\(\int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx\) [948]

Optimal result

Integrand size = 39, antiderivative size = 17 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\cos (a+b x) \sin (a+b x)}{b} \]

[Out]

-cos(b*x+a)*sin(b*x+a)/b

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {391} \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\sin (a+b x) \cos (a+b x)}{b} \]

[In]

Int[(-Csc[a + b*x]^2 + Sec[a + b*x]^2)/(Csc[a + b*x]^2 + Sec[a + b*x]^2),x]

[Out]

-((Cos[a + b*x]*Sin[a + b*x])/b)

Rule 391

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*x*((a + b*x^n)^(p + 1)/a), x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a*d - b*c*(n*(p + 1) + 1), 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-1+x^2}{\left (1+x^2\right )^2} \, dx,x,\tan (a+b x)\right )}{b} \\ & = -\frac {\cos (a+b x) \sin (a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.94 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\cos (2 b x) \sin (2 a)}{2 b}-\frac {\cos (2 a) \sin (2 b x)}{2 b} \]

[In]

Integrate[(-Csc[a + b*x]^2 + Sec[a + b*x]^2)/(Csc[a + b*x]^2 + Sec[a + b*x]^2),x]

[Out]

-1/2*(Cos[2*b*x]*Sin[2*a])/b - (Cos[2*a]*Sin[2*b*x])/(2*b)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\sin \left (2 x b +2 a \right )}{2 b}\)	\(15\)
parallelrisch	\(-\frac {\sin \left (2 x b +2 a \right )}{2 b}\)	\(15\)
derivativedivides	\(-\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{b}\)	\(18\)
default	\(-\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{b}\)	\(18\)
norman	\(\frac {\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b}-\frac {4 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{b}+\frac {2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{10}}{b}}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{4}}\)	\(92\)

[In]

int((-csc(b*x+a)^2+sec(b*x+a)^2)/(csc(b*x+a)^2+sec(b*x+a)^2),x,method=_RETURNVERBOSE)

[Out]

-1/2*sin(2*b*x+2*a)/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\cos \left (b x + a\right ) \sin \left (b x + a\right )}{b} \]

[In]

integrate((-csc(b*x+a)^2+sec(b*x+a)^2)/(csc(b*x+a)^2+sec(b*x+a)^2),x, algorithm="fricas")

[Out]

-cos(b*x + a)*sin(b*x + a)/b

Sympy [F]

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

[In]

integrate((-csc(b*x+a)**2+sec(b*x+a)**2)/(csc(b*x+a)**2+sec(b*x+a)**2),x)

[Out]

-Integral(csc(a + b*x)**2/(csc(a + b*x)**2 + sec(a + b*x)**2), x) - Integral(-sec(a + b*x)**2/(csc(a + b*x)**2
+ sec(a + b*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )} b} \]

[In]

integrate((-csc(b*x+a)^2+sec(b*x+a)^2)/(csc(b*x+a)^2+sec(b*x+a)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\sin \left (2 \, b x + 2 \, a\right )}{2 \, b} \]

[In]

integrate((-csc(b*x+a)^2+sec(b*x+a)^2)/(csc(b*x+a)^2+sec(b*x+a)^2),x, algorithm="giac")

[Out]

-1/2*sin(2*b*x + 2*a)/b

Mupad [B] (verification not implemented)

Time = 26.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-\csc ^2(a+b x)+\sec ^2(a+b x)}{\csc ^2(a+b x)+\sec ^2(a+b x)} \, dx=-\frac {\sin \left (2\,a+2\,b\,x\right )}{2\,b} \]

[In]

int((1/cos(a + b*x)^2 - 1/sin(a + b*x)^2)/(1/cos(a + b*x)^2 + 1/sin(a + b*x)^2),x)

[Out]

-sin(2*a + 2*b*x)/(2*b)