3.1.41 \(\int \frac {\text {ArcTan}(\tan (a+b x))}{x} \, dx\) [41]

Optimal. Leaf size=21 \[ b x-(b x-\text {ArcTan}(\tan (a+b x))) \log (x) \]

[Out]

b*x-(b*x-arctan(tan(b*x+a)))*ln(x)

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Rubi [A]
time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2189, 29} \begin {gather*} b x-\log (x) (b x-\text {ArcTan}(\tan (a+b x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[Tan[a + b*x]]/x,x]

[Out]

b*x - (b*x - ArcTan[Tan[a + b*x]])*Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2189

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(\tan (a+b x))}{x} \, dx &=b x-\left (b x-\tan ^{-1}(\tan (a+b x))\right ) \int \frac {1}{x} \, dx\\ &=b x-\left (b x-\tan ^{-1}(\tan (a+b x))\right ) \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.90 \begin {gather*} b x+(-b x+\text {ArcTan}(\tan (a+b x))) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Tan[a + b*x]]/x,x]

[Out]

b*x + (-(b*x) + ArcTan[Tan[a + b*x]])*Log[x]

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Maple [A]
time = 0.04, size = 23, normalized size = 1.10

method	result	size
default	\(\ln \left (x \right ) \arctan \left (\tan \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\)	\(23\)
risch	\(-i \ln \left (x \right ) \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )-\ln \left (x \right ) x b +b x -\frac {\ln \left (x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )}{4}+\frac {\ln \left (x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}}{4}-\frac {\ln \left (x \right ) \pi \mathrm {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )}{4}+\frac {\ln \left (x \right ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{2}}{2}-\frac {\ln \left (x \right ) \pi \mathrm {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{3}}{4}+\frac {\ln \left (x \right ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}}{4}-\frac {\ln \left (x \right ) \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{3}}{4}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(tan(b*x+a))/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*arctan(tan(b*x+a))-b*(x*ln(x)-x)

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Maxima [A]
time = 0.46, size = 42, normalized size = 2.00 \begin {gather*} \frac {b \arctan \left (\tan \left (b x + a\right )\right ) \log \left (b x\right ) + {\left (b x - {\left (b x + a\right )} \log \left (b x\right ) + a \log \left (b x\right ) + a\right )} b}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(tan(b*x+a))/x,x, algorithm="maxima")

[Out]

(b*arctan(tan(b*x + a))*log(b*x) + (b*x - (b*x + a)*log(b*x) + a*log(b*x) + a)*b)/b

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Fricas [A]
time = 0.61, size = 8, normalized size = 0.38 \begin {gather*} b x + a \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(tan(b*x+a))/x,x, algorithm="fricas")

[Out]

b*x + a*log(x)

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Sympy [A]
time = 0.49, size = 34, normalized size = 1.62 \begin {gather*} - b x \log {\left (x \right )} + b x + \left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \log {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(tan(b*x+a))/x,x)

[Out]

-b*x*log(x) + b*x + (atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))*log(x)

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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(arctan(tan(b*x+a))/x,x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\mathrm {atan}\left (\mathrm {tan}\left (a+b\,x\right )\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(tan(a + b*x))/x,x)

[Out]

int(atan(tan(a + b*x))/x, x)

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