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

Optimal. Leaf size=20 \[ -\frac{1}{b d (a+b \tan (c+d x))} \]

[Out]

-(1/(b*d*(a + b*Tan[c + d*x])))

________________________________________________________________________________________

Rubi [A]  time = 0.0391909, antiderivative size = 20, 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, 32} \[ -\frac{1}{b d (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[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 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 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]

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{\operatorname{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]  time = 0.0443234, size = 32, normalized size = 1.6 \[ \frac{\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[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]  time = 0.043, size = 21, normalized size = 1.1 \begin{align*} -{\frac{1}{bd \left ( a+b\tan \left ( dx+c \right ) \right ) }} \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))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.12928, size = 27, normalized size = 1.35 \begin{align*} -\frac{1}{{\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))^2,x, algorithm="maxima")

[Out]

-1/((b*tan(d*x + c) + a)*b*d)

________________________________________________________________________________________

Fricas [B]  time = 1.76802, size = 132, normalized size = 6.6 \begin{align*} -\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 )} \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))^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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.33803, size = 27, normalized size = 1.35 \begin{align*} -\frac{1}{{\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))^2,x, algorithm="giac")

[Out]

-1/((b*tan(d*x + c) + a)*b*d)