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

Optimal. Leaf size=95 \[ \frac {a \sin (c+d x)}{d^2}-\frac {a x \cos (c+d x)}{d}-\frac {24 b \cos (c+d x)}{d^5}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {12 b x^2 \cos (c+d x)}{d^3}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {b x^4 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3339, 3296, 2637, 2638} \[ \frac {a \sin (c+d x)}{d^2}-\frac {a x \cos (c+d x)}{d}+\frac {4 b x^3 \sin (c+d x)}{d^2}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {24 b x \sin (c+d x)}{d^4}-\frac {24 b \cos (c+d x)}{d^5}-\frac {b x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^3)*Sin[c + d*x],x]

[Out]

(-24*b*Cos[c + d*x])/d^5 - (a*x*Cos[c + d*x])/d + (12*b*x^2*Cos[c + d*x])/d^3 - (b*x^4*Cos[c + d*x])/d + (a*Si
n[c + d*x])/d^2 - (24*b*x*Sin[c + d*x])/d^4 + (4*b*x^3*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 \left (a+b x^3\right ) \sin (c+d x) \, dx &=\int \left (a x \sin (c+d x)+b x^4 \sin (c+d x)\right ) \, dx\\ &=a \int x \sin (c+d x) \, dx+b \int x^4 \sin (c+d x) \, dx\\ &=-\frac {a x \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}+\frac {a \int \cos (c+d x) \, dx}{d}+\frac {(4 b) \int x^3 \cos (c+d x) \, dx}{d}\\ &=-\frac {a x \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {(12 b) \int x^2 \sin (c+d x) \, dx}{d^2}\\ &=-\frac {a x \cos (c+d x)}{d}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {b x^4 \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {(24 b) \int x \cos (c+d x) \, dx}{d^3}\\ &=-\frac {a x \cos (c+d x)}{d}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {b x^4 \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d^2}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {4 b x^3 \sin (c+d x)}{d^2}+\frac {(24 b) \int \sin (c+d x) \, dx}{d^4}\\ &=-\frac {24 b \cos (c+d x)}{d^5}-\frac {a x \cos (c+d x)}{d}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {b x^4 \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d^2}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {4 b x^3 \sin (c+d x)}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 66, normalized size = 0.69 \[ \frac {d \left (a d^2+4 b x \left (d^2 x^2-6\right )\right ) \sin (c+d x)-\left (a d^4 x+b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \cos (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 68, normalized size = 0.72 \[ -\frac {{\left (b d^{4} x^{4} + a d^{4} x - 12 \, b d^{2} x^{2} + 24 \, b\right )} \cos \left (d x + c\right ) - {\left (4 \, b d^{3} x^{3} + a d^{3} - 24 \, b d x\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 69, normalized size = 0.73 \[ -\frac {{\left (b d^{4} x^{4} + a d^{4} x - 12 \, b d^{2} x^{2} + 24 \, b\right )} \cos \left (d x + c\right )}{d^{5}} + \frac {{\left (4 \, b d^{3} x^{3} + a d^{3} - 24 \, b d x\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 258, normalized size = 2.72 \[ \frac {\frac {b \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}-\frac {4 b c \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}+\frac {6 b \,c^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}+a \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )-\frac {4 b \,c^{3} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}+a c \cos \left (d x +c \right )-\frac {b \,c^{4} \cos \left (d x +c \right )}{d^{3}}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.77, size = 224, normalized size = 2.36 \[ \frac {a c \cos \left (d x + c\right ) - \frac {b c^{4} \cos \left (d x + c\right )}{d^{3}} - {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a + \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{3}}{d^{3}} - \frac {6 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c^{2}}{d^{3}} + \frac {4 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b c}{d^{3}} - \frac {{\left ({\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b}{d^{3}}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.79, size = 92, normalized size = 0.97 \[ -\frac {d^4\,\left (a\,x\,\cos \left (c+d\,x\right )+b\,x^4\,\cos \left (c+d\,x\right )\right )+24\,b\,\cos \left (c+d\,x\right )-d^3\,\left (a\,\sin \left (c+d\,x\right )+4\,b\,x^3\,\sin \left (c+d\,x\right )\right )+24\,b\,d\,x\,\sin \left (c+d\,x\right )-12\,b\,d^2\,x^2\,\cos \left (c+d\,x\right )}{d^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(c + d*x)*(a + b*x^3),x)

[Out]

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

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sympy [A]  time = 2.55, size = 116, normalized size = 1.22 \[ \begin {cases} - \frac {a x \cos {\left (c + d x \right )}}{d} + \frac {a \sin {\left (c + d x \right )}}{d^{2}} - \frac {b x^{4} \cos {\left (c + d x \right )}}{d} + \frac {4 b x^{3} \sin {\left (c + d x \right )}}{d^{2}} + \frac {12 b x^{2} \cos {\left (c + d x \right )}}{d^{3}} - \frac {24 b x \sin {\left (c + d x \right )}}{d^{4}} - \frac {24 b \cos {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{2}}{2} + \frac {b x^{5}}{5}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**3+a)*sin(d*x+c),x)

[Out]

Piecewise((-a*x*cos(c + d*x)/d + a*sin(c + d*x)/d**2 - b*x**4*cos(c + d*x)/d + 4*b*x**3*sin(c + d*x)/d**2 + 12
*b*x**2*cos(c + d*x)/d**3 - 24*b*x*sin(c + d*x)/d**4 - 24*b*cos(c + d*x)/d**5, Ne(d, 0)), ((a*x**2/2 + b*x**5/
5)*sin(c), True))

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