Integrand size = 15, antiderivative size = 64 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\frac {\cos (3 a-c+2 b x)}{4 b}-\frac {\cos ^3(a-c) \log (\cos (c+b x))}{b}+\frac {3}{4} x \sin (a-c)+\frac {1}{4} x \sin (3 (a-c)) \] Output:
1/4*cos(2*b*x+3*a-c)/b-cos(a-c)^3*ln(cos(b*x+c))/b+3/4*x*sin(a-c)+1/4*x*si n(3*a-3*c)
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\frac {\cos (3 a-c+2 b x)-3 \cos (a-c) \log (\cos (c+b x))-\cos (3 (a-c)) \log (\cos (c+b x))+3 b x \sin (a-c)+b x \sin (3 (a-c))}{4 b} \] Input:
Integrate[Sec[c + b*x]*Sin[a + b*x]^3,x]
Output:
(Cos[3*a - c + 2*b*x] - 3*Cos[a - c]*Log[Cos[c + b*x]] - Cos[3*(a - c)]*Lo g[Cos[c + b*x]] + 3*b*x*Sin[a - c] + b*x*Sin[3*(a - c)])/(4*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.
\(\displaystyle \int \sin ^3(a+b x) \sec (b x+c) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^3(a+b x) \sec (b x+c)dx\) |
Input:
Int[Sec[c + b*x]*Sin[a + b*x]^3,x]
Output:
$Aborted
Result contains complex when optimal does not.
Time = 2.32 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.48
method | result | size |
risch | \(-\frac {i x \,{\mathrm e}^{3 i \left (a -c \right )}}{4}-\frac {3 i x \,{\mathrm e}^{i \left (a -c \right )}}{4}+\frac {3 i \cos \left (a -c \right ) x}{2}+\frac {i \cos \left (3 a -3 c \right ) x}{2}+\frac {3 i \cos \left (a -c \right ) a}{2 b}+\frac {i \cos \left (3 a -3 c \right ) a}{2 b}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) \cos \left (a -c \right )}{4 b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) \cos \left (3 a -3 c \right )}{4 b}+\frac {\cos \left (2 b x +3 a -c \right )}{4 b}\) | \(159\) |
default | \(\text {Expression too large to display}\) | \(500\) |
Input:
int(sec(b*x+c)*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
-1/4*I*x*exp(3*I*(a-c))-3/4*I*x*exp(I*(a-c))+3/2*I*cos(a-c)*x+1/2*I*cos(3* a-3*c)*x+3/2*I/b*cos(a-c)*a+1/2*I/b*cos(3*a-3*c)*a-3/4*ln(exp(2*I*(b*x+a)) +exp(2*I*(a-c)))/b*cos(a-c)-1/4/b*ln(exp(2*I*(b*x+a))+exp(2*I*(a-c)))*cos( 3*a-3*c)+1/4*cos(2*b*x+3*a-c)/b
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.70 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=-\frac {2 \, \cos \left (-a + c\right )^{3} \log \left (-\cos \left (b x + c\right )\right ) - {\left (4 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) - {\left (4 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{2} + {\left (2 \, b x \cos \left (-a + c\right )^{2} + b x\right )} \sin \left (-a + c\right )}{2 \, b} \] Input:
integrate(sec(b*x+c)*sin(b*x+a)^3,x, algorithm="fricas")
Output:
-1/2*(2*cos(-a + c)^3*log(-cos(b*x + c)) - (4*cos(-a + c)^2 - 1)*cos(b*x + c)*sin(b*x + c)*sin(-a + c) - (4*cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^2 + (2*b*x*cos(-a + c)^2 + b*x)*sin(-a + c))/b
Leaf count of result is larger than twice the leaf count of optimal. 10623 vs. \(2 (0) = 0\).
Time = 67.71 (sec) , antiderivative size = 34286, normalized size of antiderivative = 535.72 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x+c)*sin(b*x+a)**3,x)
Output:
3*Piecewise((-sin(b*x)**2/(2*b), Eq(c, pi/2)), (sin(b*x)**2/(2*b), Eq(c, - pi/2)), (0, Eq(b, 0)), (-b*x*tan(c/2)**6*tan(b*x/2)**4/(2*b*tan(c/2)**6*ta n(b*x/2)**4 + 4*b*tan(c/2)**6*tan(b*x/2)**2 + 2*b*tan(c/2)**6 + 6*b*tan(c/ 2)**4*tan(b*x/2)**4 + 12*b*tan(c/2)**4*tan(b*x/2)**2 + 6*b*tan(c/2)**4 + 6 *b*tan(c/2)**2*tan(b*x/2)**4 + 12*b*tan(c/2)**2*tan(b*x/2)**2 + 6*b*tan(c/ 2)**2 + 2*b*tan(b*x/2)**4 + 4*b*tan(b*x/2)**2 + 2*b) - 2*b*x*tan(c/2)**6*t an(b*x/2)**2/(2*b*tan(c/2)**6*tan(b*x/2)**4 + 4*b*tan(c/2)**6*tan(b*x/2)** 2 + 2*b*tan(c/2)**6 + 6*b*tan(c/2)**4*tan(b*x/2)**4 + 12*b*tan(c/2)**4*tan (b*x/2)**2 + 6*b*tan(c/2)**4 + 6*b*tan(c/2)**2*tan(b*x/2)**4 + 12*b*tan(c/ 2)**2*tan(b*x/2)**2 + 6*b*tan(c/2)**2 + 2*b*tan(b*x/2)**4 + 4*b*tan(b*x/2) **2 + 2*b) - b*x*tan(c/2)**6/(2*b*tan(c/2)**6*tan(b*x/2)**4 + 4*b*tan(c/2) **6*tan(b*x/2)**2 + 2*b*tan(c/2)**6 + 6*b*tan(c/2)**4*tan(b*x/2)**4 + 12*b *tan(c/2)**4*tan(b*x/2)**2 + 6*b*tan(c/2)**4 + 6*b*tan(c/2)**2*tan(b*x/2)* *4 + 12*b*tan(c/2)**2*tan(b*x/2)**2 + 6*b*tan(c/2)**2 + 2*b*tan(b*x/2)**4 + 4*b*tan(b*x/2)**2 + 2*b) + 7*b*x*tan(c/2)**4*tan(b*x/2)**4/(2*b*tan(c/2) **6*tan(b*x/2)**4 + 4*b*tan(c/2)**6*tan(b*x/2)**2 + 2*b*tan(c/2)**6 + 6*b* tan(c/2)**4*tan(b*x/2)**4 + 12*b*tan(c/2)**4*tan(b*x/2)**2 + 6*b*tan(c/2)* *4 + 6*b*tan(c/2)**2*tan(b*x/2)**4 + 12*b*tan(c/2)**2*tan(b*x/2)**2 + 6*b* tan(c/2)**2 + 2*b*tan(b*x/2)**4 + 4*b*tan(b*x/2)**2 + 2*b) + 14*b*x*tan(c/ 2)**4*tan(b*x/2)**2/(2*b*tan(c/2)**6*tan(b*x/2)**4 + 4*b*tan(c/2)**6*ta...
Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.73 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=-\frac {2 \, {\left (3 \, b \sin \left (-a + c\right ) + b \sin \left (-3 \, a + 3 \, c\right )\right )} x + {\left (3 \, \cos \left (-a + c\right ) + \cos \left (-3 \, a + 3 \, c\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, c\right ) + \cos \left (2 \, c\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, c\right ) + \sin \left (2 \, c\right )^{2}\right ) - 2 \, \cos \left (2 \, b x + 3 \, a - c\right )}{8 \, b} \] Input:
integrate(sec(b*x+c)*sin(b*x+a)^3,x, algorithm="maxima")
Output:
-1/8*(2*(3*b*sin(-a + c) + b*sin(-3*a + 3*c))*x + (3*cos(-a + c) + cos(-3* a + 3*c))*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) - 2*cos(2*b*x + 3*a - c))/b
Leaf count of result is larger than twice the leaf count of optimal. 2306 vs. \(2 (58) = 116\).
Time = 0.16 (sec) , antiderivative size = 2306, normalized size of antiderivative = 36.03 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x+c)*sin(b*x+a)^3,x, algorithm="giac")
Output:
1/2*(2*(3*tan(1/2*a)^6*tan(1/2*c)^5 - 3*tan(1/2*a)^5*tan(1/2*c)^6 - 2*tan( 1/2*a)^6*tan(1/2*c)^3 + 21*tan(1/2*a)^5*tan(1/2*c)^4 - 21*tan(1/2*a)^4*tan (1/2*c)^5 + 2*tan(1/2*a)^3*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c) - 21*t an(1/2*a)^5*tan(1/2*c)^2 + 78*tan(1/2*a)^4*tan(1/2*c)^3 - 78*tan(1/2*a)^3* tan(1/2*c)^4 + 21*tan(1/2*a)^2*tan(1/2*c)^5 - 3*tan(1/2*a)*tan(1/2*c)^6 + 3*tan(1/2*a)^5 - 21*tan(1/2*a)^4*tan(1/2*c) + 78*tan(1/2*a)^3*tan(1/2*c)^2 - 78*tan(1/2*a)^2*tan(1/2*c)^3 + 21*tan(1/2*a)*tan(1/2*c)^4 - 3*tan(1/2*c )^5 - 2*tan(1/2*a)^3 + 21*tan(1/2*a)^2*tan(1/2*c) - 21*tan(1/2*a)*tan(1/2* c)^2 + 2*tan(1/2*c)^3 + 3*tan(1/2*a) - 3*tan(1/2*c))*(b*x + c)/(tan(1/2*a) ^6*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^4 + 3*tan(1/2*a)^4*tan(1/2*c)^ 6 + 3*tan(1/2*a)^6*tan(1/2*c)^2 + 9*tan(1/2*a)^4*tan(1/2*c)^4 + 3*tan(1/2* a)^2*tan(1/2*c)^6 + tan(1/2*a)^6 + 9*tan(1/2*a)^4*tan(1/2*c)^2 + 9*tan(1/2 *a)^2*tan(1/2*c)^4 + tan(1/2*c)^6 + 3*tan(1/2*a)^4 + 9*tan(1/2*a)^2*tan(1/ 2*c)^2 + 3*tan(1/2*c)^4 + 3*tan(1/2*a)^2 + 3*tan(1/2*c)^2 + 1) + (tan(1/2* a)^6*tan(1/2*c)^6 - 3*tan(1/2*a)^6*tan(1/2*c)^4 + 12*tan(1/2*a)^5*tan(1/2* c)^5 - 3*tan(1/2*a)^4*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^2 - 24*tan( 1/2*a)^5*tan(1/2*c)^3 + 57*tan(1/2*a)^4*tan(1/2*c)^4 - 24*tan(1/2*a)^3*tan (1/2*c)^5 + 3*tan(1/2*a)^2*tan(1/2*c)^6 - tan(1/2*a)^6 + 12*tan(1/2*a)^5*t an(1/2*c) - 57*tan(1/2*a)^4*tan(1/2*c)^2 + 112*tan(1/2*a)^3*tan(1/2*c)^3 - 57*tan(1/2*a)^2*tan(1/2*c)^4 + 12*tan(1/2*a)*tan(1/2*c)^5 - tan(1/2*c)...
Time = 17.94 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.56 \[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\frac {{\mathrm {e}}^{-a\,3{}\mathrm {i}+c\,1{}\mathrm {i}-b\,x\,2{}\mathrm {i}}}{8\,b}+\frac {{\mathrm {e}}^{a\,3{}\mathrm {i}-c\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{8\,b}+\frac {x\,{\mathrm {e}}^{-a\,3{}\mathrm {i}+c\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{c\,2{}\mathrm {i}}\,1{}\mathrm {i}\right )}{4}-\frac {{\mathrm {e}}^{-a\,6{}\mathrm {i}+c\,6{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\right )\,\left (8\,b\,{\mathrm {e}}^{a\,3{}\mathrm {i}-c\,3{}\mathrm {i}}+24\,b\,{\mathrm {e}}^{a\,5{}\mathrm {i}-c\,5{}\mathrm {i}}+24\,b\,{\mathrm {e}}^{a\,7{}\mathrm {i}-c\,7{}\mathrm {i}}+8\,b\,{\mathrm {e}}^{a\,9{}\mathrm {i}-c\,9{}\mathrm {i}}\right )}{64\,b^2} \] Input:
int(sin(a + b*x)^3/cos(c + b*x),x)
Output:
exp(c*1i - a*3i - b*x*2i)/(8*b) + exp(a*3i - c*1i + b*x*2i)/(8*b) + (x*exp (c*1i - a*3i)*(exp(a*2i)*3i + exp(c*2i)*1i))/4 - (exp(c*6i - a*6i)*log(exp (a*2i)*exp(b*x*2i) + exp(a*2i)*exp(-c*2i))*(8*b*exp(a*3i - c*3i) + 24*b*ex p(a*5i - c*5i) + 24*b*exp(a*7i - c*7i) + 8*b*exp(a*9i - c*9i)))/(64*b^2)
\[ \int \sec (c+b x) \sin ^3(a+b x) \, dx=\int \sec \left (b x +c \right ) \sin \left (b x +a \right )^{3}d x \] Input:
int(sec(b*x+c)*sin(b*x+a)^3,x)
Output:
int(sec(b*x + c)*sin(a + b*x)**3,x)