3.2.62 \(\int \frac {(c+a^2 c x^2)^2 \text {ArcTan}(a x)}{x^2} \, dx\) [162]

Optimal. Leaf size=81 \[ -\frac {1}{6} a^3 c^2 x^2-\frac {c^2 \text {ArcTan}(a x)}{x}+2 a^2 c^2 x \text {ArcTan}(a x)+\frac {1}{3} a^4 c^2 x^3 \text {ArcTan}(a x)+a c^2 \log (x)-\frac {4}{3} a c^2 \log \left (1+a^2 x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5068, 4930, 266, 4946, 272, 36, 29, 31, 45} \begin {gather*} \frac {1}{3} a^4 c^2 x^3 \text {ArcTan}(a x)-\frac {1}{6} a^3 c^2 x^2+2 a^2 c^2 x \text {ArcTan}(a x)-\frac {4}{3} a c^2 \log \left (a^2 x^2+1\right )-\frac {c^2 \text {ArcTan}(a x)}{x}+a c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 45

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 266

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 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 4930

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

Rule 4946

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

Rule 5068

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 62, normalized size = 0.77 \begin {gather*} \frac {c^2 \left (2 \left (-3+6 a^2 x^2+a^4 x^4\right ) \text {ArcTan}(a x)-a x \left (a^2 x^2-6 \log (x)+8 \log \left (1+a^2 x^2\right )\right )\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(2*(-3 + 6*a^2*x^2 + a^4*x^4)*ArcTan[a*x] - a*x*(a^2*x^2 - 6*Log[x] + 8*Log[1 + a^2*x^2])))/(6*x)

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Maple [A]
time = 0.10, size = 77, normalized size = 0.95

method	result	size
derivativedivides	\(a \left (\frac {a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}+2 a \,c^{2} x \arctan \left (a x \right )-\frac {c^{2} \arctan \left (a x \right )}{a x}-\frac {c^{2} \left (\frac {a^{2} x^{2}}{2}+4 \ln \left (a^{2} x^{2}+1\right )-3 \ln \left (a x \right )\right )}{3}\right )\)	\(77\)
default	\(a \left (\frac {a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}+2 a \,c^{2} x \arctan \left (a x \right )-\frac {c^{2} \arctan \left (a x \right )}{a x}-\frac {c^{2} \left (\frac {a^{2} x^{2}}{2}+4 \ln \left (a^{2} x^{2}+1\right )-3 \ln \left (a x \right )\right )}{3}\right )\)	\(77\)
risch	\(-\frac {i c^{2} \left (a^{4} x^{4}+6 a^{2} x^{2}-3\right ) \ln \left (i a x +1\right )}{6 x}+\frac {i c^{2} \left (x^{4} \ln \left (-i a x +1\right ) a^{4}+i a^{3} x^{3}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 i a \ln \left (x \right ) x +8 i a \ln \left (7 a^{2} x^{2}+7\right ) x -3 \ln \left (-i a x +1\right )\right )}{6 x}\)	\(119\)
meijerg	\(\frac {a \,c^{2} \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4}+\frac {a \,c^{2} \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{2}+\frac {a \,c^{2} \left (-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )+4 \ln \left (x \right )+4 \ln \left (a \right )\right )}{4}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 71, normalized size = 0.88 \begin {gather*} -\frac {1}{6} \, {\left (a^{2} c^{2} x^{2} + 8 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 6 \, c^{2} \log \left (x\right )\right )} a + \frac {1}{3} \, {\left (a^{4} c^{2} x^{3} + 6 \, a^{2} c^{2} x - \frac {3 \, c^{2}}{x}\right )} \arctan \left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.05, size = 75, normalized size = 0.93 \begin {gather*} -\frac {a^{3} c^{2} x^{3} + 8 \, a c^{2} x \log \left (a^{2} x^{2} + 1\right ) - 6 \, a c^{2} x \log \left (x\right ) - 2 \, {\left (a^{4} c^{2} x^{4} + 6 \, a^{2} c^{2} x^{2} - 3 \, c^{2}\right )} \arctan \left (a x\right )}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.48, size = 82, normalized size = 1.01 \begin {gather*} \begin {cases} \frac {a^{4} c^{2} x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {a^{3} c^{2} x^{2}}{6} + 2 a^{2} c^{2} x \operatorname {atan}{\left (a x \right )} + a c^{2} \log {\left (x \right )} - \frac {4 a c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3} - \frac {c^{2} \operatorname {atan}{\left (a x \right )}}{x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.21, size = 76, normalized size = 0.94 \begin {gather*} \frac {a^4\,c^2\,x^3\,\mathrm {atan}\left (a\,x\right )}{3}-\frac {c^2\,\mathrm {atan}\left (a\,x\right )}{x}-\frac {a^3\,c^2\,x^2}{6}-\frac {c^2\,\left (8\,a\,\ln \left (a^2\,x^2+1\right )-6\,a\,\ln \left (x\right )\right )}{6}+2\,a^2\,c^2\,x\,\mathrm {atan}\left (a\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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