3.512 \(\int \sec (c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=24 \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \sec (c+d x)}{d} \]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

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Rubi [A]  time = 0.0151662, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3486, 3770} \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Tan[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3770

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+b \tan (c+d x)) \, dx &=\frac{b \sec (c+d x)}{d}+a \int \sec (c+d x) \, dx\\ &=\frac{a \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \sec (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.0091772, size = 24, normalized size = 1. \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Tan[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

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Maple [A]  time = 0.011, size = 34, normalized size = 1.4 \begin{align*}{\frac{b}{d\cos \left ( dx+c \right ) }}+{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+b*tan(d*x+c)),x)

[Out]

1/d*b/cos(d*x+c)+1/d*a*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.3044, size = 42, normalized size = 1.75 \begin{align*} \frac{a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac{b}{\cos \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

(a*log(sec(d*x + c) + tan(d*x + c)) + b/cos(d*x + c))/d

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Fricas [B]  time = 1.8153, size = 144, normalized size = 6. \begin{align*} \frac{a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(a*cos(d*x + c)*log(sin(d*x + c) + 1) - a*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*b)/(d*cos(d*x + c))

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Sympy [A]  time = 3.8325, size = 37, normalized size = 1.54 \begin{align*} \begin{cases} \frac{a \log{\left (\tan{\left (c + d x \right )} + \sec{\left (c + d x \right )} \right )} + b \sec{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right ) \sec{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)),x)

[Out]

Piecewise(((a*log(tan(c + d*x) + sec(c + d*x)) + b*sec(c + d*x))/d, Ne(d, 0)), (x*(a + b*tan(c))*sec(c), True)
)

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Giac [B]  time = 1.27402, size = 73, normalized size = 3.04 \begin{align*} \frac{a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \, b}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

(a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*b/(tan(1/2*d*x + 1/2*c)^2 - 1
))/d