3.1.12 \(\int \frac {a+b \text {ArcTan}(c x)}{x^6} \, dx\) [12]

Optimal. Leaf size=64 \[ -\frac {b c}{20 x^4}+\frac {b c^3}{10 x^2}-\frac {a+b \text {ArcTan}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1+c^2 x^2\right ) \]

[Out]

-1/20*b*c/x^4+1/10*b*c^3/x^2+1/5*(-a-b*arctan(c*x))/x^5+1/5*b*c^5*ln(x)-1/10*b*c^5*ln(c^2*x^2+1)

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Rubi [A]
time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4946, 272, 46} \begin {gather*} -\frac {a+b \text {ArcTan}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)+\frac {b c^3}{10 x^2}-\frac {1}{10} b c^5 \log \left (c^2 x^2+1\right )-\frac {b c}{20 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/x^6,x]

[Out]

-1/20*(b*c)/x^4 + (b*c^3)/(10*x^2) - (a + b*ArcTan[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 + c^2*x^2])
/10

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]

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^6} \, dx &=-\frac {a+b \tan ^{-1}(c x)}{5 x^5}+\frac {1}{5} (b c) \int \frac {1}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {a+b \tan ^{-1}(c x)}{5 x^5}+\frac {1}{10} (b c) \text {Subst}\left (\int \frac {1}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a+b \tan ^{-1}(c x)}{5 x^5}+\frac {1}{10} (b c) \text {Subst}\left (\int \left (\frac {1}{x^3}-\frac {c^2}{x^2}+\frac {c^4}{x}-\frac {c^6}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c}{20 x^4}+\frac {b c^3}{10 x^2}-\frac {a+b \tan ^{-1}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1+c^2 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 64, normalized size = 1.00 \begin {gather*} -\frac {a}{5 x^5}-\frac {b \text {ArcTan}(c x)}{5 x^5}+\frac {1}{10} b c \left (-\frac {1}{2 x^4}+\frac {c^2}{x^2}+2 c^4 \log (x)-c^4 \log \left (1+c^2 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/x^6,x]

[Out]

-1/5*a/x^5 - (b*ArcTan[c*x])/(5*x^5) + (b*c*(-1/2*1/x^4 + c^2/x^2 + 2*c^4*Log[x] - c^4*Log[1 + c^2*x^2]))/10

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Maple [A]
time = 0.09, size = 66, normalized size = 1.03

method	result	size
derivativedivides	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}-\frac {b \arctan \left (c x \right )}{5 c^{5} x^{5}}-\frac {b \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {b}{20 c^{4} x^{4}}+\frac {b \ln \left (c x \right )}{5}+\frac {b}{10 c^{2} x^{2}}\right )\)	\(66\)
default	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}-\frac {b \arctan \left (c x \right )}{5 c^{5} x^{5}}-\frac {b \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {b}{20 c^{4} x^{4}}+\frac {b \ln \left (c x \right )}{5}+\frac {b}{10 c^{2} x^{2}}\right )\)	\(66\)
risch	\(\frac {i b \ln \left (i c x +1\right )}{10 x^{5}}-\frac {-4 b \,c^{5} \ln \left (x \right ) x^{5}+2 b \,c^{5} \ln \left (-c^{2} x^{2}-1\right ) x^{5}-2 b \,c^{3} x^{3}+2 i b \ln \left (-i c x +1\right )+x b c +4 a}{20 x^{5}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

c^5*(-1/5*a/c^5/x^5-1/5*b/c^5/x^5*arctan(c*x)-1/10*b*ln(c^2*x^2+1)-1/20*b/c^4/x^4+1/5*b*ln(c*x)+1/10*b/c^2/x^2
)

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Maxima [A]
time = 0.26, size = 62, normalized size = 0.97 \begin {gather*} -\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b - 1/5*a/x^5

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Fricas [A]
time = 1.30, size = 59, normalized size = 0.92 \begin {gather*} -\frac {2 \, b c^{5} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, b c^{5} x^{5} \log \left (x\right ) - 2 \, b c^{3} x^{3} + b c x + 4 \, b \arctan \left (c x\right ) + 4 \, a}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/20*(2*b*c^5*x^5*log(c^2*x^2 + 1) - 4*b*c^5*x^5*log(x) - 2*b*c^3*x^3 + b*c*x + 4*b*arctan(c*x) + 4*a)/x^5

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Sympy [A]
time = 0.54, size = 71, normalized size = 1.11 \begin {gather*} \begin {cases} - \frac {a}{5 x^{5}} + \frac {b c^{5} \log {\left (x \right )}}{5} - \frac {b c^{5} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3}}{10 x^{2}} - \frac {b c}{20 x^{4}} - \frac {b \operatorname {atan}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a}{5 x^{5}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/x**6,x)

[Out]

Piecewise((-a/(5*x**5) + b*c**5*log(x)/5 - b*c**5*log(x**2 + c**(-2))/10 + b*c**3/(10*x**2) - b*c/(20*x**4) -
b*atan(c*x)/(5*x**5), Ne(c, 0)), (-a/(5*x**5), 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+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 0.37, size = 56, normalized size = 0.88 \begin {gather*} \frac {b\,c^5\,\ln \left (x\right )}{5}-\frac {b\,\mathrm {atan}\left (c\,x\right )}{5\,x^5}-\frac {b\,c^5\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {-\frac {b\,c^3\,x^3}{2}+\frac {b\,c\,x}{4}+a}{5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/x^6,x)

[Out]

(b*c^5*log(x))/5 - (b*atan(c*x))/(5*x^5) - (b*c^5*log(c^2*x^2 + 1))/10 - (a - (b*c^3*x^3)/2 + (b*c*x)/4)/(5*x^
5)

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