\(\int x (a+b x)^2 \sin (c+d x) \, dx\) [11]

Optimal result

Integrand size = 15, antiderivative size = 135 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\frac {4 a b \cos (c+d x)}{d^3}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2} \] Output:

4*a*b*cos(d*x+c)/d^3+6*b^2*x*cos(d*x+c)/d^3-a^2*x*cos(d*x+c)/d-2*a*b*x^2*c 
os(d*x+c)/d-b^2*x^3*cos(d*x+c)/d-6*b^2*sin(d*x+c)/d^4+a^2*sin(d*x+c)/d^2+4 
*a*b*x*sin(d*x+c)/d^2+3*b^2*x^2*sin(d*x+c)/d^2
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\frac {-d \left (a^2 d^2 x+b^2 x \left (-6+d^2 x^2\right )+2 a b \left (-2+d^2 x^2\right )\right ) \cos (c+d x)+\left (a^2 d^2+4 a b d^2 x+3 b^2 \left (-2+d^2 x^2\right )\right ) \sin (c+d x)}{d^4} \] Input:

Integrate[x*(a + b*x)^2*Sin[c + d*x],x]
 

Output:

(-(d*(a^2*d^2*x + b^2*x*(-6 + d^2*x^2) + 2*a*b*(-2 + d^2*x^2))*Cos[c + d*x 
]) + (a^2*d^2 + 4*a*b*d^2*x + 3*b^2*(-2 + d^2*x^2))*Sin[c + d*x])/d^4
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.

\(\displaystyle \int x (a+b x)^2 \sin (c+d x) \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (a^2 x \sin (c+d x)+2 a b x^2 \sin (c+d x)+b^2 x^3 \sin (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 \sin (c+d x)}{d^2}-\frac {a^2 x \cos (c+d x)}{d}+\frac {4 a b \cos (c+d x)}{d^3}+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d}\)

Input:

Int[x*(a + b*x)^2*Sin[c + d*x],x]
 

Output:

(4*a*b*Cos[c + d*x])/d^3 + (6*b^2*x*Cos[c + d*x])/d^3 - (a^2*x*Cos[c + d*x 
])/d - (2*a*b*x^2*Cos[c + d*x])/d - (b^2*x^3*Cos[c + d*x])/d - (6*b^2*Sin[ 
c + d*x])/d^4 + (a^2*Sin[c + d*x])/d^2 + (4*a*b*x*Sin[c + d*x])/d^2 + (3*b 
^2*x^2*Sin[c + d*x])/d^2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\left (b^{2} d^{2} x^{3}+2 a b \,d^{2} x^{2}+x \,a^{2} d^{2}-6 b^{2} x -4 a b \right ) \cos \left (d x +c \right )}{d^{3}}+\frac {\left (3 x^{2} d^{2} b^{2}+4 a b \,d^{2} x +a^{2} d^{2}-6 b^{2}\right ) \sin \left (d x +c \right )}{d^{4}}\)	\(94\)
parallelrisch	\(\frac {d \left (\left (b x +a \right )^{2} d^{2}-6 b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \left (\left (3 x^{2} b^{2}+4 a b x +a^{2}\right ) d^{2}-6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-d \left (x \left (b x +a \right )^{2} d^{2}-6 b^{2} x -8 a b \right )}{d^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(116\)
orering	\(\frac {2 \left (3 b^{3} d^{2} x^{4}+7 a \,b^{2} d^{2} x^{3}+5 a^{2} b \,d^{2} x^{2}+a^{3} d^{2} x -12 x^{2} b^{3}-12 x \,b^{2} a -2 a^{2} b \right ) \sin \left (d x +c \right )}{d^{4} x \left (b x +a \right )}-\frac {\left (b^{2} d^{2} x^{3}+2 a b \,d^{2} x^{2}+x \,a^{2} d^{2}-6 b^{2} x -4 a b \right ) \left (\left (b x +a \right )^{2} \sin \left (d x +c \right )+2 x \left (b x +a \right ) \sin \left (d x +c \right ) b +x \left (b x +a \right )^{2} d \cos \left (d x +c \right )\right )}{d^{4} x \left (b x +a \right )^{2}}\)	\(188\)
parts	\(-\frac {b^{2} x^{3} \cos \left (d x +c \right )}{d}-\frac {2 a b \,x^{2} \cos \left (d x +c \right )}{d}-\frac {a^{2} x \cos \left (d x +c \right )}{d}+\frac {a^{2} \sin \left (d x +c \right )-\frac {4 a b c \sin \left (d x +c \right )}{d}+\frac {4 a b \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \sin \left (d x +c \right )}{d^{2}}-\frac {6 b^{2} c \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+\frac {3 b^{2} \left (\left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}}{d^{2}}\)	\(194\)
norman	\(\frac {\frac {b^{2} x^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (a^{2} d^{2}-6 b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d^{3}}-\frac {b^{2} x^{3}}{d}-\frac {\left (a^{2} d^{2}-6 b^{2}\right ) x}{d^{3}}+\frac {2 \left (a^{2} d^{2}-6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{4}}-\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d^{3}}+\frac {6 b^{2} x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}-\frac {2 a b \,x^{2}}{d}+\frac {8 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}+\frac {2 a b \,x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\)	\(211\)
derivativedivides	\(\frac {a^{2} c \cos \left (d x +c \right )+a^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )-\frac {2 a b \,c^{2} \cos \left (d x +c \right )}{d}-\frac {4 a b c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}+\frac {2 a b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {b^{2} c^{3} \cos \left (d x +c \right )}{d^{2}}+\frac {3 b^{2} c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}-\frac {3 b^{2} c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+\frac {b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}}{d^{2}}\)	\(281\)
default	\(\frac {a^{2} c \cos \left (d x +c \right )+a^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )-\frac {2 a b \,c^{2} \cos \left (d x +c \right )}{d}-\frac {4 a b c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}+\frac {2 a b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {b^{2} c^{3} \cos \left (d x +c \right )}{d^{2}}+\frac {3 b^{2} c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}-\frac {3 b^{2} c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+\frac {b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}}{d^{2}}\)	\(281\)
meijerg	\(\frac {8 b^{2} \sin \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 x^{2} d^{2}}{2}+3\right ) \cos \left (d x \right )}{4 \sqrt {\pi }}-\frac {x d \left (-\frac {x^{2} d^{2}}{2}+3\right ) \sin \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 b^{2} \cos \left (c \right ) \sqrt {\pi }\, \left (\frac {x d \left (-\frac {5 x^{2} d^{2}}{2}+15\right ) \cos \left (d x \right )}{20 \sqrt {\pi }}-\frac {\left (-\frac {15 x^{2} d^{2}}{2}+15\right ) \sin \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 a b \sin \left (c \right ) \sqrt {\pi }\, \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 x^{2} d^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {8 a b \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {x^{2} d^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {x d \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+\frac {2 a^{2} \sin \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {x d \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 a^{2} \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}\)	\(299\)

Input:

int(x*(b*x+a)^2*sin(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

-1/d^3*(b^2*d^2*x^3+2*a*b*d^2*x^2+a^2*d^2*x-6*b^2*x-4*a*b)*cos(d*x+c)+(3*b 
^2*d^2*x^2+4*a*b*d^2*x+a^2*d^2-6*b^2)/d^4*sin(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=-\frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} - 4 \, a b d + {\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) - {\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{4}} \] Input:

integrate(x*(b*x+a)^2*sin(d*x+c),x, algorithm="fricas")
 

Output:

-((b^2*d^3*x^3 + 2*a*b*d^3*x^2 - 4*a*b*d + (a^2*d^3 - 6*b^2*d)*x)*cos(d*x 
+ c) - (3*b^2*d^2*x^2 + 4*a*b*d^2*x + a^2*d^2 - 6*b^2)*sin(d*x + c))/d^4
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.27 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\begin {cases} - \frac {a^{2} x \cos {\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d^{2}} - \frac {2 a b x^{2} \cos {\left (c + d x \right )}}{d} + \frac {4 a b x \sin {\left (c + d x \right )}}{d^{2}} + \frac {4 a b \cos {\left (c + d x \right )}}{d^{3}} - \frac {b^{2} x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 b^{2} x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 b^{2} x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 b^{2} \sin {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a^{2} x^{2}}{2} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{4}}{4}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(x*(b*x+a)**2*sin(d*x+c),x)
 

Output:

Piecewise((-a**2*x*cos(c + d*x)/d + a**2*sin(c + d*x)/d**2 - 2*a*b*x**2*co 
s(c + d*x)/d + 4*a*b*x*sin(c + d*x)/d**2 + 4*a*b*cos(c + d*x)/d**3 - b**2* 
x**3*cos(c + d*x)/d + 3*b**2*x**2*sin(c + d*x)/d**2 + 6*b**2*x*cos(c + d*x 
)/d**3 - 6*b**2*sin(c + d*x)/d**4, Ne(d, 0)), ((a**2*x**2/2 + 2*a*b*x**3/3 
 + b**2*x**4/4)*sin(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.92 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\frac {a^{2} c \cos \left (d x + c\right ) + \frac {b^{2} c^{3} \cos \left (d x + c\right )}{d^{2}} - \frac {2 \, a b c^{2} \cos \left (d x + c\right )}{d} - {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a^{2} - \frac {3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b^{2} c^{2}}{d^{2}} + \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a b c}{d} + \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b^{2} c}{d^{2}} - \frac {2 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a b}{d} - \frac {{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b^{2}}{d^{2}}}{d^{2}} \] Input:

integrate(x*(b*x+a)^2*sin(d*x+c),x, algorithm="maxima")
 

Output:

(a^2*c*cos(d*x + c) + b^2*c^3*cos(d*x + c)/d^2 - 2*a*b*c^2*cos(d*x + c)/d 
- ((d*x + c)*cos(d*x + c) - sin(d*x + c))*a^2 - 3*((d*x + c)*cos(d*x + c) 
- sin(d*x + c))*b^2*c^2/d^2 + 4*((d*x + c)*cos(d*x + c) - sin(d*x + c))*a* 
b*c/d + 3*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b^2* 
c/d^2 - 2*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*a*b/ 
d - (((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 3*((d*x + c)^2 - 2)*sin(d* 
x + c))*b^2/d^2)/d^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=-\frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x - 6 \, b^{2} d x - 4 \, a b d\right )} \cos \left (d x + c\right )}{d^{4}} + \frac {{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{4}} \] Input:

integrate(x*(b*x+a)^2*sin(d*x+c),x, algorithm="giac")
 

Output:

-(b^2*d^3*x^3 + 2*a*b*d^3*x^2 + a^2*d^3*x - 6*b^2*d*x - 4*a*b*d)*cos(d*x + 
 c)/d^4 + (3*b^2*d^2*x^2 + 4*a*b*d^2*x + a^2*d^2 - 6*b^2)*sin(d*x + c)/d^4
 

Mupad [B] (verification not implemented)

Time = 40.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\frac {3\,b^2\,x^2\,\sin \left (c+d\,x\right )}{d^2}-\frac {b^2\,x^3\,\cos \left (c+d\,x\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (6\,b^2-a^2\,d^2\right )}{d^4}+\frac {4\,a\,b\,\cos \left (c+d\,x\right )}{d^3}+\frac {x\,\cos \left (c+d\,x\right )\,\left (6\,b^2-a^2\,d^2\right )}{d^3}-\frac {2\,a\,b\,x^2\,\cos \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,x\,\sin \left (c+d\,x\right )}{d^2} \] Input:

int(x*sin(c + d*x)*(a + b*x)^2,x)
 

Output:

(3*b^2*x^2*sin(c + d*x))/d^2 - (b^2*x^3*cos(c + d*x))/d - (sin(c + d*x)*(6 
*b^2 - a^2*d^2))/d^4 + (4*a*b*cos(c + d*x))/d^3 + (x*cos(c + d*x)*(6*b^2 - 
 a^2*d^2))/d^3 - (2*a*b*x^2*cos(c + d*x))/d + (4*a*b*x*sin(c + d*x))/d^2
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int x (a+b x)^2 \sin (c+d x) \, dx=\frac {-\cos \left (d x +c \right ) a^{2} d^{3} x -2 \cos \left (d x +c \right ) a b \,d^{3} x^{2}+4 \cos \left (d x +c \right ) a b d -\cos \left (d x +c \right ) b^{2} d^{3} x^{3}+6 \cos \left (d x +c \right ) b^{2} d x +\sin \left (d x +c \right ) a^{2} d^{2}+4 \sin \left (d x +c \right ) a b \,d^{2} x +3 \sin \left (d x +c \right ) b^{2} d^{2} x^{2}-6 \sin \left (d x +c \right ) b^{2}}{d^{4}} \] Input:

int(x*(b*x+a)^2*sin(d*x+c),x)
 

Output:

( - cos(c + d*x)*a**2*d**3*x - 2*cos(c + d*x)*a*b*d**3*x**2 + 4*cos(c + d* 
x)*a*b*d - cos(c + d*x)*b**2*d**3*x**3 + 6*cos(c + d*x)*b**2*d*x + sin(c + 
 d*x)*a**2*d**2 + 4*sin(c + d*x)*a*b*d**2*x + 3*sin(c + d*x)*b**2*d**2*x** 
2 - 6*sin(c + d*x)*b**2)/d**4