Integrand size = 18, antiderivative size = 62 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {5 \text {arctanh}(\sin (a+b x))}{32 b}-\frac {\csc (a+b x)}{8 b}-\frac {\csc ^3(a+b x)}{48 b}+\frac {\sec (a+b x) \tan (a+b x)}{32 b} \] Output:
5/32*arctanh(sin(b*x+a))/b-1/8*csc(b*x+a)/b-1/48*csc(b*x+a)^3/b+1/32*sec(b *x+a)*tan(b*x+a)/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.50 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\sin ^2(a+b x)\right )}{48 b} \] Input:
Integrate[Cos[a + b*x]*Csc[2*a + 2*b*x]^4,x]
Output:
-1/48*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, Sin[a + b*x]^2])/b
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4775, 3042, 3101, 252, 254, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^4}dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle \frac {1}{16} \int \csc ^4(a+b x) \sec ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{16} \int \csc (a+b x)^4 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{16 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\csc ^4(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)}{16 b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )}{16 b}\) |
Input:
Int[Cos[a + b*x]*Csc[2*a + 2*b*x]^4,x]
Output:
-1/16*(Csc[a + b*x]^5/(2*(1 - Csc[a + b*x]^2)) - (5*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x] - Csc[a + b*x]^3/3))/2)/b
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/e^p Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 2.65 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {5}{6 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {5}{2 \sin \left (b x +a \right )}+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{16 b}\) | \(69\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}-20 \,{\mathrm e}^{7 i \left (b x +a \right )}-22 \,{\mathrm e}^{5 i \left (b x +a \right )}-20 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{48 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{32 b}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{32 b}\) | \(126\) |
Input:
int(cos(b*x+a)*csc(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)
Output:
1/16/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)^2+5/6/sin(b*x+a)/cos(b*x+a)^2-5/2/sin (b*x+a)+5/2*ln(sec(b*x+a)+tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.10 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {30 \, \cos \left (b x + a\right )^{4} - 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{192 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:
integrate(cos(b*x+a)*csc(2*b*x+2*a)^4,x, algorithm="fricas")
Output:
-1/192*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b *x + a) + 1)*sin(b*x + a) + 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(-sin( b*x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 - b *cos(b*x + a)^2)*sin(b*x + a))
Timed out. \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)*csc(2*b*x+2*a)**4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1780 vs. \(2 (54) = 108\).
Time = 0.20 (sec) , antiderivative size = 1780, normalized size of antiderivative = 28.71 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)*csc(2*b*x+2*a)^4,x, algorithm="maxima")
Output:
1/192*(4*(15*sin(9*b*x + 9*a) - 20*sin(7*b*x + 7*a) - 22*sin(5*b*x + 5*a) - 20*sin(3*b*x + 3*a) + 15*sin(b*x + a))*cos(10*b*x + 10*a) + 60*(sin(8*b* x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos (9*b*x + 9*a) + 4*(20*sin(7*b*x + 7*a) + 22*sin(5*b*x + 5*a) + 20*sin(3*b* x + 3*a) - 15*sin(b*x + a))*cos(8*b*x + 8*a) - 80*(2*sin(6*b*x + 6*a) - 2* sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(7*b*x + 7*a) + 8*(22*sin(5*b*x + 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(6*b*x + 6*a) + 88*(2*sin (4*b*x + 4*a) + sin(2*b*x + 2*a))*cos(5*b*x + 5*a) - 40*(4*sin(3*b*x + 3*a ) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 15*(2*(cos(8*b*x + 8*a) + 2*cos(6*b *x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(10*b*x + 10*a) - cos(10*b*x + 10*a)^2 - 2*(2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos( 2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 4*(2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 4*cos(6*b*x + 6*a)^2 - 4* (cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - 4*cos(4*b*x + 4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - sin(10*b*x + 10*a)^2 - 2*(2*sin(6 *b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - si n(8*b*x + 8*a)^2 + 4*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(6*b*x + 6 *a) - 4*sin(6*b*x + 6*a)^2 - 4*sin(4*b*x + 4*a)^2 - 4*sin(4*b*x + 4*a)*sin (2*b*x + 2*a) - sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) - 1)*log((cos(b...
Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {\frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac {4 \, {\left (6 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{192 \, b} \] Input:
integrate(cos(b*x+a)*csc(2*b*x+2*a)^4,x, algorithm="giac")
Output:
-1/192*(6*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(6*sin(b*x + a)^2 + 1)/sin (b*x + a)^3 - 15*log(sin(b*x + a) + 1) + 15*log(-sin(b*x + a) + 1))/b
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{32\,b}-\frac {-\frac {5\,{\sin \left (a+b\,x\right )}^4}{32}+\frac {5\,{\sin \left (a+b\,x\right )}^2}{48}+\frac {1}{48}}{b\,\left ({\sin \left (a+b\,x\right )}^3-{\sin \left (a+b\,x\right )}^5\right )} \] Input:
int(cos(a + b*x)/sin(2*a + 2*b*x)^4,x)
Output:
(5*atanh(sin(a + b*x)))/(32*b) - ((5*sin(a + b*x)^2)/48 - (5*sin(a + b*x)^ 4)/32 + 1/48)/(b*(sin(a + b*x)^3 - sin(a + b*x)^5))
\[ \int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {-88 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right )+85 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2}+10 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-40 \cos \left (2 b x +2 a \right )+20 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}-40 \cos \left (b x +a \right )-90 \left (\int \frac {\tan \left (b x +a \right )^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) \sin \left (2 b x +2 a \right )^{3} b -60 \left (\int \frac {1}{\tan \left (b x +a \right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\tan \left (b x +a \right )^{4}}d x \right ) \sin \left (2 b x +2 a \right )^{3} b +75 \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )+90 \sin \left (2 b x +2 a \right )^{3} \tan \left (b x +a \right )-15 \sin \left (2 b x +2 a \right )^{3} b x +15 \sin \left (2 b x +2 a \right )^{2}+22 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-40}{288 \sin \left (2 b x +2 a \right )^{3} b} \] Input:
int(cos(b*x+a)*csc(2*b*x+2*a)^4,x)
Output:
( - 88*cos(2*a + 2*b*x)*cos(a + b*x) + 85*cos(2*a + 2*b*x)*sin(2*a + 2*b*x )**2 + 10*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)*sin(a + b*x) - 40*cos(2*a + 2* b*x) + 20*cos(a + b*x)*sin(2*a + 2*b*x)**2 - 40*cos(a + b*x) - 90*int((tan (a + b*x)**2*tan((a + b*x)/2)**2)/(tan((a + b*x)/2)**2 + 1),x)*sin(2*a + 2 *b*x)**3*b - 60*int(1/(tan(a + b*x)**4*tan((a + b*x)/2)**2 + tan(a + b*x)* *4),x)*sin(2*a + 2*b*x)**3*b + 75*sin(2*a + 2*b*x)**3*sin(a + b*x) + 90*si n(2*a + 2*b*x)**3*tan(a + b*x) - 15*sin(2*a + 2*b*x)**3*b*x + 15*sin(2*a + 2*b*x)**2 + 22*sin(2*a + 2*b*x)*sin(a + b*x) - 40)/(288*sin(2*a + 2*b*x)* *3*b)