3.141 \(\int \csc ^2(a+b x) \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=22 \[ \frac {\tan (a+b x)}{b}-\frac {\cot (a+b x)}{b} \]

[Out]

-cot(b*x+a)/b+tan(b*x+a)/b

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2620, 14} \[ \frac {\tan (a+b x)}{b}-\frac {\cot (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sec[a + b*x]^2,x]

[Out]

-(Cot[a + b*x]/b) + Tan[a + b*x]/b

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rubi steps

\begin {align*} \int \csc ^2(a+b x) \sec ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=-\frac {\cot (a+b x)}{b}+\frac {\tan (a+b x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 13, normalized size = 0.59 \[ -\frac {2 \cot (2 (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sec[a + b*x]^2,x]

[Out]

(-2*Cot[2*(a + b*x)])/b

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 33, normalized size = 1.50 \[ -\frac {2 \, \cos \left (b x + a\right )^{2} - 1}{b \cos \left (b x + a\right ) \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.70, size = 16, normalized size = 0.73 \[ -\frac {2}{b \tan \left (2 \, b x + 2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*tan(2*b*x + 2*a))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 31, normalized size = 1.41 \[ \frac {\frac {1}{\sin \left (b x +a \right ) \cos \left (b x +a \right )}-2 \cot \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2/sin(b*x+a)^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 22, normalized size = 1.00 \[ -\frac {\frac {1}{\tan \left (b x + a\right )} - \tan \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/tan(b*x + a) - tan(b*x + a))/b

________________________________________________________________________________________

mupad [B]  time = 0.39, size = 14, normalized size = 0.64 \[ -\frac {2\,\mathrm {cot}\left (2\,a+2\,b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*cot(2*a + 2*b*x))/b

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________