3.80 \(\int x^2 (a+b x^3) \sin (c+d x) \, dx\)

Optimal. Leaf size=126 \[ \frac {2 a \cos (c+d x)}{d^3}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {a x^2 \cos (c+d x)}{d}+\frac {120 b \sin (c+d x)}{d^6}-\frac {120 b x \cos (c+d x)}{d^5}-\frac {60 b x^2 \sin (c+d x)}{d^4}+\frac {20 b x^3 \cos (c+d x)}{d^3}+\frac {5 b x^4 \sin (c+d x)}{d^2}-\frac {b x^5 \cos (c+d x)}{d} \]

[Out]

2*a*cos(d*x+c)/d^3-120*b*x*cos(d*x+c)/d^5-a*x^2*cos(d*x+c)/d+20*b*x^3*cos(d*x+c)/d^3-b*x^5*cos(d*x+c)/d+120*b*
sin(d*x+c)/d^6+2*a*x*sin(d*x+c)/d^2-60*b*x^2*sin(d*x+c)/d^4+5*b*x^4*sin(d*x+c)/d^2

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Rubi [A]  time = 0.19, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3339, 3296, 2638, 2637} \[ \frac {2 a x \sin (c+d x)}{d^2}+\frac {2 a \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}+\frac {5 b x^4 \sin (c+d x)}{d^2}-\frac {60 b x^2 \sin (c+d x)}{d^4}+\frac {20 b x^3 \cos (c+d x)}{d^3}+\frac {120 b \sin (c+d x)}{d^6}-\frac {120 b x \cos (c+d x)}{d^5}-\frac {b x^5 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^3)*Sin[c + d*x],x]

[Out]

(2*a*Cos[c + d*x])/d^3 - (120*b*x*Cos[c + d*x])/d^5 - (a*x^2*Cos[c + d*x])/d + (20*b*x^3*Cos[c + d*x])/d^3 - (
b*x^5*Cos[c + d*x])/d + (120*b*Sin[c + d*x])/d^6 + (2*a*x*Sin[c + d*x])/d^2 - (60*b*x^2*Sin[c + d*x])/d^4 + (5
*b*x^4*Sin[c + d*x])/d^2

Rule 2637

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

Rule 2638

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

Rule 3296

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 3339

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegran
d[Sin[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^2 \left (a+b x^3\right ) \sin (c+d x) \, dx &=\int \left (a x^2 \sin (c+d x)+b x^5 \sin (c+d x)\right ) \, dx\\ &=a \int x^2 \sin (c+d x) \, dx+b \int x^5 \sin (c+d x) \, dx\\ &=-\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^5 \cos (c+d x)}{d}+\frac {(2 a) \int x \cos (c+d x) \, dx}{d}+\frac {(5 b) \int x^4 \cos (c+d x) \, dx}{d}\\ &=-\frac {a x^2 \cos (c+d x)}{d}-\frac {b x^5 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {5 b x^4 \sin (c+d x)}{d^2}-\frac {(2 a) \int \sin (c+d x) \, dx}{d^2}-\frac {(20 b) \int x^3 \sin (c+d x) \, dx}{d^2}\\ &=\frac {2 a \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}+\frac {20 b x^3 \cos (c+d x)}{d^3}-\frac {b x^5 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}+\frac {5 b x^4 \sin (c+d x)}{d^2}-\frac {(60 b) \int x^2 \cos (c+d x) \, dx}{d^3}\\ &=\frac {2 a \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}+\frac {20 b x^3 \cos (c+d x)}{d^3}-\frac {b x^5 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {60 b x^2 \sin (c+d x)}{d^4}+\frac {5 b x^4 \sin (c+d x)}{d^2}+\frac {(120 b) \int x \sin (c+d x) \, dx}{d^4}\\ &=\frac {2 a \cos (c+d x)}{d^3}-\frac {120 b x \cos (c+d x)}{d^5}-\frac {a x^2 \cos (c+d x)}{d}+\frac {20 b x^3 \cos (c+d x)}{d^3}-\frac {b x^5 \cos (c+d x)}{d}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {60 b x^2 \sin (c+d x)}{d^4}+\frac {5 b x^4 \sin (c+d x)}{d^2}+\frac {(120 b) \int \cos (c+d x) \, dx}{d^5}\\ &=\frac {2 a \cos (c+d x)}{d^3}-\frac {120 b x \cos (c+d x)}{d^5}-\frac {a x^2 \cos (c+d x)}{d}+\frac {20 b x^3 \cos (c+d x)}{d^3}-\frac {b x^5 \cos (c+d x)}{d}+\frac {120 b \sin (c+d x)}{d^6}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {60 b x^2 \sin (c+d x)}{d^4}+\frac {5 b x^4 \sin (c+d x)}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 84, normalized size = 0.67 \[ \frac {\left (2 a d^4 x+5 b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \sin (c+d x)-d \left (a d^2 \left (d^2 x^2-2\right )+b x \left (d^4 x^4-20 d^2 x^2+120\right )\right ) \cos (c+d x)}{d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^3)*Sin[c + d*x],x]

[Out]

(-(d*(a*d^2*(-2 + d^2*x^2) + b*x*(120 - 20*d^2*x^2 + d^4*x^4))*Cos[c + d*x]) + (2*a*d^4*x + 5*b*(24 - 12*d^2*x
^2 + d^4*x^4))*Sin[c + d*x])/d^6

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fricas [A]  time = 0.66, size = 87, normalized size = 0.69 \[ -\frac {{\left (b d^{5} x^{5} + a d^{5} x^{2} - 20 \, b d^{3} x^{3} - 2 \, a d^{3} + 120 \, b d x\right )} \cos \left (d x + c\right ) - {\left (5 \, b d^{4} x^{4} + 2 \, a d^{4} x - 60 \, b d^{2} x^{2} + 120 \, b\right )} \sin \left (d x + c\right )}{d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)*sin(d*x+c),x, algorithm="fricas")

[Out]

-((b*d^5*x^5 + a*d^5*x^2 - 20*b*d^3*x^3 - 2*a*d^3 + 120*b*d*x)*cos(d*x + c) - (5*b*d^4*x^4 + 2*a*d^4*x - 60*b*
d^2*x^2 + 120*b)*sin(d*x + c))/d^6

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giac [A]  time = 0.42, size = 88, normalized size = 0.70 \[ -\frac {{\left (b d^{5} x^{5} + a d^{5} x^{2} - 20 \, b d^{3} x^{3} - 2 \, a d^{3} + 120 \, b d x\right )} \cos \left (d x + c\right )}{d^{6}} + \frac {{\left (5 \, b d^{4} x^{4} + 2 \, a d^{4} x - 60 \, b d^{2} x^{2} + 120 \, b\right )} \sin \left (d x + c\right )}{d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)*sin(d*x+c),x, algorithm="giac")

[Out]

-(b*d^5*x^5 + a*d^5*x^2 - 20*b*d^3*x^3 - 2*a*d^3 + 120*b*d*x)*cos(d*x + c)/d^6 + (5*b*d^4*x^4 + 2*a*d^4*x - 60
*b*d^2*x^2 + 120*b)*sin(d*x + c)/d^6

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maple [B]  time = 0.02, size = 392, normalized size = 3.11 \[ \frac {\frac {b \left (-\left (d x +c \right )^{5} \cos \left (d x +c \right )+5 \left (d x +c \right )^{4} \sin \left (d x +c \right )+20 \left (d x +c \right )^{3} \cos \left (d x +c \right )-60 \left (d x +c \right )^{2} \sin \left (d x +c \right )+120 \sin \left (d x +c \right )-120 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-\frac {5 b c \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}+\frac {10 b \,c^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}+a \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )-\frac {10 b \,c^{3} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}-2 a c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )+\frac {5 b \,c^{4} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-a \,c^{2} \cos \left (d x +c \right )+\frac {b \,c^{5} \cos \left (d x +c \right )}{d^{3}}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^3+a)*sin(d*x+c),x)

[Out]

1/d^3*(1/d^3*b*(-(d*x+c)^5*cos(d*x+c)+5*(d*x+c)^4*sin(d*x+c)+20*(d*x+c)^3*cos(d*x+c)-60*(d*x+c)^2*sin(d*x+c)+1
20*sin(d*x+c)-120*(d*x+c)*cos(d*x+c))-5/d^3*b*c*(-(d*x+c)^4*cos(d*x+c)+4*(d*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos
(d*x+c)-24*cos(d*x+c)-24*(d*x+c)*sin(d*x+c))+10/d^3*b*c^2*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(
d*x+c)+6*(d*x+c)*cos(d*x+c))+a*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))-10/d^3*b*c^3*(-(d*x+c
)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))-2*a*c*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+5/d^3*b*c^4*(sin(d*x+c
)-(d*x+c)*cos(d*x+c))-a*c^2*cos(d*x+c)+1/d^3*b*c^5*cos(d*x+c))

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maxima [B]  time = 0.69, size = 326, normalized size = 2.59 \[ -\frac {a c^{2} \cos \left (d x + c\right ) - \frac {b c^{5} \cos \left (d x + c\right )}{d^{3}} - 2 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c + \frac {5 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{4}}{d^{3}} + {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a - \frac {10 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c^{3}}{d^{3}} + \frac {10 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b c^{2}}{d^{3}} - \frac {5 \, {\left ({\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b c}{d^{3}} + \frac {{\left ({\left ({\left (d x + c\right )}^{5} - 20 \, {\left (d x + c\right )}^{3} + 120 \, d x + 120 \, c\right )} \cos \left (d x + c\right ) - 5 \, {\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \sin \left (d x + c\right )\right )} b}{d^{3}}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)*sin(d*x+c),x, algorithm="maxima")

[Out]

-(a*c^2*cos(d*x + c) - b*c^5*cos(d*x + c)/d^3 - 2*((d*x + c)*cos(d*x + c) - sin(d*x + c))*a*c + 5*((d*x + c)*c
os(d*x + c) - sin(d*x + c))*b*c^4/d^3 + (((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*a - 10*(((
d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b*c^3/d^3 + 10*(((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x +
c) - 3*((d*x + c)^2 - 2)*sin(d*x + c))*b*c^2/d^3 - 5*(((d*x + c)^4 - 12*(d*x + c)^2 + 24)*cos(d*x + c) - 4*((
d*x + c)^3 - 6*d*x - 6*c)*sin(d*x + c))*b*c/d^3 + (((d*x + c)^5 - 20*(d*x + c)^3 + 120*d*x + 120*c)*cos(d*x +
c) - 5*((d*x + c)^4 - 12*(d*x + c)^2 + 24)*sin(d*x + c))*b/d^3)/d^3

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mupad [B]  time = 4.95, size = 121, normalized size = 0.96 \[ \frac {120\,b\,\sin \left (c+d\,x\right )+d^4\,\left (5\,b\,x^4\,\sin \left (c+d\,x\right )+2\,a\,x\,\sin \left (c+d\,x\right )\right )-d^5\,\left (a\,x^2\,\cos \left (c+d\,x\right )+b\,x^5\,\cos \left (c+d\,x\right )\right )+d^3\,\left (2\,a\,\cos \left (c+d\,x\right )+20\,b\,x^3\,\cos \left (c+d\,x\right )\right )-60\,b\,d^2\,x^2\,\sin \left (c+d\,x\right )-120\,b\,d\,x\,\cos \left (c+d\,x\right )}{d^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(c + d*x)*(a + b*x^3),x)

[Out]

(120*b*sin(c + d*x) + d^4*(5*b*x^4*sin(c + d*x) + 2*a*x*sin(c + d*x)) - d^5*(a*x^2*cos(c + d*x) + b*x^5*cos(c
+ d*x)) + d^3*(2*a*cos(c + d*x) + 20*b*x^3*cos(c + d*x)) - 60*b*d^2*x^2*sin(c + d*x) - 120*b*d*x*cos(c + d*x))
/d^6

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sympy [A]  time = 4.46, size = 151, normalized size = 1.20 \[ \begin {cases} - \frac {a x^{2} \cos {\left (c + d x \right )}}{d} + \frac {2 a x \sin {\left (c + d x \right )}}{d^{2}} + \frac {2 a \cos {\left (c + d x \right )}}{d^{3}} - \frac {b x^{5} \cos {\left (c + d x \right )}}{d} + \frac {5 b x^{4} \sin {\left (c + d x \right )}}{d^{2}} + \frac {20 b x^{3} \cos {\left (c + d x \right )}}{d^{3}} - \frac {60 b x^{2} \sin {\left (c + d x \right )}}{d^{4}} - \frac {120 b x \cos {\left (c + d x \right )}}{d^{5}} + \frac {120 b \sin {\left (c + d x \right )}}{d^{6}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{3}}{3} + \frac {b x^{6}}{6}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**3+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**5*cos(c + d*x)/d +
5*b*x**4*sin(c + d*x)/d**2 + 20*b*x**3*cos(c + d*x)/d**3 - 60*b*x**2*sin(c + d*x)/d**4 - 120*b*x*cos(c + d*x)/
d**5 + 120*b*sin(c + d*x)/d**6, Ne(d, 0)), ((a*x**3/3 + b*x**6/6)*sin(c), True))

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