3.101 \(\int \frac {a+b \tan ^{-1}(c x^3)}{x^4} \, dx\)

Optimal. Leaf size=39 \[ -\frac {a+b \tan ^{-1}\left (c x^3\right )}{3 x^3}-\frac {1}{6} b c \log \left (c^2 x^6+1\right )+b c \log (x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5033, 266, 36, 29, 31} \[ -\frac {a+b \tan ^{-1}\left (c x^3\right )}{3 x^3}-\frac {1}{6} b c \log \left (c^2 x^6+1\right )+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])/x^4,x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 44, normalized size = 1.13 \[ -\frac {a}{3 x^3}-\frac {1}{6} b c \log \left (c^2 x^6+1\right )-\frac {b \tan ^{-1}\left (c x^3\right )}{3 x^3}+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])/x^4,x]

[Out]

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

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fricas [A]  time = 0.47, size = 43, normalized size = 1.10 \[ -\frac {b c x^{3} \log \left (c^{2} x^{6} + 1\right ) - 6 \, b c x^{3} \log \relax (x) + 2 \, b \arctan \left (c x^{3}\right ) + 2 \, a}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.01, size = 60, normalized size = 1.54 \[ -\frac {b c^{3} x^{3} \log \left (c^{2} x^{6} + 1\right ) - 2 \, b c^{3} x^{3} \log \left (c x^{3}\right ) + 2 \, b c^{2} \arctan \left (c x^{3}\right ) + 2 \, a c^{2}}{6 \, c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 39, normalized size = 1.00 \[ -\frac {a}{3 x^{3}}-\frac {b \arctan \left (c \,x^{3}\right )}{3 x^{3}}-\frac {b c \ln \left (c^{2} x^{6}+1\right )}{6}+b c \ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))/x^4,x)

[Out]

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

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maxima [A]  time = 0.32, size = 41, normalized size = 1.05 \[ -\frac {1}{6} \, {\left (c {\left (\log \left (c^{2} x^{6} + 1\right ) - \log \left (x^{6}\right )\right )} + \frac {2 \, \arctan \left (c x^{3}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.38, size = 38, normalized size = 0.97 \[ b\,c\,\ln \relax (x)-\frac {a}{3\,x^3}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{3\,x^3}-\frac {b\,c\,\ln \left (c^2\,x^6+1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))/x^4,x)

[Out]

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

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sympy [A]  time = 96.09, size = 126, normalized size = 3.23 \[ \begin {cases} - \frac {a}{3 x^{3}} + b c \log {\relax (x )} - \frac {b c \log {\left (x - \sqrt [6]{-1} \sqrt [6]{\frac {1}{c^{2}}} \right )}}{3} - \frac {b c \log {\left (4 x^{2} + 4 \sqrt [6]{-1} x \sqrt [6]{\frac {1}{c^{2}}} + 4 \sqrt [3]{-1} \sqrt [3]{\frac {1}{c^{2}}} \right )}}{3} + \frac {i b \operatorname {atan}{\left (c x^{3} \right )}}{3 \sqrt {\frac {1}{c^{2}}}} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))/x**4,x)

[Out]

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

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