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

Optimal. Leaf size=88 \[ -\frac {4 a^3 \log (x)}{3 c}+\frac {a^3 \tan ^{-1}(a x)^2}{2 c}+\frac {a^2 \tan ^{-1}(a x)}{c x}+\frac {2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac {\tan ^{-1}(a x)}{3 c x^3}-\frac {a}{6 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4918, 4852, 266, 44, 36, 29, 31, 4884} \[ \frac {2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac {4 a^3 \log (x)}{3 c}+\frac {a^3 \tan ^{-1}(a x)^2}{2 c}+\frac {a^2 \tan ^{-1}(a x)}{c x}-\frac {a}{6 c x^2}-\frac {\tan ^{-1}(a x)}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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 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]

Rule 4918

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

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Mathematica [A]  time = 0.02, size = 88, normalized size = 1.00 \[ -\frac {4 a^3 \log (x)}{3 c}+\frac {a^3 \tan ^{-1}(a x)^2}{2 c}+\frac {a^2 \tan ^{-1}(a x)}{c x}+\frac {2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac {\tan ^{-1}(a x)}{3 c x^3}-\frac {a}{6 c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 71, normalized size = 0.81 \[ \frac {3 \, a^{3} x^{3} \arctan \left (a x\right )^{2} + 4 \, a^{3} x^{3} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{3} x^{3} \log \relax (x) - a x + 2 \, {\left (3 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{6 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a^3*x^3*arctan(a*x)^2 + 4*a^3*x^3*log(a^2*x^2 + 1) - 8*a^3*x^3*log(x) - a*x + 2*(3*a^2*x^2 - 1)*arctan(
a*x))/(c*x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 81, normalized size = 0.92 \[ -\frac {\arctan \left (a x \right )}{3 c \,x^{3}}+\frac {a^{2} \arctan \left (a x \right )}{c x}+\frac {a^{3} \arctan \left (a x \right )^{2}}{2 c}-\frac {a}{6 c \,x^{2}}-\frac {4 a^{3} \ln \left (a x \right )}{3 c}+\frac {2 a^{3} \ln \left (a^{2} x^{2}+1\right )}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 90, normalized size = 1.02 \[ \frac {1}{3} \, {\left (\frac {3 \, a^{3} \arctan \left (a x\right )}{c} + \frac {3 \, a^{2} x^{2} - 1}{c x^{3}}\right )} \arctan \left (a x\right ) - \frac {{\left (3 \, a^{2} x^{2} \arctan \left (a x\right )^{2} - 4 \, a^{2} x^{2} \log \left (a^{2} x^{2} + 1\right ) + 8 \, a^{2} x^{2} \log \relax (x) + 1\right )} a}{6 \, c x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^3*arctan(a*x)/c + (3*a^2*x^2 - 1)/(c*x^3))*arctan(a*x) - 1/6*(3*a^2*x^2*arctan(a*x)^2 - 4*a^2*x^2*log
(a^2*x^2 + 1) + 8*a^2*x^2*log(x) + 1)*a/(c*x^2)

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mupad [B]  time = 0.47, size = 78, normalized size = 0.89 \[ \frac {2\,a^3\,\ln \left (a^2\,x^2+1\right )}{3\,c}-\frac {\mathrm {atan}\left (a\,x\right )}{3\,c\,x^3}-\frac {a}{6\,c\,x^2}-\frac {4\,a^3\,\ln \relax (x)}{3\,c}+\frac {a^3\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,c}+\frac {a^2\,\mathrm {atan}\left (a\,x\right )}{c\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.53, size = 117, normalized size = 1.33 \[ \begin {cases} - \frac {4 a^{3} \log {\relax (x )}}{3 c} + \frac {2 a^{3} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3 c} + \frac {a^{3} \operatorname {atan}^{2}{\left (a x \right )}}{2 c} + \frac {a^{2} \operatorname {atan}{\left (a x \right )}}{c x} - \frac {a}{6 c x^{2}} - \frac {\operatorname {atan}{\left (a x \right )}}{3 c x^{3}} & \text {for}\: c \neq 0 \\\tilde {\infty } \left (- \frac {a^{3} \log {\relax (x )}}{3} + \frac {a^{3} \log {\left (a^{2} x^{2} + 1 \right )}}{6} - \frac {a}{6 x^{2}} - \frac {\operatorname {atan}{\left (a x \right )}}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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