3.54 \(\int \csc ^2(a+b x) \csc ^2(2 a+2 b x) \, dx\)
Optimal. Leaf size=42 \[ \frac{\tan (a+b x)}{4 b}-\frac{\cot ^3(a+b x)}{12 b}-\frac{\cot (a+b x)}{2 b} \]
[Out]
-Cot[a + b*x]/(2*b) - Cot[a + b*x]^3/(12*b) + Tan[a + b*x]/(4*b)
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Rubi [A] time = 0.0590592, antiderivative size = 42, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used =
{4288, 2620, 270} \[ \frac{\tan (a+b x)}{4 b}-\frac{\cot ^3(a+b x)}{12 b}-\frac{\cot (a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^2,x]
[Out]
-Cot[a + b*x]/(2*b) - Cot[a + b*x]^3/(12*b) + Tan[a + b*x]/(4*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 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]
Rule 270
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Rubi steps
\begin{align*} \int \csc ^2(a+b x) \csc ^2(2 a+2 b x) \, dx &=\frac{1}{4} \int \csc ^4(a+b x) \sec ^2(a+b x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^4} \, dx,x,\tan (a+b x)\right )}{4 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}+\frac{2}{x^2}\right ) \, dx,x,\tan (a+b x)\right )}{4 b}\\ &=-\frac{\cot (a+b x)}{2 b}-\frac{\cot ^3(a+b x)}{12 b}+\frac{\tan (a+b x)}{4 b}\\ \end{align*}
Mathematica [A] time = 0.0795716, size = 48, normalized size = 1.14 \[ \frac{\tan (a+b x)}{4 b}-\frac{5 \cot (a+b x)}{12 b}-\frac{\cot (a+b x) \csc ^2(a+b x)}{12 b} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^2,x]
[Out]
(-5*Cot[a + b*x])/(12*b) - (Cot[a + b*x]*Csc[a + b*x]^2)/(12*b) + Tan[a + b*x]/(4*b)
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Maple [A] time = 0.032, size = 51, normalized size = 1.2 \begin{align*}{\frac{1}{4\,b} \left ( -{\frac{1}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}\cos \left ( bx+a \right ) }}+{\frac{4}{3\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }}-{\frac{8\,\cot \left ( bx+a \right ) }{3}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(b*x+a)^2*csc(2*b*x+2*a)^2,x)
[Out]
1/4/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)+4/3/sin(b*x+a)/cos(b*x+a)-8/3*cot(b*x+a))
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Maxima [B] time = 1.20777, size = 416, normalized size = 9.9 \begin{align*} \frac{4 \,{\left ({\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (8 \, b x + 8 \, a\right ) - 2 \,{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (6 \, b x + 6 \, a\right ) - 2 \, \cos \left (8 \, b x + 8 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \cos \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right )\right )}}{3 \,{\left (b \cos \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \cos \left (6 \, b x + 6 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \sin \left (6 \, b x + 6 \, a\right )^{2} - 8 \, b \sin \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \,{\left (2 \, b \cos \left (6 \, b x + 6 \, a\right ) - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (8 \, b x + 8 \, a\right ) - 4 \,{\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \cos \left (6 \, b x + 6 \, a\right ) - 4 \, b \cos \left (2 \, b x + 2 \, a\right ) - 4 \,{\left (b \sin \left (6 \, b x + 6 \, a\right ) - b \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (8 \, b x + 8 \, a\right ) + b\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^2,x, algorithm="maxima")
[Out]
4/3*((2*cos(2*b*x + 2*a) - 1)*sin(8*b*x + 8*a) - 2*(2*cos(2*b*x + 2*a) - 1)*sin(6*b*x + 6*a) - 2*cos(8*b*x + 8
*a)*sin(2*b*x + 2*a) + 4*cos(6*b*x + 6*a)*sin(2*b*x + 2*a))/(b*cos(8*b*x + 8*a)^2 + 4*b*cos(6*b*x + 6*a)^2 + 4
*b*cos(2*b*x + 2*a)^2 + b*sin(8*b*x + 8*a)^2 + 4*b*sin(6*b*x + 6*a)^2 - 8*b*sin(6*b*x + 6*a)*sin(2*b*x + 2*a)
+ 4*b*sin(2*b*x + 2*a)^2 - 2*(2*b*cos(6*b*x + 6*a) - 2*b*cos(2*b*x + 2*a) + b)*cos(8*b*x + 8*a) - 4*(2*b*cos(2
*b*x + 2*a) - b)*cos(6*b*x + 6*a) - 4*b*cos(2*b*x + 2*a) - 4*(b*sin(6*b*x + 6*a) - b*sin(2*b*x + 2*a))*sin(8*b
*x + 8*a) + b)
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Fricas [A] time = 0.466911, size = 136, normalized size = 3.24 \begin{align*} -\frac{8 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} + 3}{12 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^2,x, algorithm="fricas")
[Out]
-1/12*(8*cos(b*x + a)^4 - 12*cos(b*x + a)^2 + 3)/((b*cos(b*x + a)^3 - b*cos(b*x + a))*sin(b*x + a))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc ^{2}{\left (a + b x \right )} \csc ^{2}{\left (2 a + 2 b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)**2*csc(2*b*x+2*a)**2,x)
[Out]
Integral(csc(a + b*x)**2*csc(2*a + 2*b*x)**2, x)
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Giac [B] time = 1.54552, size = 1457, normalized size = 34.69 \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+a)^2*csc(2*b*x+2*a)^2,x, algorithm="giac")
[Out]
-1/24*(3*(tan(1/2*a)^12 + 6*tan(1/2*a)^10 + 15*tan(1/2*a)^8 + 20*tan(1/2*a)^6 + 15*tan(1/2*a)^4 + 6*tan(1/2*a)
^2 + 1)/((6*tan(b*x + 4*a)*tan(1/2*a)^5 - tan(1/2*a)^6 - 20*tan(b*x + 4*a)*tan(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)*(3*tan(1/2*a)^5 - 10*tan(1/2*a)^3 + 3*tan(1/2*a))) + 2*(6*tan
(b*x + 4*a)^2*tan(1/2*a)^36 - 216*tan(b*x + 4*a)^2*tan(1/2*a)^34 + 54*tan(b*x + 4*a)*tan(1/2*a)^35 + tan(1/2*a
)^36 + 3078*tan(b*x + 4*a)^2*tan(1/2*a)^32 - 1638*tan(b*x + 4*a)*tan(1/2*a)^33 + 126*tan(1/2*a)^34 - 23328*tan
(b*x + 4*a)^2*tan(1/2*a)^30 + 19872*tan(b*x + 4*a)*tan(1/2*a)^31 - 3159*tan(1/2*a)^32 + 62856*tan(b*x + 4*a)^2
*tan(1/2*a)^28 - 100224*tan(b*x + 4*a)*tan(1/2*a)^29 + 29232*tan(1/2*a)^30 + 328608*tan(b*x + 4*a)^2*tan(1/2*a
)^26 - 127176*tan(b*x + 4*a)*tan(1/2*a)^27 - 26460*tan(1/2*a)^28 - 96504*tan(b*x + 4*a)^2*tan(1/2*a)^24 + 8084
88*tan(b*x + 4*a)*tan(1/2*a)^25 - 228600*tan(1/2*a)^26 - 879840*tan(b*x + 4*a)^2*tan(1/2*a)^22 + 833472*tan(b*
x + 4*a)*tan(1/2*a)^23 + 237588*tan(1/2*a)^24 + 30564*tan(b*x + 4*a)^2*tan(1/2*a)^20 - 1653792*tan(b*x + 4*a)*
tan(1/2*a)^21 + 944208*tan(1/2*a)^22 + 1149552*tan(b*x + 4*a)^2*tan(1/2*a)^18 - 1673388*tan(b*x + 4*a)*tan(1/2
*a)^19 - 142434*tan(1/2*a)^20 + 30564*tan(b*x + 4*a)^2*tan(1/2*a)^16 + 1673388*tan(b*x + 4*a)*tan(1/2*a)^17 -
1358860*tan(1/2*a)^18 - 879840*tan(b*x + 4*a)^2*tan(1/2*a)^14 + 1653792*tan(b*x + 4*a)*tan(1/2*a)^15 - 142434*
tan(1/2*a)^16 - 96504*tan(b*x + 4*a)^2*tan(1/2*a)^12 - 833472*tan(b*x + 4*a)*tan(1/2*a)^13 + 944208*tan(1/2*a)
^14 + 328608*tan(b*x + 4*a)^2*tan(1/2*a)^10 - 808488*tan(b*x + 4*a)*tan(1/2*a)^11 + 237588*tan(1/2*a)^12 + 628
56*tan(b*x + 4*a)^2*tan(1/2*a)^8 + 127176*tan(b*x + 4*a)*tan(1/2*a)^9 - 228600*tan(1/2*a)^10 - 23328*tan(b*x +
4*a)^2*tan(1/2*a)^6 + 100224*tan(b*x + 4*a)*tan(1/2*a)^7 - 26460*tan(1/2*a)^8 + 3078*tan(b*x + 4*a)^2*tan(1/2
*a)^4 - 19872*tan(b*x + 4*a)*tan(1/2*a)^5 + 29232*tan(1/2*a)^6 - 216*tan(b*x + 4*a)^2*tan(1/2*a)^2 + 1638*tan(
b*x + 4*a)*tan(1/2*a)^3 - 3159*tan(1/2*a)^4 + 6*tan(b*x + 4*a)^2 - 54*tan(b*x + 4*a)*tan(1/2*a) + 126*tan(1/2*
a)^2 + 1)/((tan(1/2*a)^18 - 45*tan(1/2*a)^16 + 720*tan(1/2*a)^14 - 4728*tan(1/2*a)^12 + 10890*tan(1/2*a)^10 -
10890*tan(1/2*a)^8 + 4728*tan(1/2*a)^6 - 720*tan(1/2*a)^4 + 45*tan(1/2*a)^2 - 1)*(tan(b*x + 4*a)*tan(1/2*a)^6
- 15*tan(b*x + 4*a)*tan(1/2*a)^4 + 6*tan(1/2*a)^5 + 15*tan(b*x + 4*a)*tan(1/2*a)^2 - 20*tan(1/2*a)^3 - tan(b*x
+ 4*a) + 6*tan(1/2*a))^3))/b