\(\int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [557]

Optimal result

Integrand size = 21, antiderivative size = 20 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{b d (a+b \tan (c+d x))} \]

[Out]

-1/b/d/(a+b*tan(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, 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, 32} \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{b d (a+b \tan (c+d x))} \]

[In]

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

[Out]

-(1/(b*d*(a + b*Tan[c + d*x])))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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)^2} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = -\frac {1}{b d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {1}{b d \left (a +b \tan \left (d x +c \right )\right )}\)	\(21\)
default	\(-\frac {1}{b d \left (a +b \tan \left (d x +c \right )\right )}\)	\(21\)
risch	\(\frac {2 i}{d \left (-i b +a \right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )}\)	\(47\)

[In]

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

[Out]

-1/b/d/(a+b*tan(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )}{{\left (a^{3} + a b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{2} b + b^{3}\right )} d \sin \left (d x + c\right )} \]

[In]

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

[Out]

-(b*cos(d*x + c) - a*sin(d*x + c))/((a^3 + a*b^2)*d*cos(d*x + c) + (a^2*b + b^3)*d*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )} b d} \]

[In]

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

[Out]

-1/((b*tan(d*x + c) + a)*b*d)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )} b d} \]

[In]

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

[Out]

-1/((b*tan(d*x + c) + a)*b*d)

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{b\,d\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \]

[In]

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

[Out]

-1/(b*d*(a + b*tan(c + d*x)))