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

Optimal. Leaf size=70 \[ -\frac {1}{2} a d^2 \sin (c) \text {Ci}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}-\frac {b \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 70, 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, 2638, 3297, 3303, 3299, 3302} \[ -\frac {1}{2} a d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}-\frac {b \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x^3,x]

[Out]

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

Rule 2638

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

Rule 3297

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

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^3} \, dx &=\int \left (b \sin (c+d x)+\frac {a \sin (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x^3} \, dx+b \int \sin (c+d x) \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a \sin (c+d x)}{2 x^2}+\frac {1}{2} (a d) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} \left (a d^2\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} \left (a d^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx-\frac {1}{2} \left (a d^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {1}{2} a d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 66, normalized size = 0.94 \[ \frac {1}{2} \left (-a d^2 \sin (c) \text {Ci}(d x)-a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{x^2}-\frac {a d \cos (c+d x)}{x}-\frac {2 b \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 84, normalized size = 1.20 \[ -\frac {2 \, a d^{3} x^{2} \cos \relax (c) \operatorname {Si}\left (d x\right ) + 2 \, a d \sin \left (d x + c\right ) + 2 \, {\left (a d^{2} x + 2 \, b x^{2}\right )} \cos \left (d x + c\right ) + {\left (a d^{3} x^{2} \operatorname {Ci}\left (d x\right ) + a d^{3} x^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{4 \, d x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.33, size = 564, normalized size = 8.06 \[ \frac {a d^{3} x^{2} \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} x^{2} \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} x^{2} \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} x^{2} \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} x^{2} \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 ) - a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{3} x^{2} \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} x^{2} \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (d x\right ) \right ) + a d^{3} x^{2} \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) - 4 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} + 8 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 2 \, a d^{2} x \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + 16 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, a d \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, b x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, a d \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{2} x - 4 \, b x^{2} - 4 \, a d \tan \left (\frac {1}{2} \, d x\right ) - 4 \, a d \tan \left (\frac {1}{2} \, c\right )}{4 \, {\left (d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + d x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^3*x^2*imag_part(cos_integral(-d*
x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^3*x^2*real
_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^3*x^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*
tan(1/2*c) - a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a*d^3*x^2*imag_part(cos_integral(-d*x))*t
an(1/2*d*x)^2 - 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 + a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*
c)^2 - a*d^3*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*c)^2 - 2*a
*d^3*x^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^3*x^2*real_part(cos_integral(-d*x))*tan(1/2*c) - 2*a*
d^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^3*x^2*imag_part(cos_integral(d*x)) + a*d^3*x^2*imag_part(cos_integral(
-d*x)) - 2*a*d^3*x^2*sin_integral(d*x) - 4*b*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^2*x*tan(1/2*d*x)^2 + 8*a*
d^2*x*tan(1/2*d*x)*tan(1/2*c) + 2*a*d^2*x*tan(1/2*c)^2 + 4*b*x^2*tan(1/2*d*x)^2 + 16*b*x^2*tan(1/2*d*x)*tan(1/
2*c) + 4*a*d*tan(1/2*d*x)^2*tan(1/2*c) + 4*b*x^2*tan(1/2*c)^2 + 4*a*d*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a*d^2*x -
4*b*x^2 - 4*a*d*tan(1/2*d*x) - 4*a*d*tan(1/2*c))/(d*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*x^2*tan(1/2*d*x)^2 + d
*x^2*tan(1/2*c)^2 + d*x^2)

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maple [A]  time = 0.04, size = 65, normalized size = 0.93 \[ d^{2} \left (-\frac {b \cos \left (d x +c \right )}{d^{3}}+a \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.52, size = 1151, normalized size = 16.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((I*exp_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c)^3 + (I*exp_integral_e(3, I*d*x) - I*ex
p_integral_e(3, -I*d*x))*cos(c)*sin(c)^2 + (exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c)^3 + (
I*exp_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c) + ((exp_integral_e(3, I*d*x) + exp_integral_e
(3, -I*d*x))*cos(c)^2 + exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c))*b*c^3/((d*x + c)^2*(cos(
c)^2 + sin(c)^2)*d^3 - 2*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)*d^3 + (c^2*cos(c)^2 + c^2*sin(c)^2)*d^3) - ((I*ex
p_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c)^3 + (I*exp_integral_e(3, I*d*x) - I*exp_integral_
e(3, -I*d*x))*cos(c)*sin(c)^2 + (exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c)^3 + (I*exp_integ
ral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c) + ((exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x)
)*cos(c)^2 + exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c))*a/(c^2*cos(c)^2 + c^2*sin(c)^2 + (d
*x + c)^2*(cos(c)^2 + sin(c)^2) - 2*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)) - (2*((b*cos(c)^2 + b*sin(c)^2)*(d*x
+ c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x + c)^2 + 3*(b*c^2*cos(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x
+ c)^3 - (3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)^3 + 3*b*c^3*(exp_integral_e(4,
I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)*sin(c)^2 - b*c^3*(3*I*exp_integral_e(4, I*d*x) - 3*I*exp_integral_
e(4, -I*d*x))*sin(c)^3 + 3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c) - (b*c^3*(3*I*e
xp_integral_e(4, I*d*x) - 3*I*exp_integral_e(4, -I*d*x))*cos(c)^2 + b*c^3*(3*I*exp_integral_e(4, I*d*x) - 3*I*
exp_integral_e(4, -I*d*x)))*sin(c))*cos(d*x + c)^2 - (3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I
*d*x))*cos(c)^3 + 3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)*sin(c)^2 - b*c^3*(3*I*
exp_integral_e(4, I*d*x) - 3*I*exp_integral_e(4, -I*d*x))*sin(c)^3 + 3*b*c^3*(exp_integral_e(4, I*d*x) + exp_i
ntegral_e(4, -I*d*x))*cos(c) - 2*((b*cos(c)^2 + b*sin(c)^2)*(d*x + c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x
+ c)^2 + 3*(b*c^2*cos(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x + c) - (b*c^3*(3*I*exp_integral_e(4, I*d*x) -
3*I*exp_integral_e(4, -I*d*x))*cos(c)^2 + b*c^3*(3*I*exp_integral_e(4, I*d*x) - 3*I*exp_integral_e(4, -I*d*x)
))*sin(c))*sin(d*x + c)^2 + 2*((b*cos(c)^2 + b*sin(c)^2)*(d*x + c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x +
c)^2 + 3*(b*c^2*cos(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x + c))/(((d*x + c)^3*(cos(c)^2 + sin(c)^2)*d^3 -
3*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)^2*d^3 + 3*(c^2*cos(c)^2 + c^2*sin(c)^2)*(d*x + c)*d^3 - (c^3*cos(c)^2 +
c^3*sin(c)^2)*d^3)*cos(d*x + c)^2 + ((d*x + c)^3*(cos(c)^2 + sin(c)^2)*d^3 - 3*(c*cos(c)^2 + c*sin(c)^2)*(d*x
+ c)^2*d^3 + 3*(c^2*cos(c)^2 + c^2*sin(c)^2)*(d*x + c)*d^3 - (c^3*cos(c)^2 + c^3*sin(c)^2)*d^3)*sin(d*x + c)^2
))*d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3))/x^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{3}\right ) \sin {\left (c + d x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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