\(\int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx\) [56]

Optimal result

Integrand size = 20, antiderivative size = 72 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {3 \cot (a+b x)}{8 b}-\frac {\cot ^3(a+b x)}{12 b}-\frac {\cot ^5(a+b x)}{80 b}+\frac {\tan (a+b x)}{4 b}+\frac {\tan ^3(a+b x)}{48 b} \]

[Out]

-3/8*cot(b*x+a)/b-1/12*cot(b*x+a)^3/b-1/80*cot(b*x+a)^5/b+1/4*tan(b*x+a)/b+1/48*tan(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2700, 276} \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {\tan ^3(a+b x)}{48 b}+\frac {\tan (a+b x)}{4 b}-\frac {\cot ^5(a+b x)}{80 b}-\frac {\cot ^3(a+b x)}{12 b}-\frac {3 \cot (a+b x)}{8 b} \]

[In]

Int[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^4,x]

[Out]

(-3*Cot[a + b*x])/(8*b) - Cot[a + b*x]^3/(12*b) - Cot[a + b*x]^5/(80*b) + Tan[a + b*x]/(4*b) + Tan[a + b*x]^3/
(48*b)

Rule 276

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]

Rule 2700

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 4373

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int \csc ^6(a+b x) \sec ^4(a+b x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^4}{x^6} \, dx,x,\tan (a+b x)\right )}{16 b} \\ & = \frac {\text {Subst}\left (\int \left (4+\frac {1}{x^6}+\frac {4}{x^4}+\frac {6}{x^2}+x^2\right ) \, dx,x,\tan (a+b x)\right )}{16 b} \\ & = -\frac {3 \cot (a+b x)}{8 b}-\frac {\cot ^3(a+b x)}{12 b}-\frac {\cot ^5(a+b x)}{80 b}+\frac {\tan (a+b x)}{4 b}+\frac {\tan ^3(a+b x)}{48 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {73 \cot (a+b x)}{240 b}-\frac {7 \cot (a+b x) \csc ^2(a+b x)}{120 b}-\frac {\cot (a+b x) \csc ^4(a+b x)}{80 b}+\frac {11 \tan (a+b x)}{48 b}+\frac {\sec ^2(a+b x) \tan (a+b x)}{48 b} \]

[In]

Integrate[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^4,x]

[Out]

(-73*Cot[a + b*x])/(240*b) - (7*Cot[a + b*x]*Csc[a + b*x]^2)/(120*b) - (Cot[a + b*x]*Csc[a + b*x]^4)/(80*b) +
(11*Tan[a + b*x])/(48*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(48*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {16 i \left (6 \,{\mathrm e}^{6 i \left (x b +a \right )}-2 \,{\mathrm e}^{4 i \left (x b +a \right )}-2 \,{\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{15 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{5}}\)	\(68\)
default	\(\frac {-\frac {1}{5 \sin \left (x b +a \right )^{5} \cos \left (x b +a \right )^{3}}+\frac {8}{15 \cos \left (x b +a \right )^{3} \sin \left (x b +a \right )^{3}}-\frac {16}{15 \cos \left (x b +a \right ) \sin \left (x b +a \right )^{3}}+\frac {64}{15 \sin \left (x b +a \right ) \cos \left (x b +a \right )}-\frac {128 \cot \left (x b +a \right )}{15}}{16 b}\)	\(87\)

[In]

int(csc(b*x+a)^2*csc(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)

[Out]

-16/15*I*(6*exp(6*I*(b*x+a))-2*exp(4*I*(b*x+a))-2*exp(2*I*(b*x+a))+1)/b/(exp(2*I*(b*x+a))+1)^3/(exp(2*I*(b*x+a
))-1)^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {128 \, \cos \left (b x + a\right )^{8} - 320 \, \cos \left (b x + a\right )^{6} + 240 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} - 5}{240 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )} \sin \left (b x + a\right )} \]

[In]

integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

-1/240*(128*cos(b*x + a)^8 - 320*cos(b*x + a)^6 + 240*cos(b*x + a)^4 - 40*cos(b*x + a)^2 - 5)/((b*cos(b*x + a)
^7 - 2*b*cos(b*x + a)^5 + b*cos(b*x + a)^3)*sin(b*x + a))

Sympy [F]

\[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\int \csc ^{2}{\left (a + b x \right )} \csc ^{4}{\left (2 a + 2 b x \right )}\, dx \]

[In]

integrate(csc(b*x+a)**2*csc(2*b*x+2*a)**4,x)

[Out]

Integral(csc(a + b*x)**2*csc(2*a + 2*b*x)**4, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1227 vs. \(2 (62) = 124\).

Time = 0.25 (sec) , antiderivative size = 1227, normalized size of antiderivative = 17.04 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\text {Too large to display} \]

[In]

integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

16/15*(2*(3*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(16*b*x + 16*a) - 4*(3*sin(6*b*x + 6*a)
- sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(14*b*x + 14*a) - 4*(3*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - sin(2*
b*x + 2*a))*cos(12*b*x + 12*a) + 12*(3*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(10*b*x + 10
*a) - (6*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*sin(16*b*x + 16*a) + 2*(6*cos(6*b*x +
6*a) - 2*cos(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*sin(14*b*x + 14*a) + 2*(6*cos(6*b*x + 6*a) - 2*cos(4*b*x
+ 4*a) - 2*cos(2*b*x + 2*a) + 1)*sin(12*b*x + 12*a) - 6*(6*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - 2*cos(2*b*x
+ 2*a) + 1)*sin(10*b*x + 10*a))/(b*cos(16*b*x + 16*a)^2 + 4*b*cos(14*b*x + 14*a)^2 + 4*b*cos(12*b*x + 12*a)^2
+ 36*b*cos(10*b*x + 10*a)^2 + 36*b*cos(6*b*x + 6*a)^2 + 4*b*cos(4*b*x + 4*a)^2 + 4*b*cos(2*b*x + 2*a)^2 + b*s
in(16*b*x + 16*a)^2 + 4*b*sin(14*b*x + 14*a)^2 + 4*b*sin(12*b*x + 12*a)^2 + 36*b*sin(10*b*x + 10*a)^2 + 36*b*s
in(6*b*x + 6*a)^2 + 4*b*sin(4*b*x + 4*a)^2 + 8*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b*sin(2*b*x + 2*a)^2 -
2*(2*b*cos(14*b*x + 14*a) + 2*b*cos(12*b*x + 12*a) - 6*b*cos(10*b*x + 10*a) + 6*b*cos(6*b*x + 6*a) - 2*b*cos(4
*b*x + 4*a) - 2*b*cos(2*b*x + 2*a) + b)*cos(16*b*x + 16*a) + 4*(2*b*cos(12*b*x + 12*a) - 6*b*cos(10*b*x + 10*a
) + 6*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - 2*b*cos(2*b*x + 2*a) + b)*cos(14*b*x + 14*a) - 4*(6*b*cos(10
*b*x + 10*a) - 6*b*cos(6*b*x + 6*a) + 2*b*cos(4*b*x + 4*a) + 2*b*cos(2*b*x + 2*a) - b)*cos(12*b*x + 12*a) - 12
*(6*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - 2*b*cos(2*b*x + 2*a) + b)*cos(10*b*x + 10*a) - 12*(2*b*cos(4*b
*x + 4*a) + 2*b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) + 4*(2*b*cos(2*b*x + 2*a) - b)*cos(4*b*x + 4*a) - 4*b*c
os(2*b*x + 2*a) - 4*(b*sin(14*b*x + 14*a) + b*sin(12*b*x + 12*a) - 3*b*sin(10*b*x + 10*a) + 3*b*sin(6*b*x + 6*
a) - b*sin(4*b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(16*b*x + 16*a) + 8*(b*sin(12*b*x + 12*a) - 3*b*sin(10*b*x +
10*a) + 3*b*sin(6*b*x + 6*a) - b*sin(4*b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(14*b*x + 14*a) - 8*(3*b*sin(10*b*x
+ 10*a) - 3*b*sin(6*b*x + 6*a) + b*sin(4*b*x + 4*a) + b*sin(2*b*x + 2*a))*sin(12*b*x + 12*a) - 24*(3*b*sin(6*
b*x + 6*a) - b*sin(4*b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - 24*(b*sin(4*b*x + 4*a) + b*sin(2*b*
x + 2*a))*sin(6*b*x + 6*a) + b)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {5 \, \tan \left (b x + a\right )^{3} - \frac {90 \, \tan \left (b x + a\right )^{4} + 20 \, \tan \left (b x + a\right )^{2} + 3}{\tan \left (b x + a\right )^{5}} + 60 \, \tan \left (b x + a\right )}{240 \, b} \]

[In]

integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

1/240*(5*tan(b*x + a)^3 - (90*tan(b*x + a)^4 + 20*tan(b*x + a)^2 + 3)/tan(b*x + a)^5 + 60*tan(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 20.87 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int \csc ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {-5\,{\mathrm {tan}\left (a+b\,x\right )}^8-60\,{\mathrm {tan}\left (a+b\,x\right )}^6+90\,{\mathrm {tan}\left (a+b\,x\right )}^4+20\,{\mathrm {tan}\left (a+b\,x\right )}^2+3}{240\,b\,{\mathrm {tan}\left (a+b\,x\right )}^5} \]

[In]

int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^4),x)

[Out]

-(20*tan(a + b*x)^2 + 90*tan(a + b*x)^4 - 60*tan(a + b*x)^6 - 5*tan(a + b*x)^8 + 3)/(240*b*tan(a + b*x)^5)