Integrand size = 14, antiderivative size = 50 \[ \int (c+d x)^2 \sin (a+b x) \, dx=\frac {2 d^2 \cos (a+b x)}{b^3}-\frac {(c+d x)^2 \cos (a+b x)}{b}+\frac {2 d (c+d x) \sin (a+b x)}{b^2} \] Output:
2*d^2*cos(b*x+a)/b^3-(d*x+c)^2*cos(b*x+a)/b+2*d*(d*x+c)*sin(b*x+a)/b^2
Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int (c+d x)^2 \sin (a+b x) \, dx=\frac {-\left (\left (-2 d^2+b^2 (c+d x)^2\right ) \cos (a+b x)\right )+2 b d (c+d x) \sin (a+b x)}{b^3} \] Input:
Integrate[(c + d*x)^2*Sin[a + b*x],x]
Output:
(-((-2*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x]) + 2*b*d*(c + d*x)*Sin[a + b*x] )/b^3
Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3777, 3042, 3777, 25, 3042, 3118}
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.
\(\displaystyle \int (c+d x)^2 \sin (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x)^2 \sin (a+b x)dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {2 d \int (c+d x) \cos (a+b x)dx}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \int (c+d x) \sin \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {2 d \left (\frac {d \int -\sin (a+b x)dx}{b}+\frac {(c+d x) \sin (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 d \left (\frac {(c+d x) \sin (a+b x)}{b}-\frac {d \int \sin (a+b x)dx}{b}\right )}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \left (\frac {(c+d x) \sin (a+b x)}{b}-\frac {d \int \sin (a+b x)dx}{b}\right )}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {2 d \left (\frac {d \cos (a+b x)}{b^2}+\frac {(c+d x) \sin (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \cos (a+b x)}{b}\) |
Input:
Int[(c + d*x)^2*Sin[a + b*x],x]
Output:
-(((c + d*x)^2*Cos[a + b*x])/b) + (2*d*((d*Cos[a + b*x])/b^2 + ((c + d*x)* Sin[a + b*x])/b))/b
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Time = 0.82 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22
method | result | size |
risch | \(-\frac {\left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}-2 d^{2}\right ) \cos \left (b x +a \right )}{b^{3}}+\frac {2 d \left (d x +c \right ) \sin \left (b x +a \right )}{b^{2}}\) | \(61\) |
parallelrisch | \(\frac {2 \left (\frac {d x}{2}+c \right ) d \,b^{2} x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+4 b d \left (d x +c \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+\left (-x^{2} d^{2}-2 c d x -2 c^{2}\right ) b^{2}+4 d^{2}}{b^{3} \left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\) | \(91\) |
parts | \(-\frac {\cos \left (b x +a \right ) x^{2} d^{2}}{b}-\frac {2 \cos \left (b x +a \right ) c d x}{b}-\frac {\cos \left (b x +a \right ) c^{2}}{b}+\frac {2 d \left (-\frac {d a \sin \left (b x +a \right )}{b}+c \sin \left (b x +a \right )+\frac {d \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{b}\right )}{b^{2}}\) | \(99\) |
orering | \(\frac {4 d \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}-d^{2}\right ) \sin \left (b x +a \right )}{b^{4} \left (d x +c \right )}-\frac {\left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}-2 d^{2}\right ) \left (2 \left (d x +c \right ) \sin \left (b x +a \right ) d +\left (d x +c \right )^{2} b \cos \left (b x +a \right )\right )}{b^{4} \left (d x +c \right )^{2}}\) | \(125\) |
norman | \(\frac {\frac {-2 b^{2} c^{2}+4 d^{2}}{b^{3}}+\frac {d^{2} x^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{b}-\frac {d^{2} x^{2}}{b}-\frac {2 c d x}{b}+\frac {4 c d \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b^{2}}+\frac {4 d^{2} x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b^{2}}+\frac {2 c d x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{b}}{1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\) | \(129\) |
derivativedivides | \(\frac {-\frac {a^{2} d^{2} \cos \left (b x +a \right )}{b^{2}}+\frac {2 a c d \cos \left (b x +a \right )}{b}-\frac {2 a \,d^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{2}}-c^{2} \cos \left (b x +a \right )+\frac {2 c d \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b}+\frac {d^{2} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{2}}}{b}\) | \(148\) |
default | \(\frac {-\frac {a^{2} d^{2} \cos \left (b x +a \right )}{b^{2}}+\frac {2 a c d \cos \left (b x +a \right )}{b}-\frac {2 a \,d^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{2}}-c^{2} \cos \left (b x +a \right )+\frac {2 c d \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b}+\frac {d^{2} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{2}}}{b}\) | \(148\) |
meijerg | \(\frac {4 d^{2} \sin \left (a \right ) \sqrt {\pi }\, \left (\frac {x \left (b^{2}\right )^{\frac {3}{2}} \cos \left (b x \right )}{2 \sqrt {\pi }\, b^{2}}-\frac {\left (b^{2}\right )^{\frac {3}{2}} \left (-\frac {3 x^{2} b^{2}}{2}+3\right ) \sin \left (b x \right )}{6 \sqrt {\pi }\, b^{3}}\right )}{b^{2} \sqrt {b^{2}}}+\frac {4 d^{2} \cos \left (a \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {x^{2} b^{2}}{2}+1\right ) \cos \left (b x \right )}{2 \sqrt {\pi }}+\frac {x b \sin \left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{3}}+\frac {4 d c \sin \left (a \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (b x \right )}{2 \sqrt {\pi }}+\frac {x b \sin \left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{2}}+\frac {4 d c \cos \left (a \right ) \sqrt {\pi }\, \left (-\frac {x b \cos \left (b x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{2}}+\frac {c^{2} \sin \left (a \right ) \sin \left (b x \right )}{b}+\frac {c^{2} \cos \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (b x \right )}{\sqrt {\pi }}\right )}{b}\) | \(224\) |
Input:
int((d*x+c)^2*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
-(b^2*d^2*x^2+2*b^2*c*d*x+b^2*c^2-2*d^2)/b^3*cos(b*x+a)+2*d*(d*x+c)*sin(b* x+a)/b^2
Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int (c+d x)^2 \sin (a+b x) \, dx=-\frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right ) - 2 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \] Input:
integrate((d*x+c)^2*sin(b*x+a),x, algorithm="fricas")
Output:
-((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a) - 2*(b*d^2*x + b*c*d)*sin(b*x + a))/b^3
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (48) = 96\).
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.24 \[ \int (c+d x)^2 \sin (a+b x) \, dx=\begin {cases} - \frac {c^{2} \cos {\left (a + b x \right )}}{b} - \frac {2 c d x \cos {\left (a + b x \right )}}{b} - \frac {d^{2} x^{2} \cos {\left (a + b x \right )}}{b} + \frac {2 c d \sin {\left (a + b x \right )}}{b^{2}} + \frac {2 d^{2} x \sin {\left (a + b x \right )}}{b^{2}} + \frac {2 d^{2} \cos {\left (a + b x \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin {\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**2*sin(b*x+a),x)
Output:
Piecewise((-c**2*cos(a + b*x)/b - 2*c*d*x*cos(a + b*x)/b - d**2*x**2*cos(a + b*x)/b + 2*c*d*sin(a + b*x)/b**2 + 2*d**2*x*sin(a + b*x)/b**2 + 2*d**2* cos(a + b*x)/b**3, Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a), T rue))
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (50) = 100\).
Time = 0.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.82 \[ \int (c+d x)^2 \sin (a+b x) \, dx=-\frac {c^{2} \cos \left (b x + a\right ) - \frac {2 \, a c d \cos \left (b x + a\right )}{b} + \frac {a^{2} d^{2} \cos \left (b x + a\right )}{b^{2}} + \frac {2 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} c d}{b} - \frac {2 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{b} \] Input:
integrate((d*x+c)^2*sin(b*x+a),x, algorithm="maxima")
Output:
-(c^2*cos(b*x + a) - 2*a*c*d*cos(b*x + a)/b + a^2*d^2*cos(b*x + a)/b^2 + 2 *((b*x + a)*cos(b*x + a) - sin(b*x + a))*c*d/b - 2*((b*x + a)*cos(b*x + a) - sin(b*x + a))*a*d^2/b^2 + (((b*x + a)^2 - 2)*cos(b*x + a) - 2*(b*x + a) *sin(b*x + a))*d^2/b^2)/b
Time = 0.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int (c+d x)^2 \sin (a+b x) \, dx=-\frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{b^{3}} + \frac {2 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \] Input:
integrate((d*x+c)^2*sin(b*x+a),x, algorithm="giac")
Output:
-(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a)/b^3 + 2*(b*d^2 *x + b*c*d)*sin(b*x + a)/b^3
Time = 34.96 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.68 \[ \int (c+d x)^2 \sin (a+b x) \, dx=\frac {\cos \left (a+b\,x\right )\,\left (2\,d^2-b^2\,c^2\right )}{b^3}-\frac {d^2\,x^2\,\cos \left (a+b\,x\right )}{b}+\frac {2\,c\,d\,\sin \left (a+b\,x\right )}{b^2}+\frac {2\,d^2\,x\,\sin \left (a+b\,x\right )}{b^2}-\frac {2\,c\,d\,x\,\cos \left (a+b\,x\right )}{b} \] Input:
int(sin(a + b*x)*(c + d*x)^2,x)
Output:
(cos(a + b*x)*(2*d^2 - b^2*c^2))/b^3 - (d^2*x^2*cos(a + b*x))/b + (2*c*d*s in(a + b*x))/b^2 + (2*d^2*x*sin(a + b*x))/b^2 - (2*c*d*x*cos(a + b*x))/b
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.70 \[ \int (c+d x)^2 \sin (a+b x) \, dx=\frac {-\cos \left (b x +a \right ) b^{2} c^{2}-2 \cos \left (b x +a \right ) b^{2} c d x -\cos \left (b x +a \right ) b^{2} d^{2} x^{2}+2 \cos \left (b x +a \right ) d^{2}+2 \sin \left (b x +a \right ) b c d +2 \sin \left (b x +a \right ) b \,d^{2} x}{b^{3}} \] Input:
int((d*x+c)^2*sin(b*x+a),x)
Output:
( - cos(a + b*x)*b**2*c**2 - 2*cos(a + b*x)*b**2*c*d*x - cos(a + b*x)*b**2 *d**2*x**2 + 2*cos(a + b*x)*d**2 + 2*sin(a + b*x)*b*c*d + 2*sin(a + b*x)*b *d**2*x)/b**3