Integrand size = 13, antiderivative size = 65 \[ \int x (a+b x) \sin (c+d x) \, dx=\frac {2 b \cos (c+d x)}{d^3}-\frac {a x \cos (c+d x)}{d}-\frac {b x^2 \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d^2}+\frac {2 b x \sin (c+d x)}{d^2} \] Output:
2*b*cos(d*x+c)/d^3-a*x*cos(d*x+c)/d-b*x^2*cos(d*x+c)/d+a*sin(d*x+c)/d^2+2* b*x*sin(d*x+c)/d^2
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int x (a+b x) \sin (c+d x) \, dx=\frac {-\left (\left (a d^2 x+b \left (-2+d^2 x^2\right )\right ) \cos (c+d x)\right )+d (a+2 b x) \sin (c+d x)}{d^3} \] Input:
Integrate[x*(a + b*x)*Sin[c + d*x],x]
Output:
(-((a*d^2*x + b*(-2 + d^2*x^2))*Cos[c + d*x]) + d*(a + 2*b*x)*Sin[c + d*x] )/d^3
Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {7293, 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.
\(\displaystyle \int x (a+b x) \sin (c+d x) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (a x \sin (c+d x)+b x^2 \sin (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \sin (c+d x)}{d^2}-\frac {a x \cos (c+d x)}{d}+\frac {2 b \cos (c+d x)}{d^3}+\frac {2 b x \sin (c+d x)}{d^2}-\frac {b x^2 \cos (c+d x)}{d}\) |
Input:
Int[x*(a + b*x)*Sin[c + d*x],x]
Output:
(2*b*Cos[c + d*x])/d^3 - (a*x*Cos[c + d*x])/d - (b*x^2*Cos[c + d*x])/d + ( a*Sin[c + d*x])/d^2 + (2*b*x*Sin[c + d*x])/d^2
Time = 0.84 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {\left (x^{2} d^{2} b +a \,d^{2} x -2 b \right ) \cos \left (d x +c \right )}{d^{3}}+\frac {\left (2 b x +a \right ) \sin \left (d x +c \right )}{d^{2}}\) | \(47\) |
parallelrisch | \(\frac {\left (x \left (b x +a \right ) d^{2}-4 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 d \left (2 b x +a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-x \left (b x +a \right ) d^{2}}{d^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\) | \(76\) |
parts | \(-\frac {b \,x^{2} \cos \left (d x +c \right )}{d}-\frac {a x \cos \left (d x +c \right )}{d}+\frac {a \sin \left (d x +c \right )-\frac {2 b c \sin \left (d x +c \right )}{d}+\frac {2 b \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}}{d^{2}}\) | \(81\) |
norman | \(\frac {\frac {4 b}{d^{3}}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {b \,x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}-\frac {a x}{d}-\frac {b \,x^{2}}{d}+\frac {4 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\) | \(107\) |
orering | \(\frac {2 \left (2 b x +a \right ) \left (x^{2} d^{2} b +a \,d^{2} x -b \right ) \sin \left (d x +c \right )}{d^{4} x \left (b x +a \right )}-\frac {\left (x^{2} d^{2} b +a \,d^{2} x -2 b \right ) \left (\left (b x +a \right ) \sin \left (d x +c \right )+x b \sin \left (d x +c \right )+x \left (b x +a \right ) d \cos \left (d x +c \right )\right )}{d^{4} x \left (b x +a \right )}\) | \(116\) |
derivativedivides | \(\frac {a c \cos \left (d x +c \right )+a \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )-\frac {b \,c^{2} \cos \left (d x +c \right )}{d}-\frac {2 b c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}+\frac {b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}}{d^{2}}\) | \(121\) |
default | \(\frac {a c \cos \left (d x +c \right )+a \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )-\frac {b \,c^{2} \cos \left (d x +c \right )}{d}-\frac {2 b c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}+\frac {b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}}{d^{2}}\) | \(121\) |
meijerg | \(\frac {4 b \sin \left (c \right ) \sqrt {\pi }\, \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 x^{2} d^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {4 b \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {x^{2} d^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {x d \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+\frac {2 a \sin \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {x d \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 a \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}\) | \(180\) |
Input:
int(x*(b*x+a)*sin(d*x+c),x,method=_RETURNVERBOSE)
Output:
-(b*d^2*x^2+a*d^2*x-2*b)/d^3*cos(d*x+c)+1/d^2*(2*b*x+a)*sin(d*x+c)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int x (a+b x) \sin (c+d x) \, dx=-\frac {{\left (b d^{2} x^{2} + a d^{2} x - 2 \, b\right )} \cos \left (d x + c\right ) - {\left (2 \, b d x + a d\right )} \sin \left (d x + c\right )}{d^{3}} \] Input:
integrate(x*(b*x+a)*sin(d*x+c),x, algorithm="fricas")
Output:
-((b*d^2*x^2 + a*d^2*x - 2*b)*cos(d*x + c) - (2*b*d*x + a*d)*sin(d*x + c)) /d^3
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26 \[ \int x (a+b x) \sin (c+d x) \, dx=\begin {cases} - \frac {a x \cos {\left (c + d x \right )}}{d} + \frac {a \sin {\left (c + d x \right )}}{d^{2}} - \frac {b x^{2} \cos {\left (c + d x \right )}}{d} + \frac {2 b x \sin {\left (c + d x \right )}}{d^{2}} + \frac {2 b \cos {\left (c + d x \right )}}{d^{3}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{2}}{2} + \frac {b x^{3}}{3}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(x*(b*x+a)*sin(d*x+c),x)
Output:
Piecewise((-a*x*cos(c + d*x)/d + a*sin(c + d*x)/d**2 - b*x**2*cos(c + d*x) /d + 2*b*x*sin(c + d*x)/d**2 + 2*b*cos(c + d*x)/d**3, Ne(d, 0)), ((a*x**2/ 2 + b*x**3/3)*sin(c), True))
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.80 \[ \int x (a+b x) \sin (c+d x) \, dx=\frac {a c \cos \left (d x + c\right ) - \frac {b c^{2} \cos \left (d x + c\right )}{d} - {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a + \frac {2 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c}{d} - \frac {{\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b}{d}}{d^{2}} \] Input:
integrate(x*(b*x+a)*sin(d*x+c),x, algorithm="maxima")
Output:
(a*c*cos(d*x + c) - b*c^2*cos(d*x + c)/d - ((d*x + c)*cos(d*x + c) - sin(d *x + c))*a + 2*((d*x + c)*cos(d*x + c) - sin(d*x + c))*b*c/d - (((d*x + c) ^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b/d)/d^2
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int x (a+b x) \sin (c+d x) \, dx=-\frac {{\left (b d^{2} x^{2} + a d^{2} x - 2 \, b\right )} \cos \left (d x + c\right )}{d^{3}} + \frac {{\left (2 \, b d x + a d\right )} \sin \left (d x + c\right )}{d^{3}} \] Input:
integrate(x*(b*x+a)*sin(d*x+c),x, algorithm="giac")
Output:
-(b*d^2*x^2 + a*d^2*x - 2*b)*cos(d*x + c)/d^3 + (2*b*d*x + a*d)*sin(d*x + c)/d^3
Time = 40.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int x (a+b x) \sin (c+d x) \, dx=\frac {a\,\sin \left (c+d\,x\right )+2\,b\,x\,\sin \left (c+d\,x\right )}{d^2}-\frac {a\,x\,\cos \left (c+d\,x\right )+b\,x^2\,\cos \left (c+d\,x\right )}{d}+\frac {2\,b\,\cos \left (c+d\,x\right )}{d^3} \] Input:
int(x*sin(c + d*x)*(a + b*x),x)
Output:
(a*sin(c + d*x) + 2*b*x*sin(c + d*x))/d^2 - (a*x*cos(c + d*x) + b*x^2*cos( c + d*x))/d + (2*b*cos(c + d*x))/d^3
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int x (a+b x) \sin (c+d x) \, dx=\frac {-\cos \left (d x +c \right ) a \,d^{2} x -\cos \left (d x +c \right ) b \,d^{2} x^{2}+2 \cos \left (d x +c \right ) b +\sin \left (d x +c \right ) a d +2 \sin \left (d x +c \right ) b d x}{d^{3}} \] Input:
int(x*(b*x+a)*sin(d*x+c),x)
Output:
( - cos(c + d*x)*a*d**2*x - cos(c + d*x)*b*d**2*x**2 + 2*cos(c + d*x)*b + sin(c + d*x)*a*d + 2*sin(c + d*x)*b*d*x)/d**3