3.247 \(\int \frac{\tan ^{-1}(a x)}{x^2 (c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{5 a}{3 c^2 \sqrt{a^2 c x^2+c}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{c^3 x}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )}{c^{5/2}}-\frac{a}{9 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{a^2 x \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-a/(9*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*Ar
cTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/c^(5/2)

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Rubi [A]  time = 0.33748, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {4966, 4944, 266, 63, 208, 4894, 4896} \[ -\frac{5 a}{3 c^2 \sqrt{a^2 c x^2+c}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{c^3 x}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )}{c^{5/2}}-\frac{a}{9 c \left (a^2 c x^2+c\right )^{3/2}}-\frac{a^2 x \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(9*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*Ar
cTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/c^(5/2)

Rule 4966

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]

Rule 4944

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 266

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 63

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 208

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

Rule 4894

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 4896

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]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\left (a^2 \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\right )+\frac{\int \frac{\tan ^{-1}(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 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\int \frac{\tan ^{-1}(a x)}{x^2 \sqrt{c+a^2 c x^2}} \, dx}{c^2}-\frac{\left (2 a^2\right ) \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}-\frac{a^2 \int \frac{\tan ^{-1}(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 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{c+a^2 c x^2}}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(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 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{c+a^2 c x^2}}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{c^3 x}+\frac{a \operatorname{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 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{c+a^2 c x^2}}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{c^3 x}+\frac{\operatorname{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 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt{c+a^2 c x^2}}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{c^3 x}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c+a^2 c x^2}}{\sqrt{c}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.24921, size = 151, normalized size = 0.96 \[ \frac{a x \left (-\left (15 a^2 x^2+16\right ) \sqrt{a^2 c x^2+c}+9 \sqrt{c} \left (a^2 x^2+1\right )^2 \log (x)-9 \sqrt{c} \left (a^2 x^2+1\right )^2 \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+c\right )\right )-3 \left (8 a^4 x^4+12 a^2 x^2+3\right ) \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{9 c^3 x \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[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)

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Maple [C]  time = 0.337, size = 369, normalized size = 2.3 \begin{align*}{\frac{a \left ( i+3\,\arctan \left ( ax \right ) \right ) \left ({a}^{3}{x}^{3}-3\,i{a}^{2}{x}^{2}-3\,ax+i \right ) }{72\, \left ({a}^{2}{x}^{2}+1 \right ) ^{2}{c}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{7\,a \left ( \arctan \left ( ax \right ) +i \right ) \left ( ax-i \right ) }{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{ \left ( 7\,ax+7\,i \right ) \left ( \arctan \left ( ax \right ) -i \right ) a}{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ({a}^{3}{x}^{3}+3\,i{a}^{2}{x}^{2}-3\,ax-i \right ) \left ( -i+3\,\arctan \left ( ax \right ) \right ) a}{72\,{c}^{3} \left ({a}^{4}{x}^{4}+2\,{a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{\arctan \left ( ax \right ) }{{c}^{3}x}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{a}{{c}^{3}}\ln \left ( 1+{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{a}{{c}^{3}}\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-1 \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*a*(I+3*arctan(a*x))*(a^3*x^3-3*I*a^2*x^2-3*a*x+I)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)^2/c^3-7/8*a*(arct
an(a*x)+I)*(a*x-I)*(c*(a*x-I)*(a*x+I))^(1/2)/c^3/(a^2*x^2+1)-7/8*(c*(a*x-I)*(a*x+I))^(1/2)*(a*x+I)*(arctan(a*x
)-I)*a/c^3/(a^2*x^2+1)+1/72*(c*(a*x-I)*(a*x+I))^(1/2)*(a^3*x^3+3*I*a^2*x^2-3*a*x-I)*(-I+3*arctan(a*x))*a/c^3/(
a^4*x^4+2*a^2*x^2+1)-arctan(a*x)*(c*(a*x-I)*(a*x+I))^(1/2)/x/c^3-a*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))/(a^2*x^2+
1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/c^3+a*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I
))^(1/2)/c^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.78061, size = 317, normalized size = 2.01 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atan}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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