Integrand size = 15, antiderivative size = 106 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\frac {\cos (2 (a+b x))}{4 b}-\frac {\cos (4 (a+b x))}{32 b}+\frac {\cos (4 a-2 c+2 b x)}{8 b}-\frac {\cos ^4(a-c) \log (\cos (c+b x))}{b}+\frac {3}{4} x \cos (a-c) \sin (a-c)+\frac {1}{4} x \cos (a-c) \sin (3 (a-c)) \] Output:
1/4*cos(2*b*x+2*a)/b-1/32*cos(4*b*x+4*a)/b+1/8*cos(2*b*x+4*a-2*c)/b-cos(a- c)^4*ln(cos(b*x+c))/b+3/4*x*cos(a-c)*sin(a-c)+1/4*x*cos(a-c)*sin(3*a-3*c)
Time = 0.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\frac {8 \cos (2 (a+b x))-\cos (4 (a+b x))+4 \cos (4 a-2 c+2 b x)+8 \cos (a-c) (-3 \cos (a-c) \log (\cos (c+b x))-\cos (3 (a-c)) \log (\cos (c+b x))+b x (3 \sin (a-c)+\sin (3 (a-c))))}{32 b} \] Input:
Integrate[Sin[a + b*x]^4*Tan[c + b*x],x]
Output:
(8*Cos[2*(a + b*x)] - Cos[4*(a + b*x)] + 4*Cos[4*a - 2*c + 2*b*x] + 8*Cos[ a - c]*(-3*Cos[a - c]*Log[Cos[c + b*x]] - Cos[3*(a - c)]*Log[Cos[c + b*x]] + b*x*(3*Sin[a - c] + Sin[3*(a - c)])))/(32*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.
\(\displaystyle \int \sin ^4(a+b x) \tan (b x+c) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^4(a+b x) \tan (b x+c)dx\) |
Input:
Int[Sin[a + b*x]^4*Tan[c + b*x],x]
Output:
$Aborted
Result contains complex when optimal does not.
Time = 1.15 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.29
method | result | size |
risch | \(\frac {{\mathrm e}^{-2 i \left (b x +a \right )}}{8 b}+\frac {{\mathrm e}^{2 i \left (b x +a \right )}}{8 b}+\frac {i \cos \left (2 a -2 c \right ) a}{b}-\frac {i x \,{\mathrm e}^{2 i \left (a -c \right )}}{2}+i \cos \left (2 a -2 c \right ) x -\frac {i x \,{\mathrm e}^{4 i \left (a -c \right )}}{8}+\frac {3 i x}{8}+\frac {i \cos \left (4 a -4 c \right ) a}{4 b}+\frac {3 i a}{4 b}+\frac {i \cos \left (4 a -4 c \right ) x}{4}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right )}{8 b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) \cos \left (4 a -4 c \right )}{8 b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) \cos \left (2 a -2 c \right )}{2 b}-\frac {\cos \left (4 b x +4 a \right )}{32 b}+\frac {\cos \left (2 b x +4 a -2 c \right )}{8 b}\) | \(243\) |
Input:
int(sin(b*x+a)^4*tan(b*x+c),x,method=_RETURNVERBOSE)
Output:
1/8/b*exp(-2*I*(b*x+a))+1/8/b*exp(2*I*(b*x+a))+I/b*cos(2*a-2*c)*a-1/2*I*x* exp(2*I*(a-c))+I*cos(2*a-2*c)*x-1/8*I*x*exp(4*I*(a-c))+3/8*I*x+1/4*I/b*cos (4*a-4*c)*a+3/4*I/b*a+1/4*I*cos(4*a-4*c)*x-3/8*ln(exp(2*I*(b*x+a))+exp(2*I *(a-c)))/b-1/8/b*ln(exp(2*I*(b*x+a))+exp(2*I*(a-c)))*cos(4*a-4*c)-1/2*ln(e xp(2*I*(b*x+a))+exp(2*I*(a-c)))/b*cos(2*a-2*c)-1/32*cos(4*b*x+4*a)/b+1/8*c os(2*b*x+4*a-2*c)/b
Time = 0.09 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.63 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{4} - 2 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 3\right )} \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}{\cos \left (-2 \, a + 2 \, c\right ) + 1}\right ) + 2 \, {\left (b x \cos \left (-2 \, a + 2 \, c\right ) + 2 \, b x\right )} \sin \left (-2 \, a + 2 \, c\right )}{8 \, b} \] Input:
integrate(sin(b*x+a)^4*tan(b*x+c),x, algorithm="fricas")
Output:
-1/8*(2*cos(b*x + a)^4 - 2*(cos(-2*a + 2*c) + 3)*cos(b*x + a)^2 - 2*cos(b* x + a)*sin(b*x + a)*sin(-2*a + 2*c) + (cos(-2*a + 2*c)^2 + 2*cos(-2*a + 2* c) + 1)*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) - cos(-2*a + 2*c) + 1)/(cos(-2*a + 2*c) + 1)) + 2*(b*x*c os(-2*a + 2*c) + 2*b*x)*sin(-2*a + 2*c))/b
\[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\int \sin ^{4}{\left (a + b x \right )} \tan {\left (b x + c \right )}\, dx \] Input:
integrate(sin(b*x+a)**4*tan(b*x+c),x)
Output:
Integral(sin(a + b*x)**4*tan(b*x + c), x)
Time = 0.06 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.29 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=-\frac {4 \, {\left (4 \, b \sin \left (-2 \, a + 2 \, c\right ) + b \sin \left (-4 \, a + 4 \, c\right )\right )} x + 2 \, {\left (4 \, \cos \left (-2 \, a + 2 \, c\right ) + \cos \left (-4 \, a + 4 \, c\right ) + 3\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, c\right ) + \cos \left (2 \, c\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, c\right ) + \sin \left (2 \, c\right )^{2}\right ) + \cos \left (4 \, b x + 4 \, a\right ) - 4 \, \cos \left (2 \, b x + 4 \, a - 2 \, c\right ) - 8 \, \cos \left (2 \, b x + 2 \, a\right )}{32 \, b} \] Input:
integrate(sin(b*x+a)^4*tan(b*x+c),x, algorithm="maxima")
Output:
-1/32*(4*(4*b*sin(-2*a + 2*c) + b*sin(-4*a + 4*c))*x + 2*(4*cos(-2*a + 2*c ) + cos(-4*a + 4*c) + 3)*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*c) + cos(2* c)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*c) + sin(2*c)^2) + cos(4*b*x + 4* a) - 4*cos(2*b*x + 4*a - 2*c) - 8*cos(2*b*x + 2*a))/b
Leaf count of result is larger than twice the leaf count of optimal. 454260 vs. \(2 (98) = 196\).
Time = 19.00 (sec) , antiderivative size = 454260, normalized size of antiderivative = 4285.47 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\text {Too large to display} \] Input:
integrate(sin(b*x+a)^4*tan(b*x+c),x, algorithm="giac")
Output:
1/32*(12*b*x*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^2*tan(2*a - c)^2*tan(a + c )^2*tan(a - c)^2*tan(a)^2*tan(c)^5 - 16*b*x*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^2*tan(2*a - c)^2*tan(a + c)^2*tan(a - c)^2*tan(a)*tan(c)^6 + 16*b*x*t an(b*x)^4*tan(2*a)^2*tan(2*a + c)^2*tan(2*a - c)^2*tan(a + c)^2*tan(a - c) *tan(a)^2*tan(c)^6 - 4*b*x*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^2*tan(2*a - c)*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^6 + 4*b*x*tan(b*x)^4*tan(2*a) *tan(2*a + c)^2*tan(2*a - c)^2*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^6 - 6*log(4*(tan(b*x)^2*tan(c)^2 - 2*tan(b*x)*tan(c) + 1)/(tan(b*x)^2*tan(c )^2 + tan(b*x)^2 + tan(c)^2 + 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^2*tan (2*a - c)^2*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^6 + 12*arctan((tan(b *x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^2* tan(2*a - c)^2*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^5 + 8*arctan((tan (b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^ 2*tan(2*a - c)^2*tan(a + c)^2*tan(a - c)*tan(a)^2*tan(c)^6 - 8*arctan((tan (b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^ 2*tan(2*a - c)^2*tan(a + c)*tan(a - c)^2*tan(a)^2*tan(c)^6 - 2*arctan((tan (b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)^ 2*tan(2*a - c)*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^6 + 2*arctan((tan (b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(2*a)^2*tan(2*a + c)* tan(2*a - c)^2*tan(a + c)^2*tan(a - c)^2*tan(a)^2*tan(c)^6 + 40*b*x*tan...
Time = 40.60 (sec) , antiderivative size = 12617, normalized size of antiderivative = 119.03 \[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\text {Too large to display} \] Input:
int(sin(a + b*x)^4*tan(c + b*x),x)
Output:
(log(tan((b*x)/2)^2 + 1)*(4*cos(2*a) + cos(4*a) + 6*tan(c)^2 + 3*tan(c)^4 + 8*sin(2*a)*tan(c) + 4*sin(4*a)*tan(c) - 4*cos(2*a)*tan(c)^4 - 6*cos(4*a) *tan(c)^2 + cos(4*a)*tan(c)^4 + 8*sin(2*a)*tan(c)^3 - 4*sin(4*a)*tan(c)^3 + 3))/(2*b*(8*tan(c)^2 + 4*tan(c)^4 + 4)) - ((tan((b*x)/2)*(sin(2*a) + sin (4*a)/4 - cos(4*a)*tan(c) + sin(2*a)*tan(c)^2 - (3*sin(4*a)*tan(c)^2)/4))/ (tan(c)^2 + 1) + (4*tan((b*x)/2)^4*(cos(2*a) + cos(4*a) + sin(4*a)*tan(c) + cos(2*a)*tan(c)^2))/(tan(c)^2 + 1) - (tan((b*x)/2)^7*(sin(2*a) + sin(4*a )/4 - cos(4*a)*tan(c) + sin(2*a)*tan(c)^2 - (3*sin(4*a)*tan(c)^2)/4))/(tan (c)^2 + 1) + (tan((b*x)/2)^3*(sin(2*a) + (9*sin(4*a))/4 - cos(4*a)*tan(c) + sin(2*a)*tan(c)^2 + (5*sin(4*a)*tan(c)^2)/4))/(tan(c)^2 + 1) - (tan((b*x )/2)^5*(sin(2*a) + (9*sin(4*a))/4 - cos(4*a)*tan(c) + sin(2*a)*tan(c)^2 + (5*sin(4*a)*tan(c)^2)/4))/(tan(c)^2 + 1) + (2*tan((b*x)/2)^2*(cos(2*a) + s in(4*a)*tan(c) + cos(2*a)*tan(c)^2 - cos(4*a)*tan(c)^2))/(tan(c)^2 + 1) + (2*tan((b*x)/2)^6*(cos(2*a) + sin(4*a)*tan(c) + cos(2*a)*tan(c)^2 - cos(4* a)*tan(c)^2))/(tan(c)^2 + 1))/(b*(4*tan((b*x)/2)^2 + 6*tan((b*x)/2)^4 + 4* tan((b*x)/2)^6 + tan((b*x)/2)^8 + 1)) - (log(tan((b*x)/2)^2*cos(c) - cos(c ) + 2*tan((b*x)/2)*sin(c))*(cos(2*a)/2 + cos(4*a)/8 + (3*tan(c)^2)/4 + (3* tan(c)^4)/8 + sin(2*a)*tan(c) + (sin(4*a)*tan(c))/2 - (cos(2*a)*tan(c)^4)/ 2 - (3*cos(4*a)*tan(c)^2)/4 + (cos(4*a)*tan(c)^4)/8 + sin(2*a)*tan(c)^3 - (sin(4*a)*tan(c)^3)/2 + 3/8))/(b*(2*tan(c)^2 + tan(c)^4 + 1)) + (atan((...
\[ \int \sin ^4(a+b x) \tan (c+b x) \, dx=\int \sin \left (b x +a \right )^{4} \tan \left (b x +c \right )d x \] Input:
int(sin(b*x+a)^4*tan(b*x+c),x)
Output:
int(sin(a + b*x)**4*tan(b*x + c),x)