3.18 \(\int \csc (2 a+2 b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=14 \[ -\frac{\log (\cos (a+b x))}{2 b} \]

[Out]

-Log[Cos[a + b*x]]/(2*b)

________________________________________________________________________________________

Rubi [A]  time = 0.026394, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4288, 3475} \[ -\frac{\log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[2*a + 2*b*x]*Sin[a + b*x]^2,x]

[Out]

-Log[Cos[a + b*x]]/(2*b)

Rule 4288

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \csc (2 a+2 b x) \sin ^2(a+b x) \, dx &=\frac{1}{2} \int \tan (a+b x) \, dx\\ &=-\frac{\log (\cos (a+b x))}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.011125, size = 14, normalized size = 1. \[ -\frac{\log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[Cos[a + b*x]]/(2*b)

________________________________________________________________________________________

Maple [A]  time = 0.022, size = 13, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( \cos \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.12982, size = 74, normalized size = 5.29 \begin{align*} -\frac{\log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2
)/b

________________________________________________________________________________________

Fricas [A]  time = 0.497735, size = 36, normalized size = 2.57 \begin{align*} -\frac{\log \left (-\cos \left (b x + a\right )\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-cos(b*x + a))/b

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*b*x+2*a)*sin(b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.50726, size = 128, normalized size = 9.14 \begin{align*} \frac{\log \left (\tan \left (b x + 4 \, a\right )^{2} + 1\right ) - 2 \, \log \left ({\left | 6 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{5} - \tan \left (\frac{1}{2} \, a\right )^{6} - 20 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{3} + 15 \, \tan \left (\frac{1}{2} \, a\right )^{4} + 6 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right ) - 15 \, \tan \left (\frac{1}{2} \, a\right )^{2} + 1 \right |}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(tan(b*x + 4*a)^2 + 1) - 2*log(abs(6*tan(b*x + 4*a)*tan(1/2*a)^5 - tan(1/2*a)^6 - 20*tan(b*x + 4*a)*ta
n(1/2*a)^3 + 15*tan(1/2*a)^4 + 6*tan(b*x + 4*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)))/b