3.196 \(\int \csc ^2(c+b x) \sin (a+b x) \, dx\)
Optimal. Leaf size=36 \[ -\frac{\cos (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a-c) \csc (b x+c)}{b} \]
[Out]
-((ArcTanh[Cos[c + b*x]]*Cos[a - c])/b) - (Csc[c + b*x]*Sin[a - c])/b
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Rubi [A] time = 0.0327012, antiderivative size = 36, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used =
{4582, 2606, 8, 3770} \[ -\frac{\cos (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a-c) \csc (b x+c)}{b} \]
Antiderivative was successfully verified.
[In]
Int[Csc[c + b*x]^2*Sin[a + b*x],x]
[Out]
-((ArcTanh[Cos[c + b*x]]*Cos[a - c])/b) - (Csc[c + b*x]*Sin[a - c])/b
Rule 4582
Int[Csc[w_]^(n_.)*Sin[v_], x_Symbol] :> Dist[Sin[v - w], Int[Cot[w]*Csc[w]^(n - 1), x], x] + Dist[Cos[v - w],
Int[Csc[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]
Rule 2606
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \csc ^2(c+b x) \sin (a+b x) \, dx &=\cos (a-c) \int \csc (c+b x) \, dx+\sin (a-c) \int \cot (c+b x) \csc (c+b x) \, dx\\ &=-\frac{\tanh ^{-1}(\cos (c+b x)) \cos (a-c)}{b}-\frac{\sin (a-c) \operatorname{Subst}(\int 1 \, dx,x,\csc (c+b x))}{b}\\ &=-\frac{\tanh ^{-1}(\cos (c+b x)) \cos (a-c)}{b}-\frac{\csc (c+b x) \sin (a-c)}{b}\\ \end{align*}
Mathematica [C] time = 0.0968568, size = 90, normalized size = 2.5 \[ -\frac{\sin (a-c) \csc (b x+c)}{b}-\frac{2 i \cos (a-c) \tan ^{-1}\left (\frac{(\cos (c)-i \sin (c)) \left (\cos (c) \cos \left (\frac{b x}{2}\right )-\sin (c) \sin \left (\frac{b x}{2}\right )\right )}{\sin (c) \cos \left (\frac{b x}{2}\right )+i \cos (c) \cos \left (\frac{b x}{2}\right )}\right )}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[c + b*x]^2*Sin[a + b*x],x]
[Out]
((-2*I)*ArcTan[((Cos[c] - I*Sin[c])*(Cos[c]*Cos[(b*x)/2] - Sin[c]*Sin[(b*x)/2]))/(I*Cos[c]*Cos[(b*x)/2] + Cos[
(b*x)/2]*Sin[c])]*Cos[a - c])/b - (Csc[c + b*x]*Sin[a - c])/b
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Maple [B] time = 0.446, size = 890, normalized size = 24.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(b*x+c)^2*sin(b*x+a),x)
[Out]
-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(-cos(a)*sin(c)*tan(1/
2*b*x+1/2*a)^2+cos(c)*sin(a)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*cos(c)+2*tan(1/2*b*x+1/2*a)*sin(
a)*sin(c)+cos(a)*sin(c)-sin(a)*cos(c))*tan(1/2*b*x+1/2*a)*cos(a)*cos(c)-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*s
in(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(-cos(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+cos(c)*sin(a)*tan(1/2*b*
x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*cos(c)+2*tan(1/2*b*x+1/2*a)*sin(a)*sin(c)+cos(a)*sin(c)-sin(a)*cos(c))*
tan(1/2*b*x+1/2*a)*sin(a)*sin(c)-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*
sin(c)^2)/(-cos(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+cos(c)*sin(a)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*
cos(c)+2*tan(1/2*b*x+1/2*a)*sin(a)*sin(c)+cos(a)*sin(c)-sin(a)*cos(c))*cos(a)*sin(c)+8/b/(-4*cos(a)^2*cos(c)^2
-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(-cos(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+cos(c)*sin(
a)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*cos(c)+2*tan(1/2*b*x+1/2*a)*sin(a)*sin(c)+cos(a)*sin(c)-si
n(a)*cos(c))*sin(a)*cos(c)-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)
^2)/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2)*arctan(1/2*(2*(sin(a)*cos
(c)-cos(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*cos(c)+2*sin(a)*sin(c))/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-c
os(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*cos(a)*cos(c)-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c
)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(
1/2)*arctan(1/2*(2*(sin(a)*cos(c)-cos(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*cos(c)+2*sin(a)*sin(c))/(-cos(a)^
2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*sin(a)*sin(c)
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Maxima [B] time = 1.28253, size = 613, normalized size = 17.03 \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)^2*sin(b*x+a),x, algorithm="maxima")
[Out]
-1/2*(2*(cos(b*x + 2*a) - cos(b*x + 2*c))*cos(2*b*x + a + 2*c) - 2*cos(b*x + 2*a)*cos(a) + 2*cos(b*x + 2*c)*co
s(a) + (cos(2*b*x + a + 2*c)^2*cos(-a + c) - 2*cos(2*b*x + a + 2*c)*cos(a)*cos(-a + c) + cos(-a + c)*sin(2*b*x
+ a + 2*c)^2 - 2*cos(-a + c)*sin(2*b*x + a + 2*c)*sin(a) + (cos(a)^2 + sin(a)^2)*cos(-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(2*b*x + a + 2*c)^2*cos(-a +
c) - 2*cos(2*b*x + a + 2*c)*cos(a)*cos(-a + c) + cos(-a + c)*sin(2*b*x + a + 2*c)^2 - 2*cos(-a + c)*sin(2*b*x
+ a + 2*c)*sin(a) + (cos(a)^2 + sin(a)^2)*cos(-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) + 2*(sin(b*x + 2*a) - sin(b*x + 2*c))*sin(2*b*x + a + 2*c) - 2*sin(b*x +
2*a)*sin(a) + 2*sin(b*x + 2*c)*sin(a))/(b*cos(2*b*x + a + 2*c)^2 - 2*b*cos(2*b*x + a + 2*c)*cos(a) + b*sin(2*b
*x + a + 2*c)^2 - 2*b*sin(2*b*x + a + 2*c)*sin(a) + (cos(a)^2 + sin(a)^2)*b)
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Fricas [A] time = 0.502203, size = 203, normalized size = 5.64 \begin{align*} -\frac{\cos \left (-a + c\right ) \log \left (\frac{1}{2} \, \cos \left (b x + c\right ) + \frac{1}{2}\right ) \sin \left (b x + c\right ) - \cos \left (-a + c\right ) \log \left (-\frac{1}{2} \, \cos \left (b x + c\right ) + \frac{1}{2}\right ) \sin \left (b x + c\right ) - 2 \, \sin \left (-a + c\right )}{2 \, b \sin \left (b x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+c)^2*sin(b*x+a),x, algorithm="fricas")
[Out]
-1/2*(cos(-a + c)*log(1/2*cos(b*x + c) + 1/2)*sin(b*x + c) - cos(-a + c)*log(-1/2*cos(b*x + c) + 1/2)*sin(b*x
+ c) - 2*sin(-a + c))/(b*sin(b*x + 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)**2*sin(b*x+a),x)
[Out]
Timed out
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Giac [B] time = 1.13628, size = 471, normalized size = 13.08 \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)^2*sin(b*x+a),x, algorithm="giac")
[Out]
((tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)*log(abs(tan(1/2*b*x +
1/2*c)))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) - (tan(1/2*b*x + 1/2*c)*tan(1/2*a)^2*t
an(1/2*c) - tan(1/2*b*x + 1/2*c)*tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*b*x + 1/2*c)*tan(1/2*a) - tan(1/2*b*x + 1/2
*c)*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) - (tan(1/2*b*x + 1/2*c)*tan(1/2*
a)^2*tan(1/2*c)^2 - tan(1/2*b*x + 1/2*c)*tan(1/2*a)^2 + 4*tan(1/2*b*x + 1/2*c)*tan(1/2*a)*tan(1/2*c) + tan(1/2
*a)^2*tan(1/2*c) - tan(1/2*b*x + 1/2*c)*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*b*x + 1/2*c) + 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)*tan(1/2*b*x + 1/2*c)))/b