Integrand size = 18, antiderivative size = 84 \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (c-b x)) \cos ^2(a+c)}{2 b}+\frac {\text {arctanh}(\sin (c-b x)) \cos (2 (a+c))}{b}+\frac {\sec (c-b x) \sin (2 (a+c))}{b}-\frac {\cos ^2(a+c) \sec (c-b x) \tan (c-b x)}{2 b} \] Output:
1/2*arctanh(sin(b*x-c))*cos(a+c)^2/b-arctanh(sin(b*x-c))*cos(2*a+2*c)/b+se c(b*x-c)*sin(2*a+2*c)/b+1/2*cos(a+c)^2*sec(b*x-c)*tan(b*x-c)/b
Time = 2.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.00 \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=-\frac {-\log \left (\cos \left (\frac {1}{2} (c-b x)\right )-\sin \left (\frac {1}{2} (c-b x)\right )\right )+3 \cos (2 (a+c)) \left (\log \left (\cos \left (\frac {1}{2} (c-b x)\right )-\sin \left (\frac {1}{2} (c-b x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c-b x)\right )+\sin \left (\frac {1}{2} (c-b x)\right )\right )\right )+\log \left (\cos \left (\frac {1}{2} (c-b x)\right )+\sin \left (\frac {1}{2} (c-b x)\right )\right )+4 \sec (c) \sin (2 (a+c))-4 \sec (c-b x) \sin (2 (a+c))+2 \cos ^2(a+c) \sec (c-b x) \tan (c-b x)}{4 b} \] Input:
Integrate[Sec[c - b*x]^3*Sin[a + b*x]^2,x]
Output:
-1/4*(-Log[Cos[(c - b*x)/2] - Sin[(c - b*x)/2]] + 3*Cos[2*(a + c)]*(Log[Co s[(c - b*x)/2] - Sin[(c - b*x)/2]] - Log[Cos[(c - b*x)/2] + Sin[(c - b*x)/ 2]]) + Log[Cos[(c - b*x)/2] + Sin[(c - b*x)/2]] + 4*Sec[c]*Sin[2*(a + c)] - 4*Sec[c - b*x]*Sin[2*(a + c)] + 2*Cos[a + c]^2*Sec[c - b*x]*Tan[c - b*x] )/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.
\(\displaystyle \int \sin ^2(a+b x) \sec ^3(c-b x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^2(a+b x) \sec ^3(c-b x)dx\) |
Input:
Int[Sec[c - b*x]^3*Sin[a + b*x]^2,x]
Output:
$Aborted
Result contains complex when optimal does not.
Time = 8.32 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.86
method | result | size |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{i \left (b x +6 a +5 c \right )}+5 \,{\mathrm e}^{3 i \left (b x +2 a +c \right )}-2 \,{\mathrm e}^{i \left (b x +4 a +3 c \right )}+2 \,{\mathrm e}^{i \left (3 b x +4 a +c \right )}-5 \,{\mathrm e}^{i \left (b x +2 a +c \right )}-3 \,{\mathrm e}^{i \left (3 b x +2 a -c \right )}\right )}{4 \left ({\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{2 i \left (b x +a \right )}\right )^{2} b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a +c \right )}\right )}{4 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a +c \right )}\right ) \cos \left (2 a +2 c \right )}{4 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a +c \right )}\right )}{4 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a +c \right )}\right ) \cos \left (2 a +2 c \right )}{4 b}\) | \(240\) |
default | \(\text {Expression too large to display}\) | \(1155\) |
Input:
int(sec(b*x-c)^3*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/4*I/(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*(b*x+a))+I*exp(I*(a+c))) -3/4/b*ln(exp(I*(b*x+a))+I*exp(I*(a+c)))*cos(2*a+2*c)-1/4/b*ln(exp(I*(b*x+ a))-I*exp(I*(a+c)))+3/4/b*ln(exp(I*(b*x+a))-I*exp(I*(a+c)))*cos(2*a+2*c)
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (90) = 180\).
Time = 0.09 (sec) , antiderivative size = 335, normalized size of antiderivative = 3.99 \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\frac {6 \, \cos \left (b x + a\right ) \cos \left (a + c\right )^{2} \sin \left (a + c\right ) + {\left (3 \, \cos \left (a + c\right )^{4} - 2 \, {\left (3 \, \cos \left (a + c\right )^{3} - 2 \, \cos \left (a + c\right )\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) - {\left (6 \, \cos \left (a + c\right )^{4} - 7 \, \cos \left (a + c\right )^{2} + 2\right )} \cos \left (b x + a\right )^{2} - 5 \, \cos \left (a + c\right )^{2} + 2\right )} \log \left (\frac {2 \, {\left (\cos \left (a + c\right ) \sin \left (b x + a\right ) - \cos \left (b x + a\right ) \sin \left (a + c\right ) + 1\right )}}{\cos \left (a + c\right ) + 1}\right ) - {\left (3 \, \cos \left (a + c\right )^{4} - 2 \, {\left (3 \, \cos \left (a + c\right )^{3} - 2 \, \cos \left (a + c\right )\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) - {\left (6 \, \cos \left (a + c\right )^{4} - 7 \, \cos \left (a + c\right )^{2} + 2\right )} \cos \left (b x + a\right )^{2} - 5 \, \cos \left (a + c\right )^{2} + 2\right )} \log \left (-\frac {2 \, {\left (\cos \left (a + c\right ) \sin \left (b x + a\right ) - \cos \left (b x + a\right ) \sin \left (a + c\right ) - 1\right )}}{\cos \left (a + c\right ) + 1}\right ) - 2 \, {\left (3 \, \cos \left (a + c\right )^{3} - 4 \, \cos \left (a + c\right )\right )} \sin \left (b x + a\right )}{4 \, {\left (2 \, b \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) + {\left (2 \, b \cos \left (a + c\right )^{2} - b\right )} \cos \left (b x + a\right )^{2} - b \cos \left (a + c\right )^{2} + b\right )}} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/4*(6*cos(b*x + a)*cos(a + c)^2*sin(a + c) + (3*cos(a + c)^4 - 2*(3*cos(a + c)^3 - 2*cos(a + c))*cos(b*x + a)*sin(b*x + a)*sin(a + c) - (6*cos(a + c)^4 - 7*cos(a + c)^2 + 2)*cos(b*x + a)^2 - 5*cos(a + c)^2 + 2)*log(2*(cos (a + c)*sin(b*x + a) - cos(b*x + a)*sin(a + c) + 1)/(cos(a + c) + 1)) - (3 *cos(a + c)^4 - 2*(3*cos(a + c)^3 - 2*cos(a + c))*cos(b*x + a)*sin(b*x + a )*sin(a + c) - (6*cos(a + c)^4 - 7*cos(a + c)^2 + 2)*cos(b*x + a)^2 - 5*co s(a + c)^2 + 2)*log(-2*(cos(a + c)*sin(b*x + a) - cos(b*x + a)*sin(a + c) - 1)/(cos(a + c) + 1)) - 2*(3*cos(a + c)^3 - 4*cos(a + c))*sin(b*x + a))/( 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 + b)
Timed out. \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate(sec(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. 1217 vs. \(2 (90) = 180\).
Time = 0.21 (sec) , antiderivative size = 1217, normalized size of antiderivative = 14.49 \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a)^2,x, algorithm="maxima")
Output:
-1/8*(2*(3*sin(3*b*x) - 5*sin(3*b*x + 4*a + 4*c) - 2*sin(3*b*x + 2*a + 2*c ) - 3*sin(b*x + 4*a + 6*c) + 2*sin(b*x + 2*a + 4*c) + 5*sin(b*x + 2*c))*co s(4*b*x + 2*a + c) + 10*(2*sin(2*b*x + 2*a + 3*c) + sin(2*a + 5*c))*cos(3* b*x + 4*a + 4*c) + 4*(2*sin(2*b*x + 2*a + 3*c) + sin(2*a + 5*c))*cos(3*b*x + 2*a + 2*c) + 4*(3*sin(3*b*x) - 3*sin(b*x + 4*a + 6*c) + 2*sin(b*x + 2*a + 4*c) + 5*sin(b*x + 2*c))*cos(2*b*x + 2*a + 3*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*c os(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 + cos(c)^2 - 2*cos(c)*sin(b *x) + sin(b*x)^2 + 2*cos(b*x)*sin(c) + sin(c)^2)/(cos(b*x)^2 + cos(c)^2 + 2*cos(c)*sin(b*x) + sin(b*x)^2 - 2*cos(b*x)*sin(c) + sin(c)^2)) + 6*cos(2* a + 5*c)*sin(3*b*x) - 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*co s(b*x + 2*c))*sin(4*b*x + 2*a + c) - 10*(2*cos(2*b*x + 2*a + 3*c) + cos...
Leaf count of result is larger than twice the leaf count of optimal. 1502 vs. \(2 (90) = 180\).
Time = 0.17 (sec) , antiderivative size = 1502, normalized size of antiderivative = 17.88 \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/2*((tan(1/2*a)^4*tan(1/2*c)^4 - 10*tan(1/2*a)^4*tan(1/2*c)^2 - 24*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)^4 + 24*ta n(1/2*a)^3*tan(1/2*c) + 52*tan(1/2*a)^2*tan(1/2*c)^2 + 24*tan(1/2*a)*tan(1 /2*c)^3 + tan(1/2*c)^4 - 10*tan(1/2*a)^2 - 24*tan(1/2*a)*tan(1/2*c) - 10*t an(1/2*c)^2 + 1)*log(abs(tan(1/2*b*x - 1/2*c) + 1))/(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*ta n(1/2*c)^2 + 1) - (tan(1/2*a)^4*tan(1/2*c)^4 - 10*tan(1/2*a)^4*tan(1/2*c)^ 2 - 24*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)^4 + 24*tan(1/2*a)^3*tan(1/2*c) + 52*tan(1/2*a)^2*tan(1/2*c)^2 + 24*tan( 1/2*a)*tan(1/2*c)^3 + tan(1/2*c)^4 - 10*tan(1/2*a)^2 - 24*tan(1/2*a)*tan(1 /2*c) - 10*tan(1/2*c)^2 + 1)*log(abs(tan(1/2*b*x - 1/2*c) - 1))/(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) - 2*(tan(1/2*b*x - 1/2*c)^3*tan(1/2*a)^4*tan(1 /2*c)^4 - 2*tan(1/2*b*x - 1/2*c)^3*tan(1/2*a)^4*tan(1/2*c)^2 - 8*tan(1/2*b *x - 1/2*c)^3*tan(1/2*a)^3*tan(1/2*c)^3 + 8*tan(1/2*b*x - 1/2*c)^2*tan(1/2 *a)^4*tan(1/2*c)^3 - 2*tan(1/2*b*x - 1/2*c)^3*tan(1/2*a)^2*tan(1/2*c)^4 + 8*tan(1/2*b*x - 1/2*c)^2*tan(1/2*a)^3*tan(1/2*c)^4 + tan(1/2*b*x - 1/2*c)* tan(1/2*a)^4*tan(1/2*c)^4 + tan(1/2*b*x - 1/2*c)^3*tan(1/2*a)^4 + 8*tan...
Timed out. \[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\text {Hanged} \] Input:
int(sin(a + b*x)^2/cos(c - b*x)^3,x)
Output:
\text{Hanged}
\[ \int \sec ^3(c-b x) \sin ^2(a+b x) \, dx=\int \sec \left (b x -c \right )^{3} \sin \left (b x +a \right )^{2}d x \] Input:
int(sec(b*x-c)^3*sin(b*x+a)^2,x)
Output:
int(sec(b*x - c)**3*sin(a + b*x)**2,x)