Integrand size = 15, antiderivative size = 157 \[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 d}-\frac {i e^{-2 i a-2 i b x+i (c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {b}{d},\frac {3}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (2 b-d)}+\frac {i e^{2 i a+2 i b x+i (c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {b}{d},\frac {3}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (2 b+d)} \] Output:
1/2*arctanh(sin(d*x+c))/d-1/2*I*exp(-2*I*a-2*I*b*x+I*(d*x+c))*hypergeom([1 , 1/2-b/d],[3/2-b/d],-exp(2*I*(d*x+c)))/(2*b-d)+1/2*I*exp(2*I*a+2*I*b*x+I* (d*x+c))*hypergeom([1, 1/2+b/d],[3/2+b/d],-exp(2*I*(d*x+c)))/(2*b+d)
Time = 0.51 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.22 \[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=-\frac {i e^{-i (2 a-c+2 b x-d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {b}{d},\frac {3}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (2 b-d)}+\frac {i e^{i (2 a+c+2 b x+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {b}{d},\frac {3}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (2 b+d)}+\frac {-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]*Sin[a + b*x]^2,x]
Output:
((-1/2*I)*Hypergeometric2F1[1, 1/2 - b/d, 3/2 - b/d, -E^((2*I)*(c + d*x))] )/((2*b - d)*E^(I*(2*a - c + 2*b*x - d*x))) + ((I/2)*E^(I*(2*a + c + 2*b*x + d*x))*Hypergeometric2F1[1, 1/2 + b/d, 3/2 + b/d, -E^((2*I)*(c + d*x))]) /(2*b + d) + (-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x )/2] + Sin[(c + d*x)/2]])/(2*d)
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.
\(\displaystyle \int \sin ^2(a+b x) \sec (c+d x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^2(a+b x) \sec (c+d x)dx\) |
Input:
Int[Sec[c + d*x]*Sin[a + b*x]^2,x]
Output:
$Aborted
\[\int \sec \left (d x +c \right ) \sin \left (b x +a \right )^{2}d x\]
Input:
int(sec(d*x+c)*sin(b*x+a)^2,x)
Output:
int(sec(d*x+c)*sin(b*x+a)^2,x)
\[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right ) \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
integral(-(cos(b*x + a)^2 - 1)*sec(d*x + c), x)
\[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int \sin ^{2}{\left (a + b x \right )} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*sin(b*x+a)**2,x)
Output:
Integral(sin(a + b*x)**2*sec(c + d*x), x)
\[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right ) \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
integrate(sec(d*x + c)*sin(b*x + a)^2, x)
\[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right ) \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate(sec(d*x + c)*sin(b*x + a)^2, x)
Timed out. \[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{\cos \left (c+d\,x\right )} \,d x \] Input:
int(sin(a + b*x)^2/cos(c + d*x),x)
Output:
int(sin(a + b*x)^2/cos(c + d*x), x)
\[ \int \sec (c+d x) \sin ^2(a+b x) \, dx=\int \frac {\sin \left (b x +a \right )^{2}}{\cos \left (d x +c \right )}d x \] Input:
int(sec(d*x+c)*sin(b*x+a)^2,x)
Output:
int(sin(a + b*x)**2/cos(c + d*x),x)