3.1.23 \(\int \frac {(a+b \text {ArcTan}(c x))^2}{x^5} \, dx\) [23]

Optimal. Leaf size=116 \[ -\frac {b^2 c^2}{12 x^2}-\frac {b c (a+b \text {ArcTan}(c x))}{6 x^3}+\frac {b c^3 (a+b \text {ArcTan}(c x))}{2 x}+\frac {1}{4} c^4 (a+b \text {ArcTan}(c x))^2-\frac {(a+b \text {ArcTan}(c x))^2}{4 x^4}-\frac {2}{3} b^2 c^4 \log (x)+\frac {1}{3} b^2 c^4 \log \left (1+c^2 x^2\right ) \]

[Out]

-1/12*b^2*c^2/x^2-1/6*b*c*(a+b*arctan(c*x))/x^3+1/2*b*c^3*(a+b*arctan(c*x))/x+1/4*c^4*(a+b*arctan(c*x))^2-1/4*
(a+b*arctan(c*x))^2/x^4-2/3*b^2*c^4*ln(x)+1/3*b^2*c^4*ln(c^2*x^2+1)

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Rubi [A]
time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4946, 5038, 272, 46, 36, 29, 31, 5004} \begin {gather*} \frac {1}{4} c^4 (a+b \text {ArcTan}(c x))^2+\frac {b c^3 (a+b \text {ArcTan}(c x))}{2 x}-\frac {(a+b \text {ArcTan}(c x))^2}{4 x^4}-\frac {b c (a+b \text {ArcTan}(c x))}{6 x^3}-\frac {2}{3} b^2 c^4 \log (x)-\frac {b^2 c^2}{12 x^2}+\frac {1}{3} b^2 c^4 \log \left (c^2 x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/x^5,x]

[Out]

-1/12*(b^2*c^2)/x^2 - (b*c*(a + b*ArcTan[c*x]))/(6*x^3) + (b*c^3*(a + b*ArcTan[c*x]))/(2*x) + (c^4*(a + b*ArcT
an[c*x])^2)/4 - (a + b*ArcTan[c*x])^2/(4*x^4) - (2*b^2*c^4*Log[x])/3 + (b^2*c^4*Log[1 + c^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 46

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] && Lt
Q[m + n + 2, 0])

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 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 5004

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 5038

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

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Mathematica [A]
time = 0.05, size = 128, normalized size = 1.10 \begin {gather*} \frac {-3 a^2-2 a b c x-b^2 c^2 x^2+6 a b c^3 x^3+2 b \left (b c x \left (-1+3 c^2 x^2\right )+3 a \left (-1+c^4 x^4\right )\right ) \text {ArcTan}(c x)+3 b^2 \left (-1+c^4 x^4\right ) \text {ArcTan}(c x)^2-8 b^2 c^4 x^4 \log (x)+4 b^2 c^4 x^4 \log \left (1+c^2 x^2\right )}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2 - 2*a*b*c*x - b^2*c^2*x^2 + 6*a*b*c^3*x^3 + 2*b*(b*c*x*(-1 + 3*c^2*x^2) + 3*a*(-1 + c^4*x^4))*ArcTan[c
*x] + 3*b^2*(-1 + c^4*x^4)*ArcTan[c*x]^2 - 8*b^2*c^4*x^4*Log[x] + 4*b^2*c^4*x^4*Log[1 + c^2*x^2])/(12*x^4)

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Maple [A]
time = 0.25, size = 152, normalized size = 1.31

method	result	size
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}-\frac {b^{2} \arctan \left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {b^{2} \arctan \left (c x \right )^{2}}{4}-\frac {b^{2} \arctan \left (c x \right )}{6 c^{3} x^{3}}+\frac {b^{2} \arctan \left (c x \right )}{2 c x}+\frac {b^{2} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {b^{2}}{12 c^{2} x^{2}}-\frac {2 b^{2} \ln \left (c x \right )}{3}-\frac {a b \arctan \left (c x \right )}{2 c^{4} x^{4}}+\frac {a b \arctan \left (c x \right )}{2}-\frac {a b}{6 c^{3} x^{3}}+\frac {a b}{2 c x}\right )\)	\(152\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}-\frac {b^{2} \arctan \left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {b^{2} \arctan \left (c x \right )^{2}}{4}-\frac {b^{2} \arctan \left (c x \right )}{6 c^{3} x^{3}}+\frac {b^{2} \arctan \left (c x \right )}{2 c x}+\frac {b^{2} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {b^{2}}{12 c^{2} x^{2}}-\frac {2 b^{2} \ln \left (c x \right )}{3}-\frac {a b \arctan \left (c x \right )}{2 c^{4} x^{4}}+\frac {a b \arctan \left (c x \right )}{2}-\frac {a b}{6 c^{3} x^{3}}+\frac {a b}{2 c x}\right )\)	\(152\)
risch	\(-\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (i c x +1\right )^{2}}{16 x^{4}}+\frac {i b \left (-3 i b \,c^{4} x^{4} \ln \left (-i c x +1\right )-6 b \,c^{3} x^{3}+2 x b c +6 a +3 i b \ln \left (-i c x +1\right )\right ) \ln \left (i c x +1\right )}{24 x^{4}}-\frac {-12 i \ln \left (\left (4 i b c -a c \right ) x -4 b -i a \right ) a b \,c^{4} x^{4}+12 i \ln \left (\left (-4 i b c -a c \right ) x -4 b +i a \right ) a b \,c^{4} x^{4}+3 c^{4} b^{2} x^{4} \ln \left (-i c x +1\right )^{2}-16 \ln \left (\left (4 i b c -a c \right ) x -4 b -i a \right ) b^{2} c^{4} x^{4}-16 \ln \left (\left (-4 i b c -a c \right ) x -4 b +i a \right ) b^{2} c^{4} x^{4}+32 b^{2} c^{4} \ln \left (-x \right ) x^{4}-12 i b^{2} c^{3} x^{3} \ln \left (-i c x +1\right )-24 a b \,c^{3} x^{3}+4 i b^{2} c x \ln \left (-i c x +1\right )+4 b^{2} c^{2} x^{2}+12 i \ln \left (-i c x +1\right ) a b +8 a b c x -3 b^{2} \ln \left (-i c x +1\right )^{2}+12 a^{2}}{48 x^{4}}\)	\(358\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(-1/4*a^2/c^4/x^4-1/4*b^2/c^4/x^4*arctan(c*x)^2+1/4*b^2*arctan(c*x)^2-1/6*b^2*arctan(c*x)/c^3/x^3+1/2*b^2*
arctan(c*x)/c/x+1/3*b^2*ln(c^2*x^2+1)-1/12*b^2/c^2/x^2-2/3*b^2*ln(c*x)-1/2*a*b/c^4/x^4*arctan(c*x)+1/2*a*b*arc
tan(c*x)-1/6*a*b/c^3/x^3+1/2*a*b/c/x)

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Maxima [A]
time = 0.48, size = 152, normalized size = 1.31 \begin {gather*} \frac {1}{6} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} a b + \frac {1}{12} \, {\left (2 \, {\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c \arctan \left (c x\right ) - \frac {{\left (3 \, c^{2} x^{2} \arctan \left (c x\right )^{2} - 4 \, c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) + 8 \, c^{2} x^{2} \log \left (x\right ) + 1\right )} c^{2}}{x^{2}}\right )} b^{2} - \frac {b^{2} \arctan \left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*a*b + 1/12*(2*(3*c^3*arctan(c*x) + (3*c^
2*x^2 - 1)/x^3)*c*arctan(c*x) - (3*c^2*x^2*arctan(c*x)^2 - 4*c^2*x^2*log(c^2*x^2 + 1) + 8*c^2*x^2*log(x) + 1)*
c^2/x^2)*b^2 - 1/4*b^2*arctan(c*x)^2/x^4 - 1/4*a^2/x^4

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Fricas [A]
time = 1.15, size = 135, normalized size = 1.16 \begin {gather*} \frac {4 \, b^{2} c^{4} x^{4} \log \left (c^{2} x^{2} + 1\right ) - 8 \, b^{2} c^{4} x^{4} \log \left (x\right ) + 6 \, a b c^{3} x^{3} - b^{2} c^{2} x^{2} - 2 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \arctan \left (c x\right )^{2} - 3 \, a^{2} + 2 \, {\left (3 \, a b c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} - b^{2} c x - 3 \, a b\right )} \arctan \left (c x\right )}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*b^2*c^4*x^4*log(c^2*x^2 + 1) - 8*b^2*c^4*x^4*log(x) + 6*a*b*c^3*x^3 - b^2*c^2*x^2 - 2*a*b*c*x + 3*(b^2
*c^4*x^4 - b^2)*arctan(c*x)^2 - 3*a^2 + 2*(3*a*b*c^4*x^4 + 3*b^2*c^3*x^3 - b^2*c*x - 3*a*b)*arctan(c*x))/x^4

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Sympy [A]
time = 0.51, size = 170, normalized size = 1.47 \begin {gather*} \begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{4} \operatorname {atan}{\left (c x \right )}}{2} + \frac {a b c^{3}}{2 x} - \frac {a b c}{6 x^{3}} - \frac {a b \operatorname {atan}{\left (c x \right )}}{2 x^{4}} - \frac {2 b^{2} c^{4} \log {\left (x \right )}}{3} + \frac {b^{2} c^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} + \frac {b^{2} c^{4} \operatorname {atan}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{3} \operatorname {atan}{\left (c x \right )}}{2 x} - \frac {b^{2} c^{2}}{12 x^{2}} - \frac {b^{2} c \operatorname {atan}{\left (c x \right )}}{6 x^{3}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**2/x**5,x)

[Out]

Piecewise((-a**2/(4*x**4) + a*b*c**4*atan(c*x)/2 + a*b*c**3/(2*x) - a*b*c/(6*x**3) - a*b*atan(c*x)/(2*x**4) -
2*b**2*c**4*log(x)/3 + b**2*c**4*log(x**2 + c**(-2))/3 + b**2*c**4*atan(c*x)**2/4 + b**2*c**3*atan(c*x)/(2*x)
- b**2*c**2/(12*x**2) - b**2*c*atan(c*x)/(6*x**3) - b**2*atan(c*x)**2/(4*x**4), Ne(c, 0)), (-a**2/(4*x**4), Tr
ue))

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

[Out]

sage0*x

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Mupad [B]
time = 2.26, size = 171, normalized size = 1.47 \begin {gather*} \frac {b^2\,c^4\,{\mathrm {atan}\left (c\,x\right )}^2}{4}-\frac {2\,b^2\,c^4\,\ln \left (x\right )}{3}-\frac {\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{4}+\frac {a^2}{4}+x\,\left (\frac {c\,\mathrm {atan}\left (c\,x\right )\,b^2}{6}+\frac {a\,c\,b}{6}\right )-x^3\,\left (\frac {b^2\,c^3\,\mathrm {atan}\left (c\,x\right )}{2}+\frac {a\,b\,c^3}{2}\right )+\frac {b^2\,c^2\,x^2}{12}+\frac {a\,b\,\mathrm {atan}\left (c\,x\right )}{2}}{x^4}+\frac {b^2\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )}{3}+\frac {b^2\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )}{3}+\frac {a\,b\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {a\,b\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/x^5,x)

[Out]

(b^2*c^4*atan(c*x)^2)/4 - (2*b^2*c^4*log(x))/3 - ((b^2*atan(c*x)^2)/4 + a^2/4 + x*((b^2*c*atan(c*x))/6 + (a*b*
c)/6) - x^3*((b^2*c^3*atan(c*x))/2 + (a*b*c^3)/2) + (b^2*c^2*x^2)/12 + (a*b*atan(c*x))/2)/x^4 + (b^2*c^4*log(c
*x + 1i))/3 + (b^2*c^4*log(c*x*1i + 1))/3 + (a*b*c^4*log(c*x + 1i)*1i)/4 - (a*b*c^4*log(c*x*1i + 1)*1i)/4

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