Integrand size = 20, antiderivative size = 51 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {(c+d x) \cos ^3(a+b x)}{3 b}+\frac {d \sin (a+b x)}{3 b^2}-\frac {d \sin ^3(a+b x)}{9 b^2} \] Output:
-1/3*(d*x+c)*cos(b*x+a)^3/b+1/3*d*sin(b*x+a)/b^2-1/9*d*sin(b*x+a)^3/b^2
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {c \cos ^3(a+b x)}{3 b}+\frac {d (-b x \cos (a+b x)+\sin (a+b x))}{4 b^2}+\frac {d (-3 b x \cos (3 (a+b x))+\sin (3 (a+b x)))}{36 b^2} \] Input:
Integrate[(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x],x]
Output:
-1/3*(c*Cos[a + b*x]^3)/b + (d*(-(b*x*Cos[a + b*x]) + Sin[a + b*x]))/(4*b^ 2) + (d*(-3*b*x*Cos[3*(a + b*x)] + Sin[3*(a + b*x)]))/(36*b^2)
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4905, 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.
\(\displaystyle \int (c+d x) \sin (a+b x) \cos ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 4905 |
\(\displaystyle \frac {d \int \cos ^3(a+b x)dx}{3 b}-\frac {(c+d x) \cos ^3(a+b x)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \int \sin \left (a+b x+\frac {\pi }{2}\right )^3dx}{3 b}-\frac {(c+d x) \cos ^3(a+b x)}{3 b}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle -\frac {d \int \left (1-\sin ^2(a+b x)\right )d(-\sin (a+b x))}{3 b^2}-\frac {(c+d x) \cos ^3(a+b x)}{3 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d \left (\frac {1}{3} \sin ^3(a+b x)-\sin (a+b x)\right )}{3 b^2}-\frac {(c+d x) \cos ^3(a+b x)}{3 b}\) |
Input:
Int[(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x],x]
Output:
-1/3*((c + d*x)*Cos[a + b*x]^3)/b - (d*(-Sin[a + b*x] + Sin[a + b*x]^3/3)) /(3*b^2)
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_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[a + b*x]^(n + 1)/(b*(n + 1 ))), x] + Simp[d*(m/(b*(n + 1))) Int[(c + d*x)^(m - 1)*Cos[a + b*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Time = 0.88 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.25
method | result | size |
risch | \(-\frac {\left (d x +c \right ) \cos \left (b x +a \right )}{4 b}+\frac {d \sin \left (b x +a \right )}{4 b^{2}}-\frac {\left (d x +c \right ) \cos \left (3 b x +3 a \right )}{12 b}+\frac {d \sin \left (3 b x +3 a \right )}{36 b^{2}}\) | \(64\) |
derivativedivides | \(\frac {\frac {d a \cos \left (b x +a \right )^{3}}{3 b}-\frac {c \cos \left (b x +a \right )^{3}}{3}+\frac {d \left (-\frac {\left (b x +a \right ) \cos \left (b x +a \right )^{3}}{3}+\frac {\left (2+\cos \left (b x +a \right )^{2}\right ) \sin \left (b x +a \right )}{9}\right )}{b}}{b}\) | \(71\) |
default | \(\frac {\frac {d a \cos \left (b x +a \right )^{3}}{3 b}-\frac {c \cos \left (b x +a \right )^{3}}{3}+\frac {d \left (-\frac {\left (b x +a \right ) \cos \left (b x +a \right )^{3}}{3}+\frac {\left (2+\cos \left (b x +a \right )^{2}\right ) \sin \left (b x +a \right )}{9}\right )}{b}}{b}\) | \(71\) |
parallelrisch | \(\frac {3 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6} x b +6 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}-18 \left (\frac {d x}{2}+c \right ) b \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+4 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+9 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2} x b +6 \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) d -6 \left (\frac {d x}{2}+c \right ) b}{9 b^{2} \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3}}\) | \(123\) |
norman | \(\frac {-\frac {2 c \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}}{b}+\frac {d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}}{b}-\frac {2 c}{3 b}+\frac {2 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{3 b^{2}}+\frac {4 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}}{9 b^{2}}+\frac {2 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}}{3 b^{2}}-\frac {d x}{3 b}-\frac {d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}}{b}+\frac {d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6}}{3 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3}}\) | \(150\) |
orering | \(\frac {4 d \left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+2 d^{2}\right ) \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}{9 b^{4} \left (d x +c \right )^{2}}-\frac {2 \left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+4 d^{2}\right ) \left (d \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \left (d x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b +\left (d x +c \right ) \cos \left (b x +a \right )^{3} b \right )}{9 b^{4} \left (d x +c \right )^{2}}+\frac {4 d \left (-4 d \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b +2 d \cos \left (b x +a \right )^{3} b +2 \left (d x +c \right ) b^{2} \sin \left (b x +a \right )^{3}-7 \left (d x +c \right ) \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) b^{2}\right )}{9 b^{4} \left (d x +c \right )}-\frac {6 d \,b^{2} \sin \left (b x +a \right )^{3}-21 d \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) b^{2}+20 \left (d x +c \right ) b^{3} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )-7 \left (d x +c \right ) \cos \left (b x +a \right )^{3} b^{3}}{9 b^{4}}\) | \(329\) |
Input:
int((d*x+c)*cos(b*x+a)^2*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
-1/4*(d*x+c)*cos(b*x+a)/b+1/4*d*sin(b*x+a)/b^2-1/12*(d*x+c)*cos(3*b*x+3*a) /b+1/36*d*sin(3*b*x+3*a)/b^2
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {3 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (d \cos \left (b x + a\right )^{2} + 2 \, d\right )} \sin \left (b x + a\right )}{9 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)^2*sin(b*x+a),x, algorithm="fricas")
Output:
-1/9*(3*(b*d*x + b*c)*cos(b*x + a)^3 - (d*cos(b*x + a)^2 + 2*d)*sin(b*x + a))/b^2
Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.67 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=\begin {cases} - \frac {c \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {d x \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac {2 d \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {d \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin {\left (a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*cos(b*x+a)**2*sin(b*x+a),x)
Output:
Piecewise((-c*cos(a + b*x)**3/(3*b) - d*x*cos(a + b*x)**3/(3*b) + 2*d*sin( a + b*x)**3/(9*b**2) + d*sin(a + b*x)*cos(a + b*x)**2/(3*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)*cos(a)**2, True))
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.69 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {12 \, c \cos \left (b x + a\right )^{3} - \frac {12 \, a d \cos \left (b x + a\right )^{3}}{b} + \frac {{\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 9 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) - 9 \, \sin \left (b x + a\right )\right )} d}{b}}{36 \, b} \] Input:
integrate((d*x+c)*cos(b*x+a)^2*sin(b*x+a),x, algorithm="maxima")
Output:
-1/36*(12*c*cos(b*x + a)^3 - 12*a*d*cos(b*x + a)^3/b + (3*(b*x + a)*cos(3* b*x + 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x + a)) *d/b)/b
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.35 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {{\left (b d x + b c\right )} \cos \left (3 \, b x + 3 \, a\right )}{12 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (b x + a\right )}{4 \, b^{2}} + \frac {d \sin \left (3 \, b x + 3 \, a\right )}{36 \, b^{2}} + \frac {d \sin \left (b x + a\right )}{4 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)^2*sin(b*x+a),x, algorithm="giac")
Output:
-1/12*(b*d*x + b*c)*cos(3*b*x + 3*a)/b^2 - 1/4*(b*d*x + b*c)*cos(b*x + a)/ b^2 + 1/36*d*sin(3*b*x + 3*a)/b^2 + 1/4*d*sin(b*x + a)/b^2
Time = 18.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=\frac {\frac {2\,d\,\sin \left (a+b\,x\right )}{9}-b\,\left (\frac {c\,{\cos \left (a+b\,x\right )}^3}{3}+\frac {d\,x\,{\cos \left (a+b\,x\right )}^3}{3}\right )+\frac {d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{9}}{b^2} \] Input:
int(cos(a + b*x)^2*sin(a + b*x)*(c + d*x),x)
Output:
((2*d*sin(a + b*x))/9 - b*((c*cos(a + b*x)^3)/3 + (d*x*cos(a + b*x)^3)/3) + (d*cos(a + b*x)^2*sin(a + b*x))/9)/b^2
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx=\frac {3 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b c +3 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b d x -3 \cos \left (b x +a \right ) b c -3 \cos \left (b x +a \right ) b d x -\sin \left (b x +a \right )^{3} d +3 \sin \left (b x +a \right ) d -3 a d +3 b c}{9 b^{2}} \] Input:
int((d*x+c)*cos(b*x+a)^2*sin(b*x+a),x)
Output:
(3*cos(a + b*x)*sin(a + b*x)**2*b*c + 3*cos(a + b*x)*sin(a + b*x)**2*b*d*x - 3*cos(a + b*x)*b*c - 3*cos(a + b*x)*b*d*x - sin(a + b*x)**3*d + 3*sin(a + b*x)*d - 3*a*d + 3*b*c)/(9*b**2)