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

Optimal. Leaf size=80 \[ -\frac{x^2}{6 a^3 c}+\frac{2 \log \left (a^2 x^2+1\right )}{3 a^5 c}+\frac{x^3 \tan ^{-1}(a x)}{3 a^2 c}-\frac{x \tan ^{-1}(a x)}{a^4 c}+\frac{\tan ^{-1}(a x)^2}{2 a^5 c} \]

[Out]

-x^2/(6*a^3*c) - (x*ArcTan[a*x])/(a^4*c) + (x^3*ArcTan[a*x])/(3*a^2*c) + ArcTan[a*x]^2/(2*a^5*c) + (2*Log[1 +
a^2*x^2])/(3*a^5*c)

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Rubi [A]  time = 0.154707, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4916, 4852, 266, 43, 4846, 260, 4884} \[ -\frac{x^2}{6 a^3 c}+\frac{2 \log \left (a^2 x^2+1\right )}{3 a^5 c}+\frac{x^3 \tan ^{-1}(a x)}{3 a^2 c}-\frac{x \tan ^{-1}(a x)}{a^4 c}+\frac{\tan ^{-1}(a x)^2}{2 a^5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/(6*a^3*c) - (x*ArcTan[a*x])/(a^4*c) + (x^3*ArcTan[a*x])/(3*a^2*c) + ArcTan[a*x]^2/(2*a^5*c) + (2*Log[1 +
a^2*x^2])/(3*a^5*c)

Rule 4916

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 4846

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

Rule 260

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

Rule 4884

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]

Rubi steps

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

Mathematica [A]  time = 0.0557064, size = 56, normalized size = 0.7 \[ \frac{-a^2 x^2+4 \log \left (a^2 x^2+1\right )+2 a x \left (a^2 x^2-3\right ) \tan ^{-1}(a x)+3 \tan ^{-1}(a x)^2}{6 a^5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*x^2) + 2*a*x*(-3 + a^2*x^2)*ArcTan[a*x] + 3*ArcTan[a*x]^2 + 4*Log[1 + a^2*x^2])/(6*a^5*c)

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Maple [A]  time = 0.033, size = 73, normalized size = 0.9 \begin{align*} -{\frac{{x}^{2}}{6\,{a}^{3}c}}-{\frac{x\arctan \left ( ax \right ) }{{a}^{4}c}}+{\frac{{x}^{3}\arctan \left ( ax \right ) }{3\,{a}^{2}c}}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{a}^{5}c}}+{\frac{2\,\ln \left ({a}^{2}{x}^{2}+1 \right ) }{3\,{a}^{5}c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63382, size = 100, normalized size = 1.25 \begin{align*} \frac{1}{3} \,{\left (\frac{a^{2} x^{3} - 3 \, x}{a^{4} c} + \frac{3 \, \arctan \left (a x\right )}{a^{5} c}\right )} \arctan \left (a x\right ) - \frac{a^{2} x^{2} + 3 \, \arctan \left (a x\right )^{2} - 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.77491, size = 131, normalized size = 1.64 \begin{align*} -\frac{a^{2} x^{2} - 2 \,{\left (a^{3} x^{3} - 3 \, a x\right )} \arctan \left (a x\right ) - 3 \, \arctan \left (a x\right )^{2} - 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.59707, size = 110, normalized size = 1.38 \begin{align*} \begin{cases} \frac{x^{3} \operatorname{atan}{\left (a x \right )}}{3 a^{2} c} - \frac{x^{2}}{6 a^{3} c} - \frac{x \operatorname{atan}{\left (a x \right )}}{a^{4} c} + \frac{2 \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{3 a^{5} c} + \frac{\operatorname{atan}^{2}{\left (a x \right )}}{2 a^{5} c} & \text{for}\: c \neq 0 \\\tilde{\infty } \left (\frac{x^{5} \operatorname{atan}{\left (a x \right )}}{5} - \frac{x^{4}}{20 a} + \frac{x^{2}}{10 a^{3}} - \frac{\log{\left (a^{2} x^{2} + 1 \right )}}{10 a^{5}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*atan(a*x)/(3*a**2*c) - x**2/(6*a**3*c) - x*atan(a*x)/(a**4*c) + 2*log(x**2 + a**(-2))/(3*a**5*
c) + atan(a*x)**2/(2*a**5*c), Ne(c, 0)), (zoo*(x**5*atan(a*x)/5 - x**4/(20*a) + x**2/(10*a**3) - log(a**2*x**2
+ 1)/(10*a**5)), True))

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Giac [A]  time = 1.28609, size = 77, normalized size = 0.96 \begin{align*} \frac{2 \, a^{3} x^{3} \arctan \left (a x\right ) - a^{2} x^{2} - 6 \, a x \arctan \left (a x\right ) + 3 \, \arctan \left (a x\right )^{2} + 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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