5016

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[b*((d + e*x^2)^(q + 1)/(4
*c*d*(q + 1)^2)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x] - Si
mp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d
] && LtQ[q, -1] && NeQ[q, -3/2]

Rule 5064

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Dist[b*c*(p/(f*(m + 1))), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 5086

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

Rubi steps \begin{align*} \text {integral}& = -\left (a^2 \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\right )+\frac {\int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx}{c} \\ & = -\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\int \frac {\arctan (a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c^2}-\frac {\left (2 a^2\right ) \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}-\frac {a^2 \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{c} \\ & = -\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}+\frac {a \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx}{c^2} \\ & = -\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )}{2 c^2} \\ & = -\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )}{a c^3} \\ & = -\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {c+a^2 c x^2} \left (3+12 a^2 x^2+8 a^4 x^4\right ) \arctan (a x)+a x \left (-\left (\left (16+15 a^2 x^2\right ) \sqrt {c+a^2 c x^2}\right )+9 \sqrt {c} \left (1+a^2 x^2\right )^2 \log (x)-9 \sqrt {c} \left (1+a^2 x^2\right )^2 \log \left (c+\sqrt {c} \sqrt {c+a^2 c x^2}\right )\right )}{9 c^3 x \left (1+a^2 x^2\right )^2} \]

[In]

Integrate[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)^(5/2)),x]

[Out]

(-3*Sqrt[c + a^2*c*x^2]*(3 + 12*a^2*x^2 + 8*a^4*x^4)*ArcTan[a*x] + a*x*(-((16 + 15*a^2*x^2)*Sqrt[c + a^2*c*x^2
]) + 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[x] - 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/(9*c^
3*x*(1 + a^2*x^2)^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {\left (9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a^{5} x^{5}-9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a^{5} x^{5}+24 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}+15 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+18 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a^{3} x^{3}-18 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a^{3} x^{3}+36 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+16 \sqrt {a^{2} x^{2}+1}\, a x +9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right ) a x -9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a x +9 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{9 \sqrt {a^{2} x^{2}+1}\, x \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(325\)

[In]

int(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/9*(9*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)*a^5*x^5-9*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)*a^5*x^5+24*arctan(a*x)*(
a^2*x^2+1)^(1/2)*a^4*x^4+15*(a^2*x^2+1)^(1/2)*a^3*x^3+18*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)*a^3*x^3-18*ln((1+I*
a*x)/(a^2*x^2+1)^(1/2)-1)*a^3*x^3+36*arctan(a*x)*(a^2*x^2+1)^(1/2)*a^2*x^2+16*(a^2*x^2+1)^(1/2)*a*x+9*ln((1+I*
a*x)/(a^2*x^2+1)^(1/2)+1)*a*x-9*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)*a*x+9*arctan(a*x)*(a^2*x^2+1)^(1/2))/(a^2*x^
2+1)^(1/2)*(c*(a*x-I)*(I+a*x))^(1/2)/x/c^3/(a^4*x^4+2*a^2*x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {9 \, {\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (15 \, a^{3} x^{3} + 16 \, a x + 3 \, {\left (8 \, a^{4} x^{4} + 12 \, a^{2} x^{2} + 3\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{18 \, {\left (a^{4} c^{3} x^{5} + 2 \, a^{2} c^{3} x^{3} + c^{3} x\right )}} \]

[In]

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/18*(9*(a^5*x^5 + 2*a^3*x^3 + a*x)*sqrt(c)*log(-(a^2*c*x^2 - 2*sqrt(a^2*c*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(1
5*a^3*x^3 + 16*a*x + 3*(8*a^4*x^4 + 12*a^2*x^2 + 3)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/(a^4*c^3*x^5 + 2*a^2*c^3
*x^3 + c^3*x)

Sympy [F]

\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(atan(a*x)/x**2/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(atan(a*x)/(x**2*(c*(a**2*x**2 + 1))**(5/2)), x)

Maxima [F]

\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \]

[In]

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)^(5/2)*x^2), x)

Giac [F]

\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \]

[In]

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

[In]

int(atan(a*x)/(x^2*(c + a^2*c*x^2)^(5/2)),x)

[Out]

int(atan(a*x)/(x^2*(c + a^2*c*x^2)^(5/2)), x)