Integrand size = 15, antiderivative size = 67 \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=-\frac {\text {arctanh}(\cos (c+b x)) \cos (a-c)}{2 b}-\frac {\cos (a-c) \cot (c+b x) \csc (c+b x)}{2 b}-\frac {\csc ^3(c+b x) \sin (a-c)}{3 b} \] Output:
-1/2*arctanh(cos(b*x+c))*cos(a-c)/b-1/2*cos(a-c)*cot(b*x+c)*csc(b*x+c)/b-1 /3*csc(b*x+c)^3*sin(a-c)/b
Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=-\frac {6 \text {arctanh}\left (\cos (c)-\sin (c) \tan \left (\frac {b x}{2}\right )\right ) \cos (a-c)+3 \cos (a-c) \cot (c+b x) \csc (c+b x)+2 \csc ^3(c+b x) \sin (a-c)}{6 b} \] Input:
Integrate[Csc[c + b*x]^4*Sin[a + b*x],x]
Output:
-1/6*(6*ArcTanh[Cos[c] - Sin[c]*Tan[(b*x)/2]]*Cos[a - c] + 3*Cos[a - c]*Co t[c + b*x]*Csc[c + b*x] + 2*Csc[c + b*x]^3*Sin[a - c])/b
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5093, 3042, 25, 3086, 15, 4255, 3042, 4257}
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.
\(\displaystyle \int \sin (a+b x) \csc ^4(b x+c) \, dx\) |
\(\Big \downarrow \) 5093 |
\(\displaystyle \cos (a-c) \int \csc ^3(c+b x)dx+\sin (a-c) \int \cot (c+b x) \csc ^3(c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^3dx+\sin (a-c) \int -\sec \left (c+b x-\frac {\pi }{2}\right )^3 \tan \left (c+b x-\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^3dx-\sin (a-c) \int \sec \left (\frac {1}{2} (2 c-\pi )+b x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+b x\right )dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^3dx-\frac {\sin (a-c) \int \csc ^2(c+b x)d\csc (c+b x)}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^3dx-\frac {\sin (a-c) \csc ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \cos (a-c) \left (\frac {1}{2} \int \csc (c+b x)dx-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\sin (a-c) \csc ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \left (\frac {1}{2} \int \csc (c+b x)dx-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\sin (a-c) \csc ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \cos (a-c) \left (-\frac {\text {arctanh}(\cos (b x+c))}{2 b}-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\sin (a-c) \csc ^3(b x+c)}{3 b}\) |
Input:
Int[Csc[c + b*x]^4*Sin[a + b*x],x]
Output:
Cos[a - c]*(-1/2*ArcTanh[Cos[c + b*x]]/b - (Cot[c + b*x]*Csc[c + b*x])/(2* b)) - (Csc[c + b*x]^3*Sin[a - c])/(3*b)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[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])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[Csc[w_]^(n_.)*Sin[v_], x_Symbol] :> Simp[Sin[v - w] Int[Cot[w]*Csc[w] ^(n - 1), x], x] + Simp[Cos[v - w] Int[Csc[w]^(n - 1), x], x] /; GtQ[n, 0 ] && FreeQ[v - w, x] && NeQ[w, v]
Result contains complex when optimal does not.
Time = 4.41 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.82
method | result | size |
risch | \(\frac {-3 \,{\mathrm e}^{i \left (5 b x +7 a +4 c \right )}-3 \,{\mathrm e}^{i \left (5 b x +5 a +6 c \right )}-8 \,{\mathrm e}^{i \left (3 b x +7 a +2 c \right )}+8 \,{\mathrm e}^{i \left (3 b x +5 a +4 c \right )}+3 \,{\mathrm e}^{i \left (b x +7 a \right )}+3 \,{\mathrm e}^{i \left (b x +5 a +2 c \right )}}{6 b \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}\) | \(189\) |
default | \(\text {Expression too large to display}\) | \(1779\) |
Input:
int(csc(b*x+c)^4*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/6/b/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^3*(-3*exp(I*(5*b*x+7*a+4*c))-3*exp( I*(5*b*x+5*a+6*c))-8*exp(I*(3*b*x+7*a+2*c))+8*exp(I*(3*b*x+5*a+4*c))+3*exp (I*(b*x+7*a))+3*exp(I*(b*x+5*a+2*c)))-1/2*ln(exp(I*(b*x+a))+exp(I*(a-c)))/ b*cos(a-c)+1/2*ln(exp(I*(b*x+a))-exp(I*(a-c)))/b*cos(a-c)
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.10 \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\frac {6 \, \cos \left (b x + c\right ) \cos \left (-a + c\right ) \sin \left (b x + c\right ) - 3 \, {\left (\cos \left (b x + c\right )^{2} \cos \left (-a + c\right ) - \cos \left (-a + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) + 3 \, {\left (\cos \left (b x + c\right )^{2} \cos \left (-a + c\right ) - \cos \left (-a + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) - 4 \, \sin \left (-a + c\right )}{12 \, {\left (b \cos \left (b x + c\right )^{2} - b\right )} \sin \left (b x + c\right )} \] Input:
integrate(csc(b*x+c)^4*sin(b*x+a),x, algorithm="fricas")
Output:
1/12*(6*cos(b*x + c)*cos(-a + c)*sin(b*x + c) - 3*(cos(b*x + c)^2*cos(-a + c) - cos(-a + c))*log(1/2*cos(b*x + c) + 1/2)*sin(b*x + c) + 3*(cos(b*x + c)^2*cos(-a + c) - cos(-a + c))*log(-1/2*cos(b*x + c) + 1/2)*sin(b*x + c) - 4*sin(-a + c))/((b*cos(b*x + c)^2 - b)*sin(b*x + c))
Timed out. \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+c)**4*sin(b*x+a),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1773 vs. \(2 (61) = 122\).
Time = 0.09 (sec) , antiderivative size = 1773, normalized size of antiderivative = 26.46 \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+c)^4*sin(b*x+a),x, algorithm="maxima")
Output:
1/12*(2*(3*cos(5*b*x + 2*a + 4*c) + 3*cos(5*b*x + 6*c) + 8*cos(3*b*x + 2*a + 2*c) - 8*cos(3*b*x + 4*c) - 3*cos(b*x + 2*a) - 3*cos(b*x + 2*c))*cos(6* b*x + a + 6*c) - 6*(3*cos(4*b*x + a + 4*c) - 3*cos(2*b*x + a + 2*c) + cos( a))*cos(5*b*x + 2*a + 4*c) - 6*(3*cos(4*b*x + a + 4*c) - 3*cos(2*b*x + a + 2*c) + cos(a))*cos(5*b*x + 6*c) - 6*(8*cos(3*b*x + 2*a + 2*c) - 8*cos(3*b *x + 4*c) - 3*cos(b*x + 2*a) - 3*cos(b*x + 2*c))*cos(4*b*x + a + 4*c) + 16 *(3*cos(2*b*x + a + 2*c) - cos(a))*cos(3*b*x + 2*a + 2*c) - 16*(3*cos(2*b* x + a + 2*c) - cos(a))*cos(3*b*x + 4*c) - 18*(cos(b*x + 2*a) + cos(b*x + 2 *c))*cos(2*b*x + a + 2*c) + 6*cos(b*x + 2*a)*cos(a) + 6*cos(b*x + 2*c)*cos (a) - 3*(cos(6*b*x + a + 6*c)^2*cos(-a + c) + 9*cos(4*b*x + a + 4*c)^2*cos (-a + c) + 9*cos(2*b*x + a + 2*c)^2*cos(-a + c) - 6*cos(2*b*x + a + 2*c)*c os(a)*cos(-a + c) + cos(-a + c)*sin(6*b*x + a + 6*c)^2 + 9*cos(-a + c)*sin (4*b*x + a + 4*c)^2 + 9*cos(-a + c)*sin(2*b*x + a + 2*c)^2 - 6*cos(-a + c) *sin(2*b*x + a + 2*c)*sin(a) - 2*(3*cos(4*b*x + a + 4*c)*cos(-a + c) - 3*c os(2*b*x + a + 2*c)*cos(-a + c) + cos(a)*cos(-a + c))*cos(6*b*x + a + 6*c) - 6*(3*cos(2*b*x + a + 2*c)*cos(-a + c) - cos(a)*cos(-a + c))*cos(4*b*x + a + 4*c) + (cos(a)^2 + sin(a)^2)*cos(-a + c) - 2*(3*cos(-a + c)*sin(4*b*x + a + 4*c) - 3*cos(-a + c)*sin(2*b*x + a + 2*c) + cos(-a + c)*sin(a))*sin (6*b*x + a + 6*c) - 6*(3*cos(-a + c)*sin(2*b*x + a + 2*c) - cos(-a + c)*si n(a))*sin(4*b*x + a + 4*c))*log(cos(b*x)^2 + 2*cos(b*x)*cos(c) + cos(c)...
Leaf count of result is larger than twice the leaf count of optimal. 2221 vs. \(2 (61) = 122\).
Time = 0.16 (sec) , antiderivative size = 2221, normalized size of antiderivative = 33.15 \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+c)^4*sin(b*x+a),x, algorithm="giac")
Output:
1/24*(12*(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) - (2*tan(1/2*b*x + 1/2*c)^3*tan( 1/2*a)^6*tan(1/2*c)^5 - 2*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^5*tan(1/2*c)^6 - 3*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^6*tan(1/2*c)^6 + 4*tan(1/2*b*x + 1/ 2*c)^3*tan(1/2*a)^6*tan(1/2*c)^3 - 2*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^5*t an(1/2*c)^4 - 3*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^6*tan(1/2*c)^4 + 2*tan(1 /2*b*x + 1/2*c)^3*tan(1/2*a)^4*tan(1/2*c)^5 - 12*tan(1/2*b*x + 1/2*c)^2*ta n(1/2*a)^5*tan(1/2*c)^5 + 6*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^6*tan(1/2*c)^5 - 4*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^3*tan(1/2*c)^6 - 3*tan(1/2*b*x + 1/ 2*c)^2*tan(1/2*a)^4*tan(1/2*c)^6 - 6*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^5*tan (1/2*c)^6 + 2*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^6*tan(1/2*c) + 2*tan(1/2*b *x + 1/2*c)^3*tan(1/2*a)^5*tan(1/2*c)^2 + 3*tan(1/2*b*x + 1/2*c)^2*tan(1/2 *a)^6*tan(1/2*c)^2 + 4*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^4*tan(1/2*c)^3 - 24*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^5*tan(1/2*c)^3 + 12*tan(1/2*b*x + 1/2 *c)*tan(1/2*a)^6*tan(1/2*c)^3 - 4*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^3*tan( 1/2*c)^4 - 3*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^4*tan(1/2*c)^4 - 6*tan(1/2* b*x + 1/2*c)*tan(1/2*a)^5*tan(1/2*c)^4 - 2*tan(1/2*b*x + 1/2*c)^3*tan(1/2* a)^2*tan(1/2*c)^5 - 24*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^3*tan(1/2*c)^5 + 6*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^4*tan(1/2*c)^5 - 2*tan(1/2*b*x + 1/2*...
Timed out. \[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\text {Hanged} \] Input:
int(sin(a + b*x)/sin(c + b*x)^4,x)
Output:
\text{Hanged}
\[ \int \csc ^4(c+b x) \sin (a+b x) \, dx=\frac {-2 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )-4 \cos \left (b x +c \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-2 \cos \left (b x +c \right ) \sin \left (b x +c \right )-4 \cos \left (b x +c \right ) \sin \left (b x +a \right )-4 \cos \left (b x +a \right ) \sin \left (b x +c \right )-6 \left (\int \frac {\tan \left (\frac {b x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) \sin \left (b x +c \right )^{3} b -2 \left (\int \frac {1}{\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3}}d x \right ) \sin \left (b x +c \right )^{3} b +4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (b x +c \right )^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {c}{2}\right )\right ) \sin \left (b x +c \right )^{3}+3 \sin \left (b x +c \right )^{3}+2 \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +c \right )}{12 \sin \left (b x +c \right )^{3} b} \] Input:
int(csc(b*x+c)^4*sin(b*x+a),x)
Output:
( - 2*cos(b*x + c)*cos(a + b*x)*sin(b*x + c) - 4*cos(b*x + c)*sin(b*x + c) **2*sin(a + b*x) - 2*cos(b*x + c)*sin(b*x + c) - 4*cos(b*x + c)*sin(a + b* x) - 4*cos(a + b*x)*sin(b*x + c) - 6*int(tan((b*x + c)/2)/(tan((a + b*x)/2 )**2 + 1),x)*sin(b*x + c)**3*b - 2*int(1/(tan((b*x + c)/2)**3*tan((a + b*x )/2)**2 + tan((b*x + c)/2)**3),x)*sin(b*x + c)**3*b + 4*log(tan((b*x + c)/ 2)**2 + 1)*sin(b*x + c)**3 - 2*log(tan((b*x + c)/2))*sin(b*x + c)**3 + 3*s in(b*x + c)**3 + 2*sin(b*x + c)**2*sin(a + b*x) - 2*sin(b*x + c))/(12*sin( b*x + c)**3*b)