3.546 \(\int \frac{\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=18 \[ \frac{\log (a+b \tan (c+d x))}{b d} \]

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

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Rubi [A]  time = 0.0417421, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 31} \[ \frac{\log (a+b \tan (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + b*Tan[c + d*x]),x]

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0137411, size = 18, normalized size = 1. \[ \frac{\log (a+b \tan (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + b*Tan[c + d*x]),x]

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

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Maple [A]  time = 0.029, size = 19, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{bd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.13644, size = 24, normalized size = 1.33 \begin{align*} \frac{\log \left (b \tan \left (d x + c\right ) + a\right )}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

log(b*tan(d*x + c) + a)/(b*d)

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Fricas [B]  time = 1.95648, size = 144, normalized size = 8. \begin{align*} \frac{\log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{a + b \tan{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*tan(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*tan(c + d*x)), x)

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Giac [A]  time = 1.82938, size = 26, normalized size = 1.44 \begin{align*} \frac{\log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

log(abs(b*tan(d*x + c) + a))/(b*d)