Integrand size = 20, antiderivative size = 51 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {d \cos (a+b x)}{3 b^2}-\frac {d \cos ^3(a+b x)}{9 b^2}+\frac {(c+d x) \sin ^3(a+b x)}{3 b} \] Output:
1/3*d*cos(b*x+a)/b^2-1/9*d*cos(b*x+a)^3/b^2+1/3*(d*x+c)*sin(b*x+a)^3/b
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.86 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {9 d \cos (a+b x)-d \cos (3 (a+b x))+12 b (c+d x) \sin ^3(a+b x)}{36 b^2} \] Input:
Integrate[(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^2,x]
Output:
(9*d*Cos[a + b*x] - d*Cos[3*(a + b*x)] + 12*b*(c + d*x)*Sin[a + b*x]^3)/(3 6*b^2)
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4904, 3042, 3113, 2009}
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.
\(\displaystyle \int (c+d x) \sin ^2(a+b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 4904 |
\(\displaystyle \frac {(c+d x) \sin ^3(a+b x)}{3 b}-\frac {d \int \sin ^3(a+b x)dx}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x) \sin ^3(a+b x)}{3 b}-\frac {d \int \sin (a+b x)^3dx}{3 b}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {d \int \left (1-\cos ^2(a+b x)\right )d\cos (a+b x)}{3 b^2}+\frac {(c+d x) \sin ^3(a+b x)}{3 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d \left (\cos (a+b x)-\frac {1}{3} \cos ^3(a+b x)\right )}{3 b^2}+\frac {(c+d x) \sin ^3(a+b x)}{3 b}\) |
Input:
Int[(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^2,x]
Output:
(d*(Cos[a + b*x] - Cos[a + b*x]^3/3))/(3*b^2) + ((c + d*x)*Sin[a + b*x]^3) /(3*b)
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x _)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) , x] - Simp[d*(m/(b*(n + 1))) Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Time = 0.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(\frac {\frac {8 b \left (d x +c \right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}}{3}+\frac {4 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2} d}{3}+\frac {4 d}{9}}{b^{2} \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3}}\) | \(56\) |
risch | \(\frac {d \cos \left (b x +a \right )}{4 b^{2}}+\frac {\left (d x +c \right ) \sin \left (b x +a \right )}{4 b}-\frac {d \cos \left (3 b x +3 a \right )}{36 b^{2}}-\frac {\left (d x +c \right ) \sin \left (3 b x +3 a \right )}{12 b}\) | \(64\) |
derivativedivides | \(\frac {-\frac {d a \sin \left (b x +a \right )^{3}}{3 b}+\frac {c \sin \left (b x +a \right )^{3}}{3}+\frac {d \left (\frac {\left (b x +a \right ) \sin \left (b x +a \right )^{3}}{3}+\frac {\left (2+\sin \left (b x +a \right )^{2}\right ) \cos \left (b x +a \right )}{9}\right )}{b}}{b}\) | \(71\) |
default | \(\frac {-\frac {d a \sin \left (b x +a \right )^{3}}{3 b}+\frac {c \sin \left (b x +a \right )^{3}}{3}+\frac {d \left (\frac {\left (b x +a \right ) \sin \left (b x +a \right )^{3}}{3}+\frac {\left (2+\sin \left (b x +a \right )^{2}\right ) \cos \left (b x +a \right )}{9}\right )}{b}}{b}\) | \(71\) |
norman | \(\frac {\frac {4 d}{9 b^{2}}+\frac {8 c \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}}{3 b}+\frac {4 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}}{3 b^{2}}+\frac {8 d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}}{3 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3}}\) | \(76\) |
orering | \(\frac {4 d \left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+2 d^{2}\right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}}{9 b^{4} \left (d x +c \right )^{2}}-\frac {2 \left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+4 d^{2}\right ) \left (d \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-\left (d x +c \right ) b \sin \left (b x +a \right )^{3}+2 \left (d x +c \right ) \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) b \right )}{9 b^{4} \left (d x +c \right )^{2}}+\frac {4 d \left (-2 d b \sin \left (b x +a \right )^{3}+4 d \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) b -7 \left (d x +c \right ) b^{2} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \left (d x +c \right ) \cos \left (b x +a \right )^{3} b^{2}\right )}{9 b^{4} \left (d x +c \right )}-\frac {-21 d \,b^{2} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )+6 d \cos \left (b x +a \right )^{3} b^{2}-20 \left (d x +c \right ) b^{3} \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}+7 \left (d x +c \right ) b^{3} \sin \left (b x +a \right )^{3}}{9 b^{4}}\) | \(330\) |
Input:
int((d*x+c)*cos(b*x+a)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
4/9*(6*b*(d*x+c)*tan(1/2*a+1/2*b*x)^3+3*tan(1/2*a+1/2*b*x)^2*d+d)/b^2/(1+t an(1/2*a+1/2*b*x)^2)^3
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=-\frac {d \cos \left (b x + a\right )^{3} - 3 \, d \cos \left (b x + a\right ) - 3 \, {\left (b d x - {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + b c\right )} \sin \left (b x + a\right )}{9 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-1/9*(d*cos(b*x + a)^3 - 3*d*cos(b*x + a) - 3*(b*d*x - (b*d*x + b*c)*cos(b *x + a)^2 + b*c)*sin(b*x + a))/b^2
Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.67 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\begin {cases} \frac {c \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {d x \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {d \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{3 b^{2}} + \frac {2 d \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)**2,x)
Output:
Piecewise((c*sin(a + b*x)**3/(3*b) + d*x*sin(a + b*x)**3/(3*b) + d*sin(a + b*x)**2*cos(a + b*x)/(3*b**2) + 2*d*cos(a + b*x)**3/(9*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)**2*cos(a), True))
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.67 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {12 \, c \sin \left (b x + a\right )^{3} - \frac {12 \, a d \sin \left (b x + a\right )^{3}}{b} - \frac {{\left (3 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 9 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) - 9 \, \cos \left (b x + a\right )\right )} d}{b}}{36 \, b} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
1/36*(12*c*sin(b*x + a)^3 - 12*a*d*sin(b*x + a)^3/b - (3*(b*x + a)*sin(3*b *x + 3*a) - 9*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) - 9*cos(b*x + a))* d/b)/b
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.35 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=-\frac {d \cos \left (3 \, b x + 3 \, a\right )}{36 \, b^{2}} + \frac {d \cos \left (b x + a\right )}{4 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (3 \, b x + 3 \, a\right )}{12 \, b^{2}} + \frac {{\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/36*d*cos(3*b*x + 3*a)/b^2 + 1/4*d*cos(b*x + a)/b^2 - 1/12*(b*d*x + b*c) *sin(3*b*x + 3*a)/b^2 + 1/4*(b*d*x + b*c)*sin(b*x + a)/b^2
Time = 18.65 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {\frac {2\,d\,{\cos \left (a+b\,x\right )}^3}{9}+b\,\left (\frac {c\,{\sin \left (a+b\,x\right )}^3}{3}+\frac {d\,x\,{\sin \left (a+b\,x\right )}^3}{3}\right )+\frac {d\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{3}}{b^2} \] Input:
int(cos(a + b*x)*sin(a + b*x)^2*(c + d*x),x)
Output:
((2*d*cos(a + b*x)^3)/9 + b*((c*sin(a + b*x)^3)/3 + (d*x*sin(a + b*x)^3)/3 ) + (d*cos(a + b*x)*sin(a + b*x)^2)/3)/b^2
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (c+d x) \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} d +2 \cos \left (b x +a \right ) d +3 \sin \left (b x +a \right )^{3} b c +3 \sin \left (b x +a \right )^{3} b d x +2 d}{9 b^{2}} \] Input:
int((d*x+c)*cos(b*x+a)*sin(b*x+a)^2,x)
Output:
(cos(a + b*x)*sin(a + b*x)**2*d + 2*cos(a + b*x)*d + 3*sin(a + b*x)**3*b*c + 3*sin(a + b*x)**3*b*d*x + 2*d)/(9*b**2)