\(\int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx\) [546]

Optimal result

Integrand size = 21, antiderivative size = 18 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log (a+b \tan (c+d x))}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3587, 31} \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log (a+b \tan (c+d x))}{b d} \]

[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 31

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

Rule 3587

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log (a+b \tan (c+d x))}{b d} \]

[In]

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

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b d}\)	\(19\)
default	\(\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b d}\)	\(19\)
risch	\(-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b d}\)	\(58\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.28 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, b d} \]

[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)

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log \left (b \tan \left (d x + c\right ) + a\right )}{b d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b d} \]

[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)

Mupad [B] (verification not implemented)

Time = 3.94 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{b\,d} \]

[In]

int(1/(cos(c + d*x)^2*(a + b*tan(c + d*x))),x)

[Out]

log(a + b*tan(c + d*x))/(b*d)