\(\int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [569]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \tan (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, 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} \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \tan (c+d x))^2} \]

[In]

Int[Sec[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-1/2*1/(b*d*(a + b*Tan[c + d*x])^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]

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)^3} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = -\frac {1}{2 b d (a+b \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \tan (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-1/2*1/(b*d*(a + b*Tan[c + d*x])^2)

Maple [A] (verified)

Time = 7.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {1}{2 b d \left (a +b \tan \left (d x +c \right )\right )^{2}}\)	\(21\)
default	\(-\frac {1}{2 b d \left (a +b \tan \left (d x +c \right )\right )^{2}}\)	\(21\)
risch	\(\frac {2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d \left (i a +b \right )^{2}}\)	\(77\)

[In]

int(sec(d*x+c)^2/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 6.45 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {4 \, a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b + b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} d\right )}} \]

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(4*a^2*b*cos(d*x + c)^2 - a^2*b + b^3 - 2*(a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^6 + a^4*b^2 - a^2*
b^4 - b^6)*d*cos(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d*cos(d*x + c)*sin(d*x + c) + (a^4*b^2 + 2*a^2*b^4
+ b^6)*d)

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(sec(d*x+c)**2/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**2/(a + b*tan(c + d*x))**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{2 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b d} \]

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2/((b*tan(d*x + c) + a)^2*b*d)

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{2 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b d} \]

[In]

integrate(sec(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2/((b*tan(d*x + c) + a)^2*b*d)

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {1}{d\,\left (2\,a^2\,b+4\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )+2\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )} \]

[In]

int(1/(cos(c + d*x)^2*(a + b*tan(c + d*x))^3),x)

[Out]

-1/(d*(2*a^2*b + 2*b^3*tan(c + d*x)^2 + 4*a*b^2*tan(c + d*x)))