Integrand size = 15, antiderivative size = 1 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=0 \] Output:
0
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 0.52 (sec) , antiderivative size = 80, normalized size of antiderivative = 80.00 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\frac {\sec (a-c) \sec (c+b x)-2 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \sec (a-c) \tan (a-c)+\left (-\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right ) \tan ^2(a-c)}{b} \] Input:
Integrate[Csc[a + b*x]*Tan[c + b*x]^2,x]
Output:
(Sec[a - c]*Sec[c + b*x] - 2*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Sec[a - c]*Tan[a - c] + (-Log[Cos[(a + b*x)/2]] + Log[Sin[(a + b*x)/2]])*Tan[a - c]^2)/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.
| \(\displaystyle \int \csc (a+b x) \tan ^2(b x+c) \, dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \csc (a+b x) \tan ^2(b x+c)dx\) |
Input:
Int[Csc[a + b*x]*Tan[c + b*x]^2,x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.21 (sec) , antiderivative size = 558, normalized size of antiderivative = 558.00
| method | result | size |
| risch | \(\frac {4 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}}{\left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right ) \left ({\mathrm e}^{2 i a}+{\mathrm e}^{2 i c}\right ) b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right ) {\mathrm e}^{4 i a}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}+\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right ) {\mathrm e}^{2 i \left (a +c \right )}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right ) {\mathrm e}^{4 i c}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}+\frac {2 i \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) {\mathrm e}^{i \left (3 a +c \right )}}{b \left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right )}-\frac {2 i \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) {\mathrm e}^{i \left (a +3 c \right )}}{b \left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right )}-\frac {2 i \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) {\mathrm e}^{i \left (3 a +c \right )}}{b \left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right )}+\frac {2 i \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) {\mathrm e}^{i \left (a +3 c \right )}}{b \left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) {\mathrm e}^{4 i a}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}-\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) {\mathrm e}^{2 i \left (a +c \right )}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) {\mathrm e}^{4 i c}}{\left ({\mathrm e}^{4 i a}+2 \,{\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{4 i c}\right ) b}\) | \(558\) |
Input:
int(csc(b*x+a)*tan(b*x+c)^2,x,method=_RETURNVERBOSE)
Output:
4/(exp(2*I*(b*x+a+c))+exp(2*I*a))/(exp(2*I*a)+exp(2*I*c))/b*exp(I*(b*x+3*a +2*c))-1/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c))/b*ln(exp(I*(b*x+a))-1)*e xp(4*I*a)+2/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c))/b*ln(exp(I*(b*x+a))-1 )*exp(2*I*(a+c))-1/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c))/b*ln(exp(I*(b* x+a))-1)*exp(4*I*c)+2*I*ln(exp(I*(b*x+a))+I*exp(I*(a-c)))/b/(exp(4*I*a)+2* exp(2*I*(a+c))+exp(4*I*c))*exp(I*(3*a+c))-2*I*ln(exp(I*(b*x+a))+I*exp(I*(a -c)))/b/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c))*exp(I*(a+3*c))-2*I*ln(exp (I*(b*x+a))-I*exp(I*(a-c)))/b/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c))*exp (I*(3*a+c))+2*I*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))/b/(exp(4*I*a)+2*exp(2*I* (a+c))+exp(4*I*c))*exp(I*(a+3*c))+1/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4*I*c ))/b*ln(exp(I*(b*x+a))+1)*exp(4*I*a)-2/(exp(4*I*a)+2*exp(2*I*(a+c))+exp(4* I*c))/b*ln(exp(I*(b*x+a))+1)*exp(2*I*(a+c))+1/(exp(4*I*a)+2*exp(2*I*(a+c)) +exp(4*I*c))/b*ln(exp(I*(b*x+a))+1)*exp(4*I*c)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.11 (sec) , antiderivative size = 413, normalized size of antiderivative = 413.00 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=-\frac {\frac {\sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) - 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) - 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 4 \, \cos \left (-2 \, a + 2 \, c\right ) + 4}{2 \, {\left ({\left (b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (b \cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)*tan(b*x+c)^2,x, algorithm="fricas")
Output:
-1/2*(sqrt(2)*((cos(-2*a + 2*c) + 1)*cos(b*x + a)*sin(-2*a + 2*c) + (cos(- 2*a + 2*c)^2 - 1)*sin(b*x + a))*log((2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2* cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - 2*sqrt(2)*((cos(-2*a + 2*c) + 1)*sin(b*x + a) + cos(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c) - 3)/(2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)* sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) + 1))/sqrt(cos(-2*a + 2*c) + 1) - ((cos(-2*a + 2*c) - 1)*sin(b*x + a)*sin(-2*a + 2*c) - (cos(-2*a + 2 *c)^2 - 1)*cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + ((cos(-2*a + 2*c) - 1)*sin(b*x + a)*sin(-2*a + 2*c) - (cos(-2*a + 2*c)^2 - 1)*cos(b*x + a))*l og(-1/2*cos(b*x + a) + 1/2) + 4*cos(-2*a + 2*c) + 4)/((b*cos(-2*a + 2*c) + b)*sin(b*x + a)*sin(-2*a + 2*c) - (b*cos(-2*a + 2*c)^2 + 2*b*cos(-2*a + 2 *c) + b)*cos(b*x + a))
Timed out. \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)*tan(b*x+c)**2,x)
Output:
Timed out
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.51 (sec) , antiderivative size = 14864, normalized size of antiderivative = 14864.00 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)*tan(b*x+c)^2,x, algorithm="maxima")
Output:
1/2*(8*(cos(4*a)^2 + 4*(cos(4*a) + cos(4*c))*cos(2*a + 2*c) + 4*cos(2*a + 2*c)^2 + 2*cos(4*a)*cos(4*c) + cos(4*c)^2 + sin(4*a)^2 + 4*(sin(4*a) + sin (4*c))*sin(2*a + 2*c) + 4*sin(2*a + 2*c)^2 + 2*sin(4*a)*sin(4*c) + sin(4*c )^2)*cos(2*b*x + 2*a + 2*c)*cos(b*x + a + 2*c) + 8*(cos(4*a)^2 + 4*(cos(4* a) + cos(4*c))*cos(2*a + 2*c) + 4*cos(2*a + 2*c)^2 + 2*cos(4*a)*cos(4*c) + cos(4*c)^2 + sin(4*a)^2 + 4*(sin(4*a) + sin(4*c))*sin(2*a + 2*c) + 4*sin( 2*a + 2*c)^2 + 2*sin(4*a)*sin(4*c) + sin(4*c)^2)*cos(2*b*x + 4*c)*cos(b*x + a + 2*c) + 8*(cos(4*a)^2 + 4*(cos(4*a) + cos(4*c))*cos(2*a + 2*c) + 4*co s(2*a + 2*c)^2 + 2*cos(4*a)*cos(4*c) + cos(4*c)^2 + sin(4*a)^2 + 4*(sin(4* a) + sin(4*c))*sin(2*a + 2*c) + 4*sin(2*a + 2*c)^2 + 2*sin(4*a)*sin(4*c) + sin(4*c)^2)*sin(2*b*x + 2*a + 2*c)*sin(b*x + a + 2*c) + 8*(cos(4*a)^2 + 4 *(cos(4*a) + cos(4*c))*cos(2*a + 2*c) + 4*cos(2*a + 2*c)^2 + 2*cos(4*a)*co s(4*c) + cos(4*c)^2 + sin(4*a)^2 + 4*(sin(4*a) + sin(4*c))*sin(2*a + 2*c) + 4*sin(2*a + 2*c)^2 + 2*sin(4*a)*sin(4*c) + sin(4*c)^2)*sin(2*b*x + 4*c)* sin(b*x + a + 2*c) - 4*(((cos(4*a) + cos(4*c))*cos(3*a + c) + 2*cos(3*a + c)*cos(2*a + 2*c) - (cos(4*a) + 2*cos(2*a + 2*c) + cos(4*c))*cos(a + 3*c) + (sin(4*a) + sin(4*c))*sin(3*a + c) + 2*sin(3*a + c)*sin(2*a + 2*c) - (si n(4*a) + 2*sin(2*a + 2*c) + sin(4*c))*sin(a + 3*c))*cos(2*b*x + 2*a + 2*c) ^2 + ((cos(4*a) + cos(4*c))*cos(3*a + c) + 2*cos(3*a + c)*cos(2*a + 2*c) - (cos(4*a) + 2*cos(2*a + 2*c) + cos(4*c))*cos(a + 3*c) + (sin(4*a) + si...
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.69 (sec) , antiderivative size = 2953, normalized size of antiderivative = 2953.00 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)*tan(b*x+c)^2,x, algorithm="giac")
Output:
2*((tan(1/2*a)^5*tan(1/2*c)^4 - tan(1/2*a)^4*tan(1/2*c)^5 - tan(1/2*a)^5*t an(1/2*c)^3 + 2*tan(1/2*a)^4*tan(1/2*c)^4 - tan(1/2*a)^3*tan(1/2*c)^5 + ta n(1/2*a)^5*tan(1/2*c)^2 + tan(1/2*a)^4*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2 *c)^4 - tan(1/2*a)^2*tan(1/2*c)^5 - tan(1/2*a)^5*tan(1/2*c) + tan(1/2*a)^4 *tan(1/2*c)^2 + tan(1/2*a)^2*tan(1/2*c)^4 - tan(1/2*a)*tan(1/2*c)^5 + 2*ta n(1/2*a)^4*tan(1/2*c) - 2*tan(1/2*a)*tan(1/2*c)^4 - tan(1/2*a)^4 + tan(1/2 *a)^3*tan(1/2*c) + tan(1/2*a)*tan(1/2*c)^3 - tan(1/2*c)^4 + tan(1/2*a)^3 + tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*c)^3 - tan(1/ 2*a)^2 + 2*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c)) *log(abs(-tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c) + tan(1/2*b*x + 1/2*a )*tan(1/2*a) - tan(1/2*b*x + 1/2*a)*tan(1/2*c) + tan(1/2*a)*tan(1/2*c) - t an(1/2*b*x + 1/2*a) + tan(1/2*a) - tan(1/2*c) + 1))/(tan(1/2*a)^5*tan(1/2* c)^5 - tan(1/2*a)^5*tan(1/2*c)^4 + tan(1/2*a)^4*tan(1/2*c)^5 - 2*tan(1/2*a )^5*tan(1/2*c)^3 + 9*tan(1/2*a)^4*tan(1/2*c)^4 - 2*tan(1/2*a)^3*tan(1/2*c) ^5 + 2*tan(1/2*a)^5*tan(1/2*c)^2 - 10*tan(1/2*a)^4*tan(1/2*c)^3 + 10*tan(1 /2*a)^3*tan(1/2*c)^4 - 2*tan(1/2*a)^2*tan(1/2*c)^5 + tan(1/2*a)^5*tan(1/2* c) - 10*tan(1/2*a)^4*tan(1/2*c)^2 + 28*tan(1/2*a)^3*tan(1/2*c)^3 - 10*tan( 1/2*a)^2*tan(1/2*c)^4 + tan(1/2*a)*tan(1/2*c)^5 - tan(1/2*a)^5 + 9*tan(1/2 *a)^4*tan(1/2*c) - 28*tan(1/2*a)^3*tan(1/2*c)^2 + 28*tan(1/2*a)^2*tan(1/2* c)^3 - 9*tan(1/2*a)*tan(1/2*c)^4 + tan(1/2*c)^5 + tan(1/2*a)^4 - 10*tan...
Time = 40.24 (sec) , antiderivative size = 14459, normalized size of antiderivative = 14459.00 \[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\text {Too large to display} \] Input:
int(tan(c + b*x)^2/sin(a + b*x),x)
Output:
- ((2*(tan(c)^2 + 1))/(cos(a) + sin(a)*tan(c)) - (2*tan((b*x)/2)*tan(c)*(t an(c)^2 + 1))/(cos(a) + sin(a)*tan(c)))/(b*(tan((b*x)/2)^2 + 2*tan((b*x)/2 )*tan(c) - 1)) - (atan((((sin(a) - cos(a)*tan(c))*(tan(c)^2 + 1)^(1/2)*((3 2*(cos(a)^2*sin(a)^4 - sin(a)^6*tan(c)^2 - 2*cos(a)^3*sin(a)^3*tan(c) + 6* cos(a)*sin(a)^5*tan(c)^3 + cos(a)*sin(a)^5*tan(c)^5 - cos(a)^5*sin(a)*tan( c)^5 + 4*cos(a)^2*sin(a)^4*tan(c)^2 + cos(a)^4*sin(a)^2*tan(c)^2 - 9*cos(a )^3*sin(a)^3*tan(c)^3 - 9*cos(a)^2*sin(a)^4*tan(c)^4 + 6*cos(a)^4*sin(a)^2 *tan(c)^4 + 4*cos(a)^3*sin(a)^3*tan(c)^5 - 2*cos(a)^2*sin(a)^4*tan(c)^6 + cos(a)^3*sin(a)^3*tan(c)^7))/(cos(a)^3 + sin(a)^3*tan(c)^3 + 3*cos(a)^2*si n(a)*tan(c) + 3*cos(a)*sin(a)^2*tan(c)^2) + (32*tan((b*x)/2)*(2*cos(a)*sin (a)^5 + 2*sin(a)^6*tan(c) + 2*cos(a)^3*sin(a)^3 + 4*sin(a)^6*tan(c)^3 + si n(a)^6*tan(c)^5 - 4*cos(a)^2*sin(a)^4*tan(c) - 4*cos(a)^4*sin(a)^2*tan(c) - 2*cos(a)*sin(a)^5*tan(c)^2 + 2*cos(a)^5*sin(a)*tan(c)^2 - 11*cos(a)*sin( a)^5*tan(c)^4 + 5*cos(a)^5*sin(a)*tan(c)^4 - 2*cos(a)*sin(a)^5*tan(c)^6 + 4*cos(a)^5*sin(a)*tan(c)^6 + 7*cos(a)^3*sin(a)^3*tan(c)^2 - cos(a)^2*sin(a )^4*tan(c)^3 - 10*cos(a)^4*sin(a)^2*tan(c)^3 + 4*cos(a)^3*sin(a)^3*tan(c)^ 4 + 14*cos(a)^2*sin(a)^4*tan(c)^5 - 7*cos(a)^4*sin(a)^2*tan(c)^5 - 11*cos( a)^3*sin(a)^3*tan(c)^6 + cos(a)^2*sin(a)^4*tan(c)^7 + 4*cos(a)^4*sin(a)^2* tan(c)^7))/(cos(a)^3 + sin(a)^3*tan(c)^3 + 3*cos(a)^2*sin(a)*tan(c) + 3*co s(a)*sin(a)^2*tan(c)^2) - ((sin(a) - cos(a)*tan(c))*(tan(c)^2 + 1)^(1/2...
\[ \int \csc (a+b x) \tan ^2(c+b x) \, dx=\int \csc \left (b x +a \right ) \tan \left (b x +c \right )^{2}d x \] Input:
int(csc(b*x+a)*tan(b*x+c)^2,x)
Output:
int(csc(a + b*x)*tan(b*x + c)**2,x)