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

Optimal. Leaf size=68 \[ -\frac {a \cos (c+d x)}{d}-\frac {6 b \sin (c+d x)}{d^4}+\frac {6 b x \cos (c+d x)}{d^3}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {b x^3 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3329, 2638, 3296, 2637} \[ -\frac {a \cos (c+d x)}{d}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {6 b \sin (c+d x)}{d^4}+\frac {6 b x \cos (c+d x)}{d^3}-\frac {b x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)*Sin[c + d*x],x]

[Out]

-((a*Cos[c + d*x])/d) + (6*b*x*Cos[c + d*x])/d^3 - (b*x^3*Cos[c + d*x])/d - (6*b*Sin[c + d*x])/d^4 + (3*b*x^2*
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 3329

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 50, normalized size = 0.74 \[ \frac {3 b \left (d^2 x^2-2\right ) \sin (c+d x)-d \left (a d^2+b x \left (d^2 x^2-6\right )\right ) \cos (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)*Sin[c + d*x],x]

[Out]

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

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fricas [A]  time = 0.69, size = 52, normalized size = 0.76 \[ -\frac {{\left (b d^{3} x^{3} + a d^{3} - 6 \, b d x\right )} \cos \left (d x + c\right ) - 3 \, {\left (b d^{2} x^{2} - 2 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.64, size = 54, normalized size = 0.79 \[ -\frac {{\left (b d^{3} x^{3} + a d^{3} - 6 \, b d x\right )} \cos \left (d x + c\right )}{d^{4}} + \frac {3 \, {\left (b d^{2} x^{2} - 2 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 159, normalized size = 2.34 \[ \frac {\frac {b \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-\frac {3 b c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}+\frac {3 b \,c^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-a \cos \left (d x +c \right )+\frac {b \,c^{3} \cos \left (d x +c \right )}{d^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.73, size = 141, normalized size = 2.07 \[ -\frac {a \cos \left (d x + c\right ) - \frac {b c^{3} \cos \left (d x + c\right )}{d^{3}} + \frac {3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{2}}{d^{3}} - \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c}{d^{3}} + \frac {{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b}{d^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 65, normalized size = 0.96 \[ -\frac {6\,b\,\sin \left (c+d\,x\right )+d^3\,\left (a\,\cos \left (c+d\,x\right )+b\,x^3\,\cos \left (c+d\,x\right )\right )-3\,b\,d^2\,x^2\,\sin \left (c+d\,x\right )-6\,b\,d\,x\,\cos \left (c+d\,x\right )}{d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)*(a + b*x^3),x)

[Out]

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

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sympy [A]  time = 1.21, size = 82, normalized size = 1.21 \[ \begin {cases} - \frac {a \cos {\left (c + d x \right )}}{d} - \frac {b x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 b x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 b \sin {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (a x + \frac {b x^{4}}{4}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*cos(c + d*x)/d - 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 + b*x**4/4)*sin(c), True))

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