3.85 \(\int \frac{(a+b x^3) \sin (c+d x)}{x^3} \, dx\)
Optimal. Leaf size=70 \[ -\frac{1}{2} a d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a d^2 \cos (c) \text{Si}(d x)-\frac{a \sin (c+d x)}{2 x^2}-\frac{a d \cos (c+d x)}{2 x}-\frac{b \cos (c+d x)}{d} \]
[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
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Rubi [A] time = 0.126812, antiderivative size = 70, normalized size of antiderivative = 1.,
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 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]
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 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 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]
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*}
Mathematica [A] time = 0.154956, size = 66, normalized size = 0.94 \[ \frac{1}{2} \left (-a d^2 \sin (c) \text{CosIntegral}(d x)-a d^2 \cos (c) \text{Si}(d x)-\frac{a \sin (c+d x)}{x^2}-\frac{a d \cos (c+d x)}{x}-\frac{2 b \cos (c+d x)}{d}\right ) \]
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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Maple [A] time = 0.015, size = 65, normalized size = 0.9 \begin{align*}{d}^{2} \left ( -{\frac{b\cos \left ( dx+c \right ) }{{d}^{3}}}+a \left ( -{\frac{\sin \left ( dx+c \right ) }{2\,{d}^{2}{x}^{2}}}-{\frac{\cos \left ( dx+c \right ) }{2\,dx}}-{\frac{{\it Si} \left ( dx \right ) \cos \left ( c \right ) }{2}}-{\frac{{\it Ci} \left ( dx \right ) \sin \left ( c \right ) }{2}} \right ) \right ) \end{align*}
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 = 2.0571, size = 1554, normalized size = 22.2 \begin{align*} \text{result too large to display} \end{align*}
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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Fricas [A] time = 1.71873, size = 244, normalized size = 3.49 \begin{align*} -\frac{2 \, a d^{3} x^{2} \cos \left (c\right ) \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 \left (c\right )}{4 \, d x^{2}} \end{align*}
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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right ) \sin{\left (c + d x \right )}}{x^{3}}\, dx \end{align*}
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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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
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]
Exception raised: TypeError