3.65 \(\int \frac {a+b \tan ^{-1}(c x^2)}{x^3} \, dx\)

Optimal. Leaf size=39 \[ -\frac {a+b \tan ^{-1}\left (c x^2\right )}{2 x^2}-\frac {1}{4} b c \log \left (c^2 x^4+1\right )+b c \log (x) \]

[Out]

1/2*(-a-b*arctan(c*x^2))/x^2+b*c*ln(x)-1/4*b*c*ln(c^2*x^4+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} \[ -\frac {a+b \tan ^{-1}\left (c x^2\right )}{2 x^2}-\frac {1}{4} b c \log \left (c^2 x^4+1\right )+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/x^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 44, normalized size = 1.13 \[ -\frac {a}{2 x^2}-\frac {1}{4} b c \log \left (c^2 x^4+1\right )-\frac {b \tan ^{-1}\left (c x^2\right )}{2 x^2}+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^2])/x^3,x]

[Out]

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

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fricas [A]  time = 0.44, size = 43, normalized size = 1.10 \[ -\frac {b c x^{2} \log \left (c^{2} x^{4} + 1\right ) - 4 \, b c x^{2} \log \relax (x) + 2 \, b \arctan \left (c x^{2}\right ) + 2 \, a}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 60, normalized size = 1.54 \[ -\frac {b c^{3} x^{2} \log \left (c^{2} x^{4} + 1\right ) - 2 \, b c^{3} x^{2} \log \left (c x^{2}\right ) + 2 \, b c^{2} \arctan \left (c x^{2}\right ) + 2 \, a c^{2}}{4 \, c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 39, normalized size = 1.00 \[ -\frac {a}{2 x^{2}}-\frac {b \arctan \left (c \,x^{2}\right )}{2 x^{2}}+b c \ln \relax (x )-\frac {b c \ln \left (c^{2} x^{4}+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))/x^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 41, normalized size = 1.05 \[ -\frac {1}{4} \, {\left (c {\left (\log \left (c^{2} x^{4} + 1\right ) - \log \left (x^{4}\right )\right )} + \frac {2 \, \arctan \left (c x^{2}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.34, size = 38, normalized size = 0.97 \[ b\,c\,\ln \relax (x)-\frac {a}{2\,x^2}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{2\,x^2}-\frac {b\,c\,\ln \left (c^2\,x^4+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))/x^3,x)

[Out]

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

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sympy [A]  time = 30.28, size = 75, normalized size = 1.92 \[ \begin {cases} - \frac {a}{2 x^{2}} + b c \log {\relax (x )} - \frac {b c \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{2} - \frac {i b \operatorname {atan}{\left (c x^{2} \right )}}{2 \sqrt {\frac {1}{c^{2}}}} - \frac {b \operatorname {atan}{\left (c x^{2} \right )}}{2 x^{2}} & \text {for}\: c \neq 0 \\- \frac {a}{2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))/x**3,x)

[Out]

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

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