Integrand size = 17, antiderivative size = 88 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {x}{4}-\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 (a-c)+2 (b-d) x)}{16 (b-d)}-\frac {\sin (2 c+2 d x)}{8 d}+\frac {\sin (2 (a+c)+2 (b+d) x)}{16 (b+d)} \]
[Out]
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4665, 2717} \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {\sin (2 (a-c)+2 x (b-d))}{16 (b-d)}+\frac {\sin (2 (a+c)+2 x (b+d))}{16 (b+d)}-\frac {\sin (2 a+2 b x)}{8 b}-\frac {\sin (2 c+2 d x)}{8 d}+\frac {x}{4} \]
[In]
[Out]
Rule 2717
Rule 4665
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4}-\frac {1}{4} \cos (2 a+2 b x)+\frac {1}{8} \cos (2 (a-c)+2 (b-d) x)-\frac {1}{4} \cos (2 c+2 d x)+\frac {1}{8} \cos (2 (a+c)+2 (b+d) x)\right ) \, dx \\ & = \frac {x}{4}+\frac {1}{8} \int \cos (2 (a-c)+2 (b-d) x) \, dx+\frac {1}{8} \int \cos (2 (a+c)+2 (b+d) x) \, dx-\frac {1}{4} \int \cos (2 a+2 b x) \, dx-\frac {1}{4} \int \cos (2 c+2 d x) \, dx \\ & = \frac {x}{4}-\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 (a-c)+2 (b-d) x)}{16 (b-d)}-\frac {\sin (2 c+2 d x)}{8 d}+\frac {\sin (2 (a+c)+2 (b+d) x)}{16 (b+d)} \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {\left (-2 b^2 d+2 d^3\right ) \sin (2 (a+b x))+b d (b+d) \sin (2 (a-c+(b-d) x))+b (b-d) (-2 (b+d) \sin (2 (c+d x))+d (4 (b+d) x+\sin (2 (a+c+(b+d) x))))}{16 b (b-d) d (b+d)} \]
[In]
[Out]
Time = 0.82 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {x}{4}-\frac {\sin \left (2 x b +2 a \right )}{8 b}-\frac {\sin \left (2 d x +2 c \right )}{8 d}+\frac {\sin \left (\left (2 b -2 d \right ) x +2 a -2 c \right )}{16 b -16 d}+\frac {\sin \left (\left (2 b +2 d \right ) x +2 a +2 c \right )}{16 b +16 d}\) | \(89\) |
parallelrisch | \(\frac {b d \left (b +d \right ) \sin \left (\left (2 b -2 d \right ) x +2 a -2 c \right )+4 \left (\frac {b d \sin \left (\left (2 b +2 d \right ) x +2 a +2 c \right )}{4}+\left (b +d \right ) \left (-\frac {d \sin \left (2 x b +2 a \right )}{2}+b \left (d x -\frac {\sin \left (2 d x +2 c \right )}{2}\right )\right )\right ) \left (b -d \right )}{16 b^{3} d -16 b \,d^{3}}\) | \(104\) |
risch | \(\frac {x}{4}-\frac {\sin \left (2 x b +2 a \right )}{8 b}-\frac {\sin \left (2 d x +2 c \right ) b^{2}}{8 d \left (b -d \right ) \left (b +d \right )}+\frac {d \sin \left (2 d x +2 c \right )}{8 \left (b -d \right ) \left (b +d \right )}+\frac {\sin \left (2 x b -2 d x +2 a -2 c \right ) b}{16 \left (b -d \right ) \left (b +d \right )}+\frac {d \sin \left (2 x b -2 d x +2 a -2 c \right )}{16 \left (b -d \right ) \left (b +d \right )}+\frac {\sin \left (2 x b +2 d x +2 a +2 c \right ) b}{16 \left (b -d \right ) \left (b +d \right )}-\frac {d \sin \left (2 x b +2 d x +2 a +2 c \right )}{16 \left (b -d \right ) \left (b +d \right )}\) | \(196\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.34 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=-\frac {{\left (2 \, b d^{2} \cos \left (b x + a\right )^{2} + b^{3} - 2 \, b d^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b^{3} d - b d^{3}\right )} x - {\left (2 \, b^{2} d \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} - {\left (2 \, b^{2} d - d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{4 \, {\left (b^{3} d - b d^{3}\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (76) = 152\).
Time = 1.53 (sec) , antiderivative size = 1027, normalized size of antiderivative = 11.67 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\text {Too large to display} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (78) = 156\).
Time = 0.25 (sec) , antiderivative size = 620, normalized size of antiderivative = 7.05 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {8 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )} x + {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) - {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) - {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) + {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) - 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a - 2 \, c\right ) + 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x\right ) - 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x + 4 \, c\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) + 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a - 2 \, c\right ) + 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x\right ) + 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x + 4 \, c\right )}{32 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )}} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {1}{4} \, x + \frac {\sin \left (2 \, b x + 2 \, d x + 2 \, a + 2 \, c\right )}{16 \, {\left (b + d\right )}} + \frac {\sin \left (2 \, b x - 2 \, d x + 2 \, a - 2 \, c\right )}{16 \, {\left (b - d\right )}} - \frac {\sin \left (2 \, b x + 2 \, a\right )}{8 \, b} - \frac {\sin \left (2 \, d x + 2 \, c\right )}{8 \, d} \]
[In]
[Out]
Time = 20.80 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.01 \[ \int \sin ^2(a+b x) \sin ^2(c+d x) \, dx=\frac {2\,d^3\,\sin \left (2\,a+2\,b\,x\right )-2\,b^3\,\sin \left (2\,c+2\,d\,x\right )+b\,d^2\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )-b\,d^2\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )+b^2\,d\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )+b^2\,d\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )-2\,b^2\,d\,\sin \left (2\,a+2\,b\,x\right )+2\,b\,d^2\,\sin \left (2\,c+2\,d\,x\right )-4\,b\,d^3\,x+4\,b^3\,d\,x}{16\,b\,d\,\left (b^2-d^2\right )} \]
[In]
[Out]