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

Optimal result

Integrand size = 15, antiderivative size = 96 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=\frac {2 a \cos (c+d x)}{d^3}+\frac {6 b x \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^3 \cos (c+d x)}{d}-\frac {6 b \sin (c+d x)}{d^4}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {3 b x^2 \sin (c+d x)}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6874, 3377, 2718, 2717} \[ \int x^2 (a+b x) \sin (c+d x) \, dx=\frac {2 a \cos (c+d x)}{d^3}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {a x^2 \cos (c+d x)}{d}-\frac {6 b \sin (c+d x)}{d^4}+\frac {6 b x \cos (c+d x)}{d^3}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {b x^3 \cos (c+d x)}{d} \]

[In]

Int[x^2*(a + b*x)*Sin[c + d*x],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a x^2 \sin (c+d x)+b x^3 \sin (c+d x)\right ) \, dx \\ & = a \int x^2 \sin (c+d x) \, dx+b \int x^3 \sin (c+d x) \, dx \\ & = -\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^3 \cos (c+d x)}{d}+\frac {(2 a) \int x \cos (c+d x) \, dx}{d}+\frac {(3 b) \int x^2 \cos (c+d x) \, dx}{d} \\ & = -\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^3 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {(2 a) \int \sin (c+d x) \, dx}{d^2}-\frac {(6 b) \int x \sin (c+d x) \, dx}{d^2} \\ & = \frac {2 a \cos (c+d x)}{d^3}+\frac {6 b x \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^3 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {(6 b) \int \cos (c+d x) \, dx}{d^3} \\ & = \frac {2 a \cos (c+d x)}{d^3}+\frac {6 b x \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^3 \cos (c+d x)}{d}-\frac {6 b \sin (c+d x)}{d^4}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {3 b x^2 \sin (c+d x)}{d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=\frac {-d \left (b x \left (-6+d^2 x^2\right )+a \left (-2+d^2 x^2\right )\right ) \cos (c+d x)+\left (2 a d^2 x+3 b \left (-2+d^2 x^2\right )\right ) \sin (c+d x)}{d^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\left (b \,d^{2} x^{3}+a \,d^{2} x^{2}-6 b x -2 a \right ) \cos \left (d x +c \right )}{d^{3}}+\frac {\left (3 d^{2} x^{2} b +2 a \,d^{2} x -6 b \right ) \sin \left (d x +c \right )}{d^{4}}\)	\(67\)
parallelrisch	\(\frac {x \left (x \left (b x +a \right ) d^{2}-6 b \right ) d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (6 b \,x^{2}+4 a x \right ) d^{2}-12 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (x^{2} \left (b x +a \right ) d^{2}-6 b x -4 a \right ) d}{d^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(101\)
norman	\(\frac {\frac {4 a}{d^{3}}+\frac {a \,x^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \,x^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \,x^{2}}{d}-\frac {12 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{4}}+\frac {6 b x}{d^{3}}-\frac {b \,x^{3}}{d}+\frac {4 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}-\frac {6 b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d^{3}}+\frac {6 b \,x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\)	\(154\)
parts	\(-\frac {b \,x^{3} \cos \left (d x +c \right )}{d}-\frac {a \,x^{2} \cos \left (d x +c \right )}{d}+\frac {-\frac {2 a c \sin \left (d x +c \right )}{d}+\frac {2 a \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {3 b \,c^{2} \sin \left (d x +c \right )}{d^{2}}-\frac {6 b c \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+\frac {3 b \left (\left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}}{d^{2}}\)	\(158\)
meijerg	\(\frac {8 b \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \cos \left (d x \right )}{4 \sqrt {\pi }}-\frac {d x \left (-\frac {d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {x d \left (-\frac {5 d^{2} x^{2}}{2}+15\right ) \cos \left (d x \right )}{20 \sqrt {\pi }}-\frac {\left (-\frac {15 d^{2} x^{2}}{2}+15\right ) \sin \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {4 a \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {4 a \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {d^{2} x^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}\)	\(220\)
derivativedivides	\(\frac {-a \,c^{2} \cos \left (d x +c \right )-2 a c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )+a \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )+\frac {b \,c^{3} \cos \left (d x +c \right )}{d}+\frac {3 b \,c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}-\frac {3 b c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {b \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}}{d^{3}}\)	\(225\)
default	\(\frac {-a \,c^{2} \cos \left (d x +c \right )-2 a c \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )+a \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )+\frac {b \,c^{3} \cos \left (d x +c \right )}{d}+\frac {3 b \,c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}-\frac {3 b c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+\frac {b \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d}}{d^{3}}\)	\(225\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=-\frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} - 6 \, b d x - 2 \, a d\right )} \cos \left (d x + c\right ) - {\left (3 \, b d^{2} x^{2} + 2 \, a d^{2} x - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.22 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=\begin {cases} - \frac {a x^{2} \cos {\left (c + d x \right )}}{d} + \frac {2 a x \sin {\left (c + d x \right )}}{d^{2}} + \frac {2 a \cos {\left (c + d x \right )}}{d^{3}} - \frac {b x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 b x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 b \sin {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{3}}{3} + \frac {b x^{4}}{4}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (96) = 192\).

Time = 0.21 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.09 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=-\frac {a c^{2} \cos \left (d x + c\right ) - \frac {b c^{3} \cos \left (d x + c\right )}{d} - 2 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c + \frac {3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{2}}{d} + {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a - \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c}{d} + \frac {{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b}{d}}{d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=-\frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} - 6 \, b d x - 2 \, a d\right )} \cos \left (d x + c\right )}{d^{4}} + \frac {{\left (3 \, b d^{2} x^{2} + 2 \, a d^{2} x - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int x^2 (a+b x) \sin (c+d x) \, dx=\frac {3\,b\,x^2\,\sin \left (c+d\,x\right )+2\,a\,x\,\sin \left (c+d\,x\right )}{d^2}+\frac {2\,a\,\cos \left (c+d\,x\right )+6\,b\,x\,\cos \left (c+d\,x\right )}{d^3}-\frac {a\,x^2\,\cos \left (c+d\,x\right )+b\,x^3\,\cos \left (c+d\,x\right )}{d}-\frac {6\,b\,\sin \left (c+d\,x\right )}{d^4} \]

[In]

int(x^2*sin(c + d*x)*(a + b*x),x)

[Out]

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