3.103 \(\int \frac {a+b \tan ^{-1}(c x^3)}{x^{10}} \, dx\)

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

[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 = 0.04, 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 = {5033, 266, 44} \[ -\frac {a+b \tan ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{18} b c^3 \log \left (c^2 x^6+1\right )-\frac {1}{3} b c^3 \log (x)-\frac {b c}{18 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])/x^10,x]

[Out]

-(b*c)/(18*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 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 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && 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) \operatorname {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) \operatorname {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 \[ -\frac {a}{9 x^9}-\frac {1}{3} b c^3 \log (x)+\frac {1}{18} b c^3 \log \left (c^2 x^6+1\right )-\frac {b c}{18 x^6}-\frac {b \tan ^{-1}\left (c x^3\right )}{9 x^9} \]

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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fricas [A]  time = 0.44, size = 54, normalized size = 0.98 \[ \frac {b c^{3} x^{9} \log \left (c^{2} x^{6} + 1\right ) - 6 \, b c^{3} x^{9} \log \relax (x) - b c x^{3} - 2 \, b \arctan \left (c x^{3}\right ) - 2 \, a}{18 \, x^{9}} \]

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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giac [A]  time = 0.16, size = 69, normalized size = 1.25 \[ \frac {b c^{7} x^{9} \log \left (c^{2} x^{6} + 1\right ) - 2 \, b c^{7} x^{9} \log \left (c x^{3}\right ) - b c^{5} x^{3} - 2 \, b c^{4} \arctan \left (c x^{3}\right ) - 2 \, a c^{4}}{18 \, c^{4} x^{9}} \]

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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maple [A]  time = 0.03, size = 51, normalized size = 0.93 \[ -\frac {a}{9 x^{9}}-\frac {b \arctan \left (c \,x^{3}\right )}{9 x^{9}}+\frac {b \,c^{3} \ln \left (c^{2} x^{6}+1\right )}{18}-\frac {b c}{18 x^{6}}-\frac {b \,c^{3} \ln \relax (x )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 53, normalized size = 0.96 \[ \frac {1}{18} \, {\left ({\left (c^{2} \log \left (c^{2} x^{6} + 1\right ) - c^{2} \log \left (x^{6}\right ) - \frac {1}{x^{6}}\right )} c - \frac {2 \, \arctan \left (c x^{3}\right )}{x^{9}}\right )} b - \frac {a}{9 \, x^{9}} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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