Integrand size = 20, antiderivative size = 89 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=-\frac {c d x}{2 b}-\frac {d^2 x^2}{4 b}+\frac {d (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac {d^2 \sin ^2(a+b x)}{4 b^3}+\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b} \]
-1/2*c*d*x/b-1/4*d^2*x^2/b+1/2*d*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b^2-1/4*d^2 *sin(b*x+a)^2/b^3+1/2*(d*x+c)^2*sin(b*x+a)^2/b
Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.56 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=\frac {\left (d^2-2 b^2 (c+d x)^2\right ) \cos (2 (a+b x))+2 b d (c+d x) \sin (2 (a+b x))}{8 b^3} \]
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4904, 3042, 3791, 17}
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.
\(\displaystyle \int (c+d x)^2 \sin (a+b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 4904 |
\(\displaystyle \frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \sin ^2(a+b x)dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \sin (a+b x)^2dx}{b}\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {d \left (\frac {1}{2} \int (c+d x)dx+\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}\right )}{b}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {d \left (\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^2}{4 d}\right )}{b}\) |
((c + d*x)^2*Sin[a + b*x]^2)/(2*b) - (d*((c + d*x)^2/(4*d) - ((c + d*x)*Co s[a + b*x]*Sin[a + b*x])/(2*b) + (d*Sin[a + b*x]^2)/(4*b^2)))/b
3.1.4.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
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 = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}-d^{2}\right ) \cos \left (2 x b +2 a \right )}{8 b^{3}}+\frac {d \left (d x +c \right ) \sin \left (2 x b +2 a \right )}{4 b^{2}}\) | \(69\) |
parallelrisch | \(\frac {-b^{2} x d \left (\frac {d x}{2}+c \right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-2 b d \left (d x +c \right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}+\left (\left (3 x^{2} d^{2}+6 c d x +4 c^{2}\right ) b^{2}-2 d^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+2 b d \left (d x +c \right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )-b^{2} x d \left (\frac {d x}{2}+c \right )}{2 b^{3} \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{2}}\) | \(138\) |
derivativedivides | \(\frac {-\frac {a^{2} d^{2} \cos \left (x b +a \right )^{2}}{2 b^{2}}+\frac {a c d \cos \left (x b +a \right )^{2}}{b}-\frac {2 a \,d^{2} \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b^{2}}-\frac {c^{2} \cos \left (x b +a \right )^{2}}{2}+\frac {2 c d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b}+\frac {d^{2} \left (-\frac {\left (x b +a \right )^{2} \cos \left (x b +a \right )^{2}}{2}+\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}\right )}{b^{2}}}{b}\) | \(215\) |
default | \(\frac {-\frac {a^{2} d^{2} \cos \left (x b +a \right )^{2}}{2 b^{2}}+\frac {a c d \cos \left (x b +a \right )^{2}}{b}-\frac {2 a \,d^{2} \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b^{2}}-\frac {c^{2} \cos \left (x b +a \right )^{2}}{2}+\frac {2 c d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b}+\frac {d^{2} \left (-\frac {\left (x b +a \right )^{2} \cos \left (x b +a \right )^{2}}{2}+\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}\right )}{b^{2}}}{b}\) | \(215\) |
norman | \(\frac {\frac {\left (2 b^{2} c^{2}-d^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b^{3}}+\frac {c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{b^{2}}+\frac {d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{b^{2}}-\frac {d^{2} x^{2}}{4 b}-\frac {c d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{b^{2}}-\frac {c d x}{2 b}-\frac {d^{2} x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{b^{2}}+\frac {3 d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{2 b}-\frac {d^{2} x^{2} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{4 b}+\frac {3 c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b}-\frac {c d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{2 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{2}}\) | \(218\) |
-1/8*(2*b^2*d^2*x^2+4*b^2*c*d*x+2*b^2*c^2-d^2)/b^3*cos(2*b*x+2*a)+1/4*d*(d *x+c)*sin(2*b*x+2*a)/b^2
Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{3}} \]
1/4*(b^2*d^2*x^2 + 2*b^2*c*d*x - (2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - d^2)*cos(b*x + a)^2 + 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a))/b^3
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (78) = 156\).
Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.97 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=\begin {cases} \frac {c^{2} \sin ^{2}{\left (a + b x \right )}}{2 b} + \frac {c d x \sin ^{2}{\left (a + b x \right )}}{2 b} - \frac {c d x \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {c d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{2}} + \frac {d^{2} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{2}} - \frac {d^{2} \sin ^{2}{\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin {\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]
Piecewise((c**2*sin(a + b*x)**2/(2*b) + c*d*x*sin(a + b*x)**2/(2*b) - c*d* x*cos(a + b*x)**2/(2*b) + d**2*x**2*sin(a + b*x)**2/(4*b) - d**2*x**2*cos( a + b*x)**2/(4*b) + c*d*sin(a + b*x)*cos(a + b*x)/(2*b**2) + d**2*x*sin(a + b*x)*cos(a + b*x)/(2*b**2) - d**2*sin(a + b*x)**2/(4*b**3), Ne(b, 0)), ( (c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)*cos(a), True))
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (79) = 158\).
Time = 0.23 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.92 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=-\frac {4 \, c^{2} \cos \left (b x + a\right )^{2} - \frac {8 \, a c d \cos \left (b x + a\right )^{2}}{b} + \frac {4 \, a^{2} d^{2} \cos \left (b x + a\right )^{2}}{b^{2}} + \frac {2 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} c d}{b} - \frac {2 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{8 \, b} \]
-1/8*(4*c^2*cos(b*x + a)^2 - 8*a*c*d*cos(b*x + a)^2/b + 4*a^2*d^2*cos(b*x + a)^2/b^2 + 2*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*c*d/b - 2 *(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a*d^2/b^2 + ((2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*d^2/b^2)/b
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=-\frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{3}} \]
-1/8*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - d^2)*cos(2*b*x + 2*a)/b^3 + 1/4*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a)/b^3
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int (c+d x)^2 \cos (a+b x) \sin (a+b x) \, dx=\frac {\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {d^2}{4}-\frac {b^2\,c^2}{2}\right )}{2\,b^3}+\frac {d^2\,x\,\sin \left (2\,a+2\,b\,x\right )}{4\,b^2}-\frac {d^2\,x^2\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}+\frac {c\,d\,\sin \left (2\,a+2\,b\,x\right )}{4\,b^2}-\frac {c\,d\,x\,\cos \left (2\,a+2\,b\,x\right )}{2\,b} \]