3.45 \(\int \tan (a+b x) \, dx\)

Optimal. Leaf size=12 \[ -\frac {\log (\cos (a+b x))}{b} \]

[Out]

-ln(cos(b*x+a))/b

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3475} \[ -\frac {\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tan[a + b*x],x]

[Out]

-(Log[Cos[a + b*x]]/b)

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.00 \[ -\frac {\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + b*x],x]

[Out]

-(Log[Cos[a + b*x]]/b)

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fricas [A]  time = 0.45, size = 14, normalized size = 1.17 \[ -\frac {\log \left (-\cos \left (b x + a\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

-log(-cos(b*x + a))/b

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giac [A]  time = 0.21, size = 18, normalized size = 1.50 \[ -\frac {\log \left (\frac {{\left | \cos \left (b x + a\right ) \right |}}{{\left | b \right |}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-log(abs(cos(b*x + a))/abs(b))/b

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maple [A]  time = 0.01, size = 12, normalized size = 1.00 \[ \frac {\ln \left (\sec \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)*sin(b*x+a),x)

[Out]

1/b*ln(sec(b*x+a))

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maxima [A]  time = 0.37, size = 18, normalized size = 1.50 \[ -\frac {\log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/2*log(-sin(b*x + a)^2 + 1)/b

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mupad [B]  time = 0.51, size = 16, normalized size = 1.33 \[ \frac {\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/cos(a + b*x),x)

[Out]

log(tan(a + b*x)^2 + 1)/(2*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sin(b*x+a),x)

[Out]

Integral(sin(a + b*x)*sec(a + b*x), x)

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