Integrand size = 31, antiderivative size = 81 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {1}{2} (b B+a (2 A+C)) x-\frac {(A b+a B+b C) \cos (c+d x)}{d}+\frac {b C \cos ^3(c+d x)}{3 d}-\frac {(b B+a C) \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*(b*B+a*(2*A+C))*x-(A*b+B*a+C*b)*cos(d*x+c)/d+1/3*b*C*cos(d*x+c)^3/d-1/ 2*(B*b+C*a)*cos(d*x+c)*sin(d*x+c)/d
Time = 0.48 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {6 b B c+6 a c C+12 a A d x+6 b B d x+6 a C d x-3 (4 A b+4 a B+3 b C) \cos (c+d x)+b C \cos (3 (c+d x))-3 b B \sin (2 (c+d x))-3 a C \sin (2 (c+d x))}{12 d} \] Input:
Integrate[(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x] + C*Sin[c + d*x]^2),x]
Output:
(6*b*B*c + 6*a*c*C + 12*a*A*d*x + 6*b*B*d*x + 6*a*C*d*x - 3*(4*A*b + 4*a*B + 3*b*C)*Cos[c + d*x] + b*C*Cos[3*(c + d*x)] - 3*b*B*Sin[2*(c + d*x)] - 3 *a*C*Sin[2*(c + d*x)])/(12*d)
Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.46, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 3502, 3042, 3213}
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.
\(\displaystyle \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\int (a+b \sin (c+d x)) (b (3 A+2 C)+(3 b B-a C) \sin (c+d x))dx}{3 b}-\frac {C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (a+b \sin (c+d x)) (b (3 A+2 C)+(3 b B-a C) \sin (c+d x))dx}{3 b}-\frac {C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {-\frac {\cos (c+d x) \left (a (3 b B-a C)+b^2 (3 A+2 C)\right )}{d}+\frac {3}{2} b x (a (2 A+C)+b B)-\frac {b (3 b B-a C) \sin (c+d x) \cos (c+d x)}{2 d}}{3 b}-\frac {C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d}\) |
Input:
Int[(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x] + C*Sin[c + d*x]^2),x]
Output:
-1/3*(C*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(b*d) + ((3*b*(b*B + a*(2*A + C))*x)/2 - ((b^2*(3*A + 2*C) + a*(3*b*B - a*C))*Cos[c + d*x])/d - (b*(3*b *B - a*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/(3*b)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 2.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99
method | result | size |
parts | \(x A a -\frac {\left (A b +B a \right ) \cos \left (d x +c \right )}{d}+\frac {\left (B b +C a \right ) \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {C b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}\) | \(80\) |
parallelrisch | \(\frac {\left (-3 B b -3 C a \right ) \sin \left (2 d x +2 c \right )+b C \cos \left (3 d x +3 c \right )+\left (\left (-12 A -9 C \right ) b -12 B a \right ) \cos \left (d x +c \right )+\left (6 B d x -12 A -8 C \right ) b +12 \left (A d x +\frac {1}{2} C d x -B \right ) a}{12 d}\) | \(89\) |
risch | \(x A a +\frac {x B b}{2}+\frac {x C a}{2}-\frac {\cos \left (d x +c \right ) A b}{d}-\frac {\cos \left (d x +c \right ) B a}{d}-\frac {3 \cos \left (d x +c \right ) C b}{4 d}+\frac {b C \cos \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}-\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(103\) |
derivativedivides | \(\frac {-\frac {C b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+B b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A b \cos \left (d x +c \right )-B a \cos \left (d x +c \right )+A a \left (d x +c \right )}{d}\) | \(104\) |
default | \(\frac {-\frac {C b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+B b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A b \cos \left (d x +c \right )-B a \cos \left (d x +c \right )+A a \left (d x +c \right )}{d}\) | \(104\) |
norman | \(\frac {\left (A a +\frac {1}{2} B b +\frac {1}{2} C a \right ) x +\left (A a +\frac {1}{2} B b +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 A a +\frac {3}{2} B b +\frac {3}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 A a +\frac {3}{2} B b +\frac {3}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {6 A b +6 B a +4 C b}{3 d}-\frac {\left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 A b +2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {\left (4 A b +4 B a +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(224\) |
orering | \(\text {Expression too large to display}\) | \(1202\) |
Input:
int((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x,method=_RETURNVERBO SE)
Output:
x*A*a-(A*b+B*a)/d*cos(d*x+c)+(B*b+C*a)/d*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d *x+1/2*c)-1/3*C*b/d*(2+sin(d*x+c)^2)*cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {2 \, C b \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} d x - 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, {\left (B a + {\left (A + C\right )} b\right )} \cos \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x, algorithm="f ricas")
Output:
1/6*(2*C*b*cos(d*x + c)^3 + 3*((2*A + C)*a + B*b)*d*x - 3*(C*a + B*b)*cos( d*x + c)*sin(d*x + c) - 6*(B*a + (A + C)*b)*cos(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (73) = 146\).
Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.33 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\begin {cases} A a x - \frac {A b \cos {\left (c + d x \right )}}{d} - \frac {B a \cos {\left (c + d x \right )}}{d} + \frac {B b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {B b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {C a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {C b \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 C b \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \left (A + B \sin {\left (c \right )} + C \sin ^{2}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)**2),x)
Output:
Piecewise((A*a*x - A*b*cos(c + d*x)/d - B*a*cos(c + d*x)/d + B*b*x*sin(c + d*x)**2/2 + B*b*x*cos(c + d*x)**2/2 - B*b*sin(c + d*x)*cos(c + d*x)/(2*d) + C*a*x*sin(c + d*x)**2/2 + C*a*x*cos(c + d*x)**2/2 - C*a*sin(c + d*x)*co s(c + d*x)/(2*d) - C*b*sin(c + d*x)**2*cos(c + d*x)/d - 2*C*b*cos(c + d*x) **3/(3*d), Ne(d, 0)), (x*(a + b*sin(c))*(A + B*sin(c) + C*sin(c)**2), True ))
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.26 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (d x + c\right )} A a + 3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} B b + 4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} C b - 12 \, B a \cos \left (d x + c\right ) - 12 \, A b \cos \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x, algorithm="m axima")
Output:
1/12*(12*(d*x + c)*A*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*C*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*B*b + 4*(cos(d*x + c)^3 - 3*cos(d*x + c))*C*b - 12*B*a*cos(d*x + c) - 12*A*b*cos(d*x + c))/d
Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {1}{2} \, {\left (2 \, A a + C a + B b\right )} x + \frac {C b \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {{\left (4 \, B a + 4 \, A b + 3 \, C b\right )} \cos \left (d x + c\right )}{4 \, d} - \frac {{\left (C a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \] Input:
integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x, algorithm="g iac")
Output:
1/2*(2*A*a + C*a + B*b)*x + 1/12*C*b*cos(3*d*x + 3*c)/d - 1/4*(4*B*a + 4*A *b + 3*C*b)*cos(d*x + c)/d - 1/4*(C*a + B*b)*sin(2*d*x + 2*c)/d
Time = 38.76 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=-\frac {6\,A\,b\,\cos \left (c+d\,x\right )+6\,B\,a\,\cos \left (c+d\,x\right )+\frac {9\,C\,b\,\cos \left (c+d\,x\right )}{2}-\frac {C\,b\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,B\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,\sin \left (2\,c+2\,d\,x\right )}{2}-6\,A\,a\,d\,x-3\,B\,b\,d\,x-3\,C\,a\,d\,x}{6\,d} \] Input:
int((a + b*sin(c + d*x))*(A + B*sin(c + d*x) + C*sin(c + d*x)^2),x)
Output:
-(6*A*b*cos(c + d*x) + 6*B*a*cos(c + d*x) + (9*C*b*cos(c + d*x))/2 - (C*b* cos(3*c + 3*d*x))/2 + (3*B*b*sin(2*c + 2*d*x))/2 + (3*C*a*sin(2*c + 2*d*x) )/2 - 6*A*a*d*x - 3*B*b*d*x - 3*C*a*d*x)/(6*d)
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-12 \cos \left (d x +c \right ) a b -4 \cos \left (d x +c \right ) b c +6 a^{2} d x +12 a b +3 a c d x +3 b^{2} d x +4 b c}{6 d} \] Input:
int((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**2*b*c - 3*cos(c + d*x)*sin(c + d*x)*a*c - 3*cos(c + d*x)*sin(c + d*x)*b**2 - 12*cos(c + d*x)*a*b - 4*cos(c + d*x)*b *c + 6*a**2*d*x + 12*a*b + 3*a*c*d*x + 3*b**2*d*x + 4*b*c)/(6*d)