3.83 \(\int \frac {(a+b x^3) \sin (c+d x)}{x} \, dx\)

Optimal. Leaf size=57 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)+\frac {2 b \cos (c+d x)}{d^3}+\frac {2 b x \sin (c+d x)}{d^2}-\frac {b x^2 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3339, 3303, 3299, 3302, 3296, 2638} \[ a \sin (c) \text {CosIntegral}(d x)+a \cos (c) \text {Si}(d x)+\frac {2 b x \sin (c+d x)}{d^2}+\frac {2 b \cos (c+d x)}{d^3}-\frac {b x^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x,x]

[Out]

(2*b*Cos[c + d*x])/d^3 - (b*x^2*Cos[c + d*x])/d + a*CosIntegral[d*x]*Sin[c] + (2*b*x*Sin[c + d*x])/d^2 + a*Cos
[c]*SinIntegral[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 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 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 \frac {\left (a+b x^3\right ) \sin (c+d x)}{x} \, dx &=\int \left (\frac {a \sin (c+d x)}{x}+b x^2 \sin (c+d x)\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x} \, dx+b \int x^2 \sin (c+d x) \, dx\\ &=-\frac {b x^2 \cos (c+d x)}{d}+\frac {(2 b) \int x \cos (c+d x) \, dx}{d}+(a \cos (c)) \int \frac {\sin (d x)}{x} \, dx+(a \sin (c)) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {b x^2 \cos (c+d x)}{d}+a \text {Ci}(d x) \sin (c)+\frac {2 b x \sin (c+d x)}{d^2}+a \cos (c) \text {Si}(d x)-\frac {(2 b) \int \sin (c+d x) \, dx}{d^2}\\ &=\frac {2 b \cos (c+d x)}{d^3}-\frac {b x^2 \cos (c+d x)}{d}+a \text {Ci}(d x) \sin (c)+\frac {2 b x \sin (c+d x)}{d^2}+a \cos (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 50, normalized size = 0.88 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)+\frac {b \left (\left (2-d^2 x^2\right ) \cos (c+d x)+2 d x \sin (c+d x)\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 72, normalized size = 1.26 \[ \frac {2 \, a d^{3} \cos \relax (c) \operatorname {Si}\left (d x\right ) + 4 \, b d x \sin \left (d x + c\right ) - 2 \, {\left (b d^{2} x^{2} - 2 \, b\right )} \cos \left (d x + c\right ) + {\left (a d^{3} \operatorname {Ci}\left (d x\right ) + a d^{3} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*d^3*cos(c)*sin_integral(d*x) + 4*b*d*x*sin(d*x + c) - 2*(b*d^2*x^2 - 2*b)*cos(d*x + c) + (a*d^3*cos_i
ntegral(d*x) + a*d^3*cos_integral(-d*x))*sin(c))/d^3

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giac [C]  time = 0.38, size = 510, normalized size = 8.95 \[ -\frac {2 \, b d^{2} x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + a d^{3} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{3} \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d^{2} x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} - a d^{3} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{3} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 8 \, b d^{2} x^{2} \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d^{2} x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + a d^{3} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{3} \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) + 8 \, b d x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 8 \, b d x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b d^{2} x^{2} - a d^{3} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) + a d^{3} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d^{3} \operatorname {Si}\left (d x\right ) - 4 \, b \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 8 \, b d x \tan \left (\frac {1}{2} \, d x\right ) - 8 \, b d x \tan \left (\frac {1}{2} \, c\right ) + 4 \, b \tan \left (\frac {1}{2} \, d x\right )^{2} + 16 \, b \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, b \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, b}{2 \, {\left (d^{3} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{3} \tan \left (\frac {1}{2} \, d x\right )^{2} + d^{3} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*d^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2
- a*d^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*sin_integral(d*x)*tan(1/2*d*x)^2*
tan(1/2*c)^2 - 2*a*d^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^3*real_part(cos_integral
(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*b*d^2*x^2*tan(1/2*d*x)^2 - a*d^3*imag_part(cos_integral(d*x))*tan(1/2*d*
x)^2 + a*d^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a*d^3*sin_integral(d*x)*tan(1/2*d*x)^2 - 8*b*d^2
*x^2*tan(1/2*d*x)*tan(1/2*c) - 2*b*d^2*x^2*tan(1/2*c)^2 + a*d^3*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a*
d^3*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^3*sin_integral(d*x)*tan(1/2*c)^2 - 2*a*d^3*real_part(co
s_integral(d*x))*tan(1/2*c) - 2*a*d^3*real_part(cos_integral(-d*x))*tan(1/2*c) + 8*b*d*x*tan(1/2*d*x)^2*tan(1/
2*c) + 8*b*d*x*tan(1/2*d*x)*tan(1/2*c)^2 + 2*b*d^2*x^2 - a*d^3*imag_part(cos_integral(d*x)) + a*d^3*imag_part(
cos_integral(-d*x)) - 2*a*d^3*sin_integral(d*x) - 4*b*tan(1/2*d*x)^2*tan(1/2*c)^2 - 8*b*d*x*tan(1/2*d*x) - 8*b
*d*x*tan(1/2*c) + 4*b*tan(1/2*d*x)^2 + 16*b*tan(1/2*d*x)*tan(1/2*c) + 4*b*tan(1/2*c)^2 - 4*b)/(d^3*tan(1/2*d*x
)^2*tan(1/2*c)^2 + d^3*tan(1/2*d*x)^2 + d^3*tan(1/2*c)^2 + d^3)

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maple [A]  time = 0.03, size = 112, normalized size = 1.96 \[ \frac {\left (c^{2}+c +1\right ) b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}-\frac {3 c b \left (1+c \right ) \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-\frac {3 c^{2} b \cos \left (d x +c \right )}{d^{3}}+a \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 2.27, size = 76, normalized size = 1.33 \[ \frac {{\left (a {\left (-i \, {\rm Ei}\left (i \, d x\right ) + i \, {\rm Ei}\left (-i \, d x\right )\right )} \cos \relax (c) + a {\left ({\rm Ei}\left (i \, d x\right ) + {\rm Ei}\left (-i \, d x\right )\right )} \sin \relax (c)\right )} d^{3} + 4 \, b d x \sin \left (d x + c\right ) - 2 \, {\left (b d^{2} x^{2} - 2 \, b\right )} \cos \left (d x + c\right )}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ a\,\mathrm {cosint}\left (d\,x\right )\,\sin \relax (c)+a\,\mathrm {sinint}\left (d\,x\right )\,\cos \relax (c)+\frac {b\,\left (2\,\cos \left (c+d\,x\right )-d^2\,x^2\,\cos \left (c+d\,x\right )+2\,d\,x\,\sin \left (c+d\,x\right )\right )}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3))/x,x)

[Out]

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

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sympy [A]  time = 6.38, size = 85, normalized size = 1.49 \[ a \sin {\relax (c )} \operatorname {Ci}{\left (d x \right )} + a \cos {\relax (c )} \operatorname {Si}{\left (d x \right )} + b x^{2} \left (\begin {cases} - \cos {\relax (c )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - 2 b \left (\begin {cases} - \frac {x^{2} \cos {\relax (c )}}{2} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {x \sin {\left (c + d x \right )}}{d} + \frac {\cos {\left (c + d x \right )}}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{2} \cos {\relax (c )}}{2} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sin(c)*Ci(d*x) + a*cos(c)*Si(d*x) + b*x**2*Piecewise((-cos(c), Eq(d, 0)), (-cos(c + d*x)/d, True)) - 2*b*Pie
cewise((-x**2*cos(c)/2, Eq(d, 0)), (-Piecewise((x*sin(c + d*x)/d + cos(c + d*x)/d**2, Ne(d, 0)), (x**2*cos(c)/
2, True))/d, True))

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