Integrand size = 18, antiderivative size = 85 \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\frac {\text {arctanh}(\cos (c-b x)) \cos (2 (a+c))}{b}+\frac {\text {arctanh}(\cos (c-b x)) \sin ^2(a+c)}{2 b}+\frac {\cot (c-b x) \csc (c-b x) \sin ^2(a+c)}{2 b}-\frac {\csc (c-b x) \sin (2 (a+c))}{b} \] Output:
arctanh(cos(b*x-c))*cos(2*a+2*c)/b+1/2*arctanh(cos(b*x-c))*sin(a+c)^2/b+1/ 2*cot(b*x-c)*csc(b*x-c)*sin(a+c)^2/b+csc(b*x-c)*sin(2*a+2*c)/b
Time = 5.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.91 \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\frac {4 (1+3 \cos (2 (a+c))) \log \left (\cos \left (\frac {1}{2} (c-b x)\right )\right )-4 (1+3 \cos (2 (a+c))) \log \left (\sin \left (\frac {1}{2} (c-b x)\right )\right )+2 \csc ^2\left (\frac {1}{2} (c-b x)\right ) \sin ^2(a+c)-2 \sec ^2\left (\frac {1}{2} (c-b x)\right ) \sin ^2(a+c)-8 \csc \left (\frac {c}{2}\right ) \csc \left (\frac {1}{2} (c-b x)\right ) \sin (2 (a+c)) \sin \left (\frac {b x}{2}\right )+8 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c-b x)\right ) \sin (2 (a+c)) \sin \left (\frac {b x}{2}\right )}{16 b} \] Input:
Integrate[Csc[c - b*x]^3*Sin[a + b*x]^2,x]
Output:
(4*(1 + 3*Cos[2*(a + c)])*Log[Cos[(c - b*x)/2]] - 4*(1 + 3*Cos[2*(a + c)]) *Log[Sin[(c - b*x)/2]] + 2*Csc[(c - b*x)/2]^2*Sin[a + c]^2 - 2*Sec[(c - b* x)/2]^2*Sin[a + c]^2 - 8*Csc[c/2]*Csc[(c - b*x)/2]*Sin[2*(a + c)]*Sin[(b*x )/2] + 8*Sec[c/2]*Sec[(c - b*x)/2]*Sin[2*(a + c)]*Sin[(b*x)/2])/(16*b)
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.
\(\displaystyle \int \sin ^2(a+b x) \csc ^3(c-b x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^2(a+b x) \csc ^3(c-b x)dx\) |
Input:
Int[Csc[c - b*x]^3*Sin[a + b*x]^2,x]
Output:
$Aborted
Result contains complex when optimal does not.
Time = 2.70 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.74
method | result | size |
risch | \(-\frac {3 \,{\mathrm e}^{i \left (b x +6 a +5 c \right )}-5 \,{\mathrm e}^{3 i \left (b x +2 a +c \right )}+2 \,{\mathrm e}^{i \left (b x +4 a +3 c \right )}+2 \,{\mathrm e}^{i \left (3 b x +4 a +c \right )}-5 \,{\mathrm e}^{i \left (b x +2 a +c \right )}+3 \,{\mathrm e}^{i \left (3 b x +2 a -c \right )}}{4 \left ({\mathrm e}^{2 i \left (a +c \right )}-{\mathrm e}^{2 i \left (b x +a \right )}\right )^{2} b}+\frac {\ln \left ({\mathrm e}^{i \left (a +c \right )}+{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (a +c \right )}+{\mathrm e}^{i \left (b x +a \right )}\right ) \cos \left (2 a +2 c \right )}{4 b}-\frac {\ln \left (-{\mathrm e}^{i \left (a +c \right )}+{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b}-\frac {3 \ln \left (-{\mathrm e}^{i \left (a +c \right )}+{\mathrm e}^{i \left (b x +a \right )}\right ) \cos \left (2 a +2 c \right )}{4 b}\) | \(233\) |
default | \(\text {Expression too large to display}\) | \(1167\) |
Input:
int(-csc(b*x-c)^3*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/4/(exp(2*I*(a+c))-exp(2*I*(b*x+a)))^2/b*(3*exp(I*(b*x+6*a+5*c))-5*exp(3 *I*(b*x+2*a+c))+2*exp(I*(b*x+4*a+3*c))+2*exp(I*(3*b*x+4*a+c))-5*exp(I*(b*x +2*a+c))+3*exp(I*(3*b*x+2*a-c)))+1/4/b*ln(exp(I*(a+c))+exp(I*(b*x+a)))+3/4 /b*ln(exp(I*(a+c))+exp(I*(b*x+a)))*cos(2*a+2*c)-1/4/b*ln(-exp(I*(a+c))+exp (I*(b*x+a)))-3/4/b*ln(-exp(I*(a+c))+exp(I*(b*x+a)))*cos(2*a+2*c)
Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (89) = 178\).
Time = 0.09 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.89 \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (3 \, \cos \left (a + c\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (a + c\right ) + 6 \, {\left (\cos \left (a + c\right )^{3} - \cos \left (a + c\right )\right )} \cos \left (b x + a\right ) + {\left (3 \, \cos \left (a + c\right )^{4} - 2 \, {\left (3 \, \cos \left (a + c\right )^{3} - \cos \left (a + c\right )\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) - {\left (6 \, \cos \left (a + c\right )^{4} - 5 \, \cos \left (a + c\right )^{2} + 1\right )} \cos \left (b x + a\right )^{2} - \cos \left (a + c\right )^{2}\right )} \log \left (\frac {\cos \left (b x + a\right ) \cos \left (a + c\right ) + \sin \left (b x + a\right ) \sin \left (a + c\right ) + 1}{\cos \left (a + c\right ) + 1}\right ) - {\left (3 \, \cos \left (a + c\right )^{4} - 2 \, {\left (3 \, \cos \left (a + c\right )^{3} - \cos \left (a + c\right )\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) - {\left (6 \, \cos \left (a + c\right )^{4} - 5 \, \cos \left (a + c\right )^{2} + 1\right )} \cos \left (b x + a\right )^{2} - \cos \left (a + c\right )^{2}\right )} \log \left (-\frac {\cos \left (b x + a\right ) \cos \left (a + c\right ) + \sin \left (b x + a\right ) \sin \left (a + c\right ) - 1}{\cos \left (a + c\right ) + 1}\right )}{4 \, {\left (2 \, b \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) + {\left (2 \, b \cos \left (a + c\right )^{2} - b\right )} \cos \left (b x + a\right )^{2} - b \cos \left (a + c\right )^{2}\right )}} \] Input:
integrate(-csc(b*x-c)^3*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-1/4*(2*(3*cos(a + c)^2 + 1)*sin(b*x + a)*sin(a + c) + 6*(cos(a + c)^3 - c os(a + c))*cos(b*x + a) + (3*cos(a + c)^4 - 2*(3*cos(a + c)^3 - cos(a + c) )*cos(b*x + a)*sin(b*x + a)*sin(a + c) - (6*cos(a + c)^4 - 5*cos(a + c)^2 + 1)*cos(b*x + a)^2 - cos(a + c)^2)*log((cos(b*x + a)*cos(a + c) + sin(b*x + a)*sin(a + c) + 1)/(cos(a + c) + 1)) - (3*cos(a + c)^4 - 2*(3*cos(a + c )^3 - cos(a + c))*cos(b*x + a)*sin(b*x + a)*sin(a + c) - (6*cos(a + c)^4 - 5*cos(a + c)^2 + 1)*cos(b*x + a)^2 - cos(a + c)^2)*log(-(cos(b*x + a)*cos (a + c) + sin(b*x + a)*sin(a + c) - 1)/(cos(a + c) + 1)))/(2*b*cos(b*x + a )*cos(a + c)*sin(b*x + a)*sin(a + c) + (2*b*cos(a + c)^2 - b)*cos(b*x + a) ^2 - b*cos(a + c)^2)
Timed out. \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate(-csc(b*x-c)**3*sin(b*x+a)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1571 vs. \(2 (89) = 178\).
Time = 0.10 (sec) , antiderivative size = 1571, normalized size of antiderivative = 18.48 \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(-csc(b*x-c)^3*sin(b*x+a)^2,x, algorithm="maxima")
Output:
-1/8*(2*(3*cos(3*b*x) - 5*cos(3*b*x + 4*a + 4*c) + 2*cos(3*b*x + 2*a + 2*c ) + 3*cos(b*x + 4*a + 6*c) + 2*cos(b*x + 2*a + 4*c) - 5*cos(b*x + 2*c))*co s(4*b*x + 2*a + c) + 10*(2*cos(2*b*x + 2*a + 3*c) - cos(2*a + 5*c))*cos(3* b*x + 4*a + 4*c) - 4*(2*cos(2*b*x + 2*a + 3*c) - cos(2*a + 5*c))*cos(3*b*x + 2*a + 2*c) - 4*(3*cos(3*b*x) + 3*cos(b*x + 4*a + 6*c) + 2*cos(b*x + 2*a + 4*c) - 5*cos(b*x + 2*c))*cos(2*b*x + 2*a + 3*c) + 6*cos(3*b*x)*cos(2*a + 5*c) + 6*cos(b*x + 4*a + 6*c)*cos(2*a + 5*c) + 4*cos(b*x + 2*a + 4*c)*co s(2*a + 5*c) - 10*cos(b*x + 2*c)*cos(2*a + 5*c) - ((3*cos(2*a + 2*c) + 1)* cos(4*b*x + 2*a + c)^2 + 4*(3*cos(2*a + 2*c) + 1)*cos(2*b*x + 2*a + 3*c)^2 - 4*(3*cos(2*a + 2*c) + 1)*cos(2*b*x + 2*a + 3*c)*cos(2*a + 5*c) + (3*cos (2*a + 2*c) + 1)*cos(2*a + 5*c)^2 + (3*cos(2*a + 2*c) + 1)*sin(4*b*x + 2*a + c)^2 + 4*(3*cos(2*a + 2*c) + 1)*sin(2*b*x + 2*a + 3*c)^2 - 4*(3*cos(2*a + 2*c) + 1)*sin(2*b*x + 2*a + 3*c)*sin(2*a + 5*c) + (3*cos(2*a + 2*c) + 1 )*sin(2*a + 5*c)^2 - 2*(2*(3*cos(2*a + 2*c) + 1)*cos(2*b*x + 2*a + 3*c) - (3*cos(2*a + 2*c) + 1)*cos(2*a + 5*c))*cos(4*b*x + 2*a + c) - 2*(2*(3*cos( 2*a + 2*c) + 1)*sin(2*b*x + 2*a + 3*c) - (3*cos(2*a + 2*c) + 1)*sin(2*a + 5*c))*sin(4*b*x + 2*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) + ((3*cos(2*a + 2*c) + 1)*cos (4*b*x + 2*a + c)^2 + 4*(3*cos(2*a + 2*c) + 1)*cos(2*b*x + 2*a + 3*c)^2 - 4*(3*cos(2*a + 2*c) + 1)*cos(2*b*x + 2*a + 3*c)*cos(2*a + 5*c) + (3*cos...
Leaf count of result is larger than twice the leaf count of optimal. 2867 vs. \(2 (89) = 178\).
Time = 0.17 (sec) , antiderivative size = 2867, normalized size of antiderivative = 33.73 \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(-csc(b*x-c)^3*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/2*(2*(tan(1/2*a)^4*tan(1/2*c)^4 - 4*tan(1/2*a)^4*tan(1/2*c)^2 - 12*tan( 1/2*a)^3*tan(1/2*c)^3 - 4*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*a)^4 + 12*ta n(1/2*a)^3*tan(1/2*c) + 28*tan(1/2*a)^2*tan(1/2*c)^2 + 12*tan(1/2*a)*tan(1 /2*c)^3 + tan(1/2*c)^4 - 4*tan(1/2*a)^2 - 12*tan(1/2*a)*tan(1/2*c) - 4*tan (1/2*c)^2 + 1)*log(abs(tan(1/2*b*x - 1/2*c)))/(tan(1/2*a)^4*tan(1/2*c)^4 + 2*tan(1/2*a)^4*tan(1/2*c)^2 + 2*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*a)^4 + 4*tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*c)^4 + 2*tan(1/2*a)^2 + 2*tan(1/2* c)^2 + 1) + (tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^8*tan(1/2*c)^6 + 2*tan(1/2* b*x - 1/2*c)^2*tan(1/2*a)^7*tan(1/2*c)^7 + 4*tan(1/2*b*x - 1/2*c)*tan(1/2* a)^8*tan(1/2*c)^7 + tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^6*tan(1/2*c)^8 + 4*t an(1/2*b*x - 1/2*c)*tan(1/2*a)^7*tan(1/2*c)^8 + 2*tan(1/2*b*x - 1/2*c)^2*t an(1/2*a)^8*tan(1/2*c)^4 + 2*tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^7*tan(1/2*c )^5 + 4*tan(1/2*b*x - 1/2*c)*tan(1/2*a)^8*tan(1/2*c)^5 - 16*tan(1/2*b*x - 1/2*c)*tan(1/2*a)^7*tan(1/2*c)^6 + 2*tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^5*t an(1/2*c)^7 - 16*tan(1/2*b*x - 1/2*c)*tan(1/2*a)^6*tan(1/2*c)^7 + 2*tan(1/ 2*b*x - 1/2*c)^2*tan(1/2*a)^4*tan(1/2*c)^8 + 4*tan(1/2*b*x - 1/2*c)*tan(1/ 2*a)^5*tan(1/2*c)^8 + tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^8*tan(1/2*c)^2 - 2 *tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^7*tan(1/2*c)^3 - 4*tan(1/2*b*x - 1/2*c) *tan(1/2*a)^8*tan(1/2*c)^3 - 2*tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^6*tan(1/2 *c)^4 - 40*tan(1/2*b*x - 1/2*c)*tan(1/2*a)^7*tan(1/2*c)^4 + 2*tan(1/2*b...
Timed out. \[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Hanged} \] Input:
int(sin(a + b*x)^2/sin(c - b*x)^3,x)
Output:
\text{Hanged}
\[ \int \csc ^3(c-b x) \sin ^2(a+b x) \, dx=-\left (\int \csc \left (b x -c \right )^{3} \sin \left (b x +a \right )^{2}d x \right ) \] Input:
int(-csc(b*x-c)^3*sin(b*x+a)^2,x)
Output:
- int(csc(b*x - c)**3*sin(a + b*x)**2,x)