3.146 \(\int \csc (c-b x) \csc (a+b x) \, dx\)

Optimal. Leaf size=33 \[ \frac{\csc (a+c) \log (\sin (a+b x))}{b}-\frac{\csc (a+c) \log (\sin (c-b x))}{b} \]

[Out]

-((Csc[a + c]*Log[Sin[c - b*x]])/b) + (Csc[a + c]*Log[Sin[a + b*x]])/b

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Rubi [A]  time = 0.0184097, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4611, 3475} \[ \frac{\csc (a+c) \log (\sin (a+b x))}{b}-\frac{\csc (a+c) \log (\sin (c-b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c - b*x]*Csc[a + b*x],x]

[Out]

-((Csc[a + c]*Log[Sin[c - b*x]])/b) + (Csc[a + c]*Log[Sin[a + b*x]])/b

Rule 4611

Int[Csc[(a_.) + (b_.)*(x_)]*Csc[(c_) + (d_.)*(x_)], x_Symbol] :> Dist[Csc[(b*c - a*d)/b], Int[Cot[a + b*x], x]
, x] - Dist[Csc[(b*c - a*d)/d], Int[Cot[c + d*x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2 - d^2, 0] && NeQ
[b*c - a*d, 0]

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 (c-b x) \csc (a+b x) \, dx &=\csc (a+c) \int \cot (c-b x) \, dx+\csc (a+c) \int \cot (a+b x) \, dx\\ &=-\frac{\csc (a+c) \log (\sin (c-b x))}{b}+\frac{\csc (a+c) \log (\sin (a+b x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.21866, size = 29, normalized size = 0.88 \[ -\frac{\csc (a+c) (\log (\sin (c-b x))-\log (-\sin (a+b x)))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c - b*x]*Csc[a + b*x],x]

[Out]

-((Csc[a + c]*(Log[Sin[c - b*x]] - Log[-Sin[a + b*x]]))/b)

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Maple [B]  time = 0.232, size = 81, normalized size = 2.5 \begin{align*}{\frac{\ln \left ( \tan \left ( bx+a \right ) \right ) }{b \left ( \sin \left ( a \right ) \cos \left ( c \right ) +\cos \left ( a \right ) \sin \left ( c \right ) \right ) }}-{\frac{\ln \left ( \tan \left ( bx+a \right ) \cos \left ( a \right ) \cos \left ( c \right ) -\tan \left ( bx+a \right ) \sin \left ( a \right ) \sin \left ( c \right ) -\cos \left ( a \right ) \sin \left ( c \right ) -\sin \left ( a \right ) \cos \left ( c \right ) \right ) }{b \left ( \sin \left ( a \right ) \cos \left ( c \right ) +\cos \left ( a \right ) \sin \left ( c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-csc(b*x-c)*csc(b*x+a),x)

[Out]

1/b/(sin(a)*cos(c)+cos(a)*sin(c))*ln(tan(b*x+a))-1/b/(sin(a)*cos(c)+cos(a)*sin(c))*ln(tan(b*x+a)*cos(a)*cos(c)
-tan(b*x+a)*sin(a)*sin(c)-cos(a)*sin(c)-sin(a)*cos(c))

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Maxima [B]  time = 1.11456, size = 724, normalized size = 21.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-csc(b*x-c)*csc(b*x+a),x, algorithm="maxima")

[Out]

-(2*(cos(2*a + 2*c)*cos(a + c) + sin(2*a + 2*c)*sin(a + c) - cos(a + c))*arctan2(sin(b*x) + sin(a), cos(b*x) -
cos(a)) + 2*(cos(2*a + 2*c)*cos(a + c) + sin(2*a + 2*c)*sin(a + c) - cos(a + c))*arctan2(sin(b*x) - sin(a), c
os(b*x) + cos(a)) - 2*(cos(2*a + 2*c)*cos(a + c) + sin(2*a + 2*c)*sin(a + c) - cos(a + c))*arctan2(sin(b*x) +
sin(c), cos(b*x) + cos(c)) - 2*(cos(2*a + 2*c)*cos(a + c) + sin(2*a + 2*c)*sin(a + c) - cos(a + c))*arctan2(si
n(b*x) - sin(c), cos(b*x) - cos(c)) - (cos(a + c)*sin(2*a + 2*c) - cos(2*a + 2*c)*sin(a + c) + sin(a + c))*log
(cos(b*x)^2 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - (cos(a + c)*sin(2*a
+ 2*c) - cos(2*a + 2*c)*sin(a + c) + sin(a + c))*log(cos(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 +
2*sin(b*x)*sin(a) + sin(a)^2) + (cos(a + c)*sin(2*a + 2*c) - cos(2*a + 2*c)*sin(a + c) + sin(a + c))*log(cos(b
*x)^2 + 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(c) + sin(c)^2) + (cos(a + c)*sin(2*a + 2*c)
- cos(2*a + 2*c)*sin(a + c) + sin(a + c))*log(cos(b*x)^2 - 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 - 2*sin(
b*x)*sin(c) + sin(c)^2))/(b*cos(2*a + 2*c)^2 + b*sin(2*a + 2*c)^2 - 2*b*cos(2*a + 2*c) + b)

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Fricas [B]  time = 2.66961, size = 270, normalized size = 8.18 \begin{align*} \frac{\log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - \log \left (-\frac{2 \, \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) +{\left (2 \, \cos \left (a + c\right )^{2} - 1\right )} \cos \left (b x + a\right )^{2} - \cos \left (a + c\right )^{2}}{\cos \left (a + c\right )^{2} + 2 \, \cos \left (a + c\right ) + 1}\right )}{2 \, b \sin \left (a + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-csc(b*x-c)*csc(b*x+a),x, algorithm="fricas")

[Out]

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

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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(b*x-c)*csc(b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.24936, size = 536, normalized size = 16.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-csc(b*x-c)*csc(b*x+a),x, algorithm="giac")

[Out]

1/2*((tan(1/2*a)^4*tan(1/2*c)^4 - 4*tan(1/2*a)^3*tan(1/2*c)^3 - tan(1/2*a)^4 - 4*tan(1/2*a)^3*tan(1/2*c) - 4*t
an(1/2*a)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*a)*tan(1/2*c) + 1)*log(abs(tan(b*x + a)*tan(1/2*a)^2*tan(1/2
*c)^2 - tan(b*x + a)*tan(1/2*a)^2 - 4*tan(b*x + a)*tan(1/2*a)*tan(1/2*c) + 2*tan(1/2*a)^2*tan(1/2*c) - tan(b*x
+ a)*tan(1/2*c)^2 + 2*tan(1/2*a)*tan(1/2*c)^2 + tan(b*x + a) - 2*tan(1/2*a) - 2*tan(1/2*c)))/(tan(1/2*a)^4*ta
n(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c)^4 - tan(1/2*a)^4*tan(1/2*c) - 6*tan(1/2*a)^3*tan(1/2*c)^2 - 6*tan(1/2*a)^
2*tan(1/2*c)^3 - tan(1/2*a)*tan(1/2*c)^4 + tan(1/2*a)^3 + 6*tan(1/2*a)^2*tan(1/2*c) + 6*tan(1/2*a)*tan(1/2*c)^
2 + tan(1/2*c)^3 - tan(1/2*a) - tan(1/2*c)) - (tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*lo
g(abs(tan(b*x + a)))/(tan(1/2*a)^2*tan(1/2*c) + tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*a) - tan(1/2*c)))/b