3.2.3 \(\int \frac {a+b \text {ArcTan}(c x^3)}{x^{10}} \, dx\) [103]

Optimal. Leaf size=55 \[ -\frac {b c}{18 x^6}-\frac {a+b \text {ArcTan}\left (c x^3\right )}{9 x^9}-\frac {1}{3} b c^3 \log (x)+\frac {1}{18} b c^3 \log \left (1+c^2 x^6\right ) \]

[Out]

-1/18*b*c/x^6+1/9*(-a-b*arctan(c*x^3))/x^9-1/3*b*c^3*ln(x)+1/18*b*c^3*ln(c^2*x^6+1)

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Rubi [A]
time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4946, 272, 46} \begin {gather*} -\frac {a+b \text {ArcTan}\left (c x^3\right )}{9 x^9}-\frac {1}{3} b c^3 \log (x)+\frac {1}{18} b c^3 \log \left (c^2 x^6+1\right )-\frac {b c}{18 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])/x^10,x]

[Out]

-1/18*(b*c)/x^6 - (a + b*ArcTan[c*x^3])/(9*x^9) - (b*c^3*Log[x])/3 + (b*c^3*Log[1 + c^2*x^6])/18

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}\left (c x^3\right )}{x^{10}} \, dx &=-\frac {a+b \tan ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} (b c) \int \frac {1}{x^7 \left (1+c^2 x^6\right )} \, dx\\ &=-\frac {a+b \tan ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{18} (b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^6\right )\\ &=-\frac {a+b \tan ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{18} (b c) \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c^2}{x}+\frac {c^4}{1+c^2 x}\right ) \, dx,x,x^6\right )\\ &=-\frac {b c}{18 x^6}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{9 x^9}-\frac {1}{3} b c^3 \log (x)+\frac {1}{18} b c^3 \log \left (1+c^2 x^6\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 60, normalized size = 1.09 \begin {gather*} -\frac {a}{9 x^9}-\frac {b c}{18 x^6}-\frac {b \text {ArcTan}\left (c x^3\right )}{9 x^9}-\frac {1}{3} b c^3 \log (x)+\frac {1}{18} b c^3 \log \left (1+c^2 x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])/x^10,x]

[Out]

-1/9*a/x^9 - (b*c)/(18*x^6) - (b*ArcTan[c*x^3])/(9*x^9) - (b*c^3*Log[x])/3 + (b*c^3*Log[1 + c^2*x^6])/18

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Maple [A]
time = 0.05, size = 51, normalized size = 0.93

method	result	size
default	\(-\frac {a}{9 x^{9}}-\frac {b \arctan \left (c \,x^{3}\right )}{9 x^{9}}-\frac {b c}{18 x^{6}}-\frac {b \,c^{3} \ln \left (x \right )}{3}+\frac {b \,c^{3} \ln \left (c^{2} x^{6}+1\right )}{18}\)	\(51\)
risch	\(\frac {i b \ln \left (i c \,x^{3}+1\right )}{18 x^{9}}-\frac {6 b \,c^{3} \ln \left (x \right ) x^{9}-b \,c^{3} \ln \left (c^{2} x^{6}+1\right ) x^{9}+b c \,x^{3}+i b \ln \left (-i c \,x^{3}+1\right )+2 a}{18 x^{9}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/9*a/x^9-1/9*b/x^9*arctan(c*x^3)-1/18*b*c/x^6-1/3*b*c^3*ln(x)+1/18*b*c^3*ln(c^2*x^6+1)

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Maxima [A]
time = 0.26, size = 53, normalized size = 0.96 \begin {gather*} \frac {1}{18} \, {\left ({\left (c^{2} \log \left (c^{2} x^{6} + 1\right ) - c^{2} \log \left (x^{6}\right ) - \frac {1}{x^{6}}\right )} c - \frac {2 \, \arctan \left (c x^{3}\right )}{x^{9}}\right )} b - \frac {a}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^10,x, algorithm="maxima")

[Out]

1/18*((c^2*log(c^2*x^6 + 1) - c^2*log(x^6) - 1/x^6)*c - 2*arctan(c*x^3)/x^9)*b - 1/9*a/x^9

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Fricas [A]
time = 1.46, size = 54, normalized size = 0.98 \begin {gather*} \frac {b c^{3} x^{9} \log \left (c^{2} x^{6} + 1\right ) - 6 \, b c^{3} x^{9} \log \left (x\right ) - b c x^{3} - 2 \, b \arctan \left (c x^{3}\right ) - 2 \, a}{18 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^10,x, algorithm="fricas")

[Out]

1/18*(b*c^3*x^9*log(c^2*x^6 + 1) - 6*b*c^3*x^9*log(x) - b*c*x^3 - 2*b*arctan(c*x^3) - 2*a)/x^9

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (53) = 106\).
time = 131.39, size = 129, normalized size = 2.35 \begin {gather*} \begin {cases} - \frac {a}{9 x^{9}} - \frac {b c^{4} \sqrt {- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{3} \right )}}{9} - \frac {b c^{3} \log {\left (x \right )}}{3} + \frac {b c^{3} \log {\left (x - \sqrt [6]{- \frac {1}{c^{2}}} \right )}}{9} + \frac {b c^{3} \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{9} - \frac {b c}{18 x^{6}} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{9 x^{9}} & \text {for}\: c \neq 0 \\- \frac {a}{9 x^{9}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))/x**10,x)

[Out]

Piecewise((-a/(9*x**9) - b*c**4*sqrt(-1/c**2)*atan(c*x**3)/9 - b*c**3*log(x)/3 + b*c**3*log(x - (-1/c**2)**(1/
6))/9 + b*c**3*log(4*x**2 + 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3))/9 - b*c/(18*x**6) - b*atan(c*x**3)/(9*x
**9), Ne(c, 0)), (-a/(9*x**9), True))

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Giac [A]
time = 0.43, size = 69, normalized size = 1.25 \begin {gather*} \frac {b c^{7} x^{9} \log \left (c^{2} x^{6} + 1\right ) - 2 \, b c^{7} x^{9} \log \left (c x^{3}\right ) - b c^{5} x^{3} - 2 \, b c^{4} \arctan \left (c x^{3}\right ) - 2 \, a c^{4}}{18 \, c^{4} x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^10,x, algorithm="giac")

[Out]

1/18*(b*c^7*x^9*log(c^2*x^6 + 1) - 2*b*c^7*x^9*log(c*x^3) - b*c^5*x^3 - 2*b*c^4*arctan(c*x^3) - 2*a*c^4)/(c^4*
x^9)

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Mupad [B]
time = 0.40, size = 50, normalized size = 0.91 \begin {gather*} \frac {b\,c^3\,\ln \left (c^2\,x^6+1\right )}{18}-\frac {a}{9\,x^9}-\frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{9\,x^9}-\frac {b\,c}{18\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))/x^10,x)

[Out]

(b*c^3*log(c^2*x^6 + 1))/18 - a/(9*x^9) - (b*c^3*log(x))/3 - (b*atan(c*x^3))/(9*x^9) - (b*c)/(18*x^6)

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