3.231 \(\int \frac{\tan ^{-1}(a x)}{x^4 \sqrt{c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{a \sqrt{a^2 c x^2+c}}{6 c x^2}+\frac{2 a^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x^3}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )}{6 \sqrt{c}} \]

[Out]

-(a*Sqrt[c + a^2*c*x^2])/(6*c*x^2) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*c*x^3) + (2*a^2*Sqrt[c + a^2*c*x^2]*
ArcTan[a*x])/(3*c*x) + (5*a^3*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/(6*Sqrt[c])

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Rubi [A]  time = 0.200355, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4962, 266, 51, 63, 208, 4944} \[ -\frac{a \sqrt{a^2 c x^2+c}}{6 c x^2}+\frac{2 a^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x}-\frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x^3}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )}{6 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[c + a^2*c*x^2])/(6*c*x^2) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*c*x^3) + (2*a^2*Sqrt[c + a^2*c*x^2]*
ArcTan[a*x])/(3*c*x) + (5*a^3*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/(6*Sqrt[c])

Rule 4962

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

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 51

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

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 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]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)}{x^4 \sqrt{c+a^2 c x^2}} \, dx &=-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac{1}{3} a \int \frac{1}{x^3 \sqrt{c+a^2 c x^2}} \, dx-\frac{1}{3} \left (2 a^2\right ) \int \frac{\tan ^{-1}(a x)}{x^2 \sqrt{c+a^2 c x^2}} \, dx\\ &=-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac{2 a^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{c+a^2 c x}} \, dx,x,x^2\right )-\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{x \sqrt{c+a^2 c x^2}} \, dx\\ &=-\frac{a \sqrt{c+a^2 c x^2}}{6 c x^2}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac{2 a^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}-\frac{1}{12} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+a^2 c x}} \, dx,x,x^2\right )-\frac{1}{3} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+a^2 c x}} \, dx,x,x^2\right )\\ &=-\frac{a \sqrt{c+a^2 c x^2}}{6 c x^2}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac{2 a^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}-\frac{a \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 )}{6 c}-\frac{(2 a) \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 )}{3 c}\\ &=-\frac{a \sqrt{c+a^2 c x^2}}{6 c x^2}-\frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac{2 a^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c+a^2 c x^2}}{\sqrt{c}}\right )}{6 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.118508, size = 110, normalized size = 0.93 \[ \frac{-a x \sqrt{a^2 c x^2+c}-5 a^3 \sqrt{c} x^3 \log (x)+5 a^3 \sqrt{c} x^3 \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+c\right )+2 \left (2 a^2 x^2-1\right ) \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{6 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + 2*(-1 + 2*a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] - 5*a^3*Sqrt[c]*x^3*Log[x] +
5*a^3*Sqrt[c]*x^3*Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(6*c*x^3)

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Maple [C]  time = 0.801, size = 163, normalized size = 1.4 \begin{align*}{\frac{4\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-ax-2\,\arctan \left ( ax \right ) }{6\,c{x}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{5\,{a}^{3}}{6\,c}\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}}}}+{\frac{5\,{a}^{3}}{6\,c}\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x)

[Out]

1/6*(4*arctan(a*x)*a^2*x^2-a*x-2*arctan(a*x))*(c*(a*x-I)*(a*x+I))^(1/2)/c/x^3-5/6*a^3*ln((1+I*a*x)/(a^2*x^2+1)
^(1/2)-1)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)^(1/2)/c+5/6*a^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*(c*(a*x-I)*(
a*x+I))^(1/2)/(a^2*x^2+1)^(1/2)/c

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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^4/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.44953, size = 211, normalized size = 1.79 \begin{align*} \frac{5 \, a^{3} \sqrt{c} x^{3} \log \left (-\frac{a^{2} c x^{2} + 2 \, \sqrt{a^{2} c x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt{a^{2} c x^{2} + c}{\left (a x - 2 \,{\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )\right )}}{12 \, c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

1/12*(5*a^3*sqrt(c)*x^3*log(-(a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*sqrt(a^2*c*x^2 + c)*(a
*x - 2*(2*a^2*x^2 - 1)*arctan(a*x)))/(c*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atan}{\left (a x \right )}}{x^{4} \sqrt{c \left (a^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)/x**4/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(atan(a*x)/(x**4*sqrt(c*(a**2*x**2 + 1))), x)

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Giac [B]  time = 1.41381, size = 381, normalized size = 3.23 \begin{align*} \frac{4 \,{\left (3 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{2} - c\right )} a^{2} c^{\frac{3}{2}}{\left | a \right |} \arctan \left (a x\right )}{3 \,{\left ({\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}} - \frac{{\left (\frac{8 \, a^{3} \arctan \left (x{\left | a \right |}\right )}{{\left | a \right |}} - 5 \, a^{2} \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1} - \frac{1}{x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}} + 2 \right |}\right ) + 5 \, a^{2} \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1} - \frac{1}{x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}} - 2 \right |}\right ) - \frac{4 \,{\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1} + \frac{1}{x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}}\right )} a^{2}}{{\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1} + \frac{1}{x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}}\right )}^{2} - 4}\right )}{\left | a \right |}}{12 \, \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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