3.84 \(\int \frac{(a+b x^3) \sin (c+d x)}{x^2} \, dx\)

Optimal. Leaf size=56 \[ a d \cos (c) \text{CosIntegral}(d x)-a d \sin (c) \text{Si}(d x)-\frac{a \sin (c+d x)}{x}+\frac{b \sin (c+d x)}{d^2}-\frac{b x \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.116717, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {3339, 3297, 3303, 3299, 3302, 3296, 2637} \[ a d \cos (c) \text{CosIntegral}(d x)-a d \sin (c) \text{Si}(d x)-\frac{a \sin (c+d x)}{x}+\frac{b \sin (c+d x)}{d^2}-\frac{b x \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x^2,x]

[Out]

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

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 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]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.135205, size = 56, normalized size = 1. \[ a d \cos (c) \text{CosIntegral}(d x)-a d \sin (c) \text{Si}(d x)-\frac{a \sin (c+d x)}{x}+\frac{b \sin (c+d x)}{d^2}-\frac{b x \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)*Sin[c + d*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.016, size = 79, normalized size = 1.4 \begin{align*} d \left ({\frac{ \left ( 1+2\,c \right ) b \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{3}}}+3\,{\frac{cb\cos \left ( dx+c \right ) }{{d}^{3}}}+a \left ( -{\frac{\sin \left ( dx+c \right ) }{dx}}-{\it Si} \left ( dx \right ) \sin \left ( c \right ) +{\it Ci} \left ( dx \right ) \cos \left ( c \right ) \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^2,x)

[Out]

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

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Maxima [C]  time = 2.54833, size = 93, normalized size = 1.66 \begin{align*} \frac{{\left (a{\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) + a{\left (-i \, \Gamma \left (-1, i \, d x\right ) + i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3} - 2 \, b d x \cos \left (d x + c\right ) + 2 \, b \sin \left (d x + c\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7162, size = 234, normalized size = 4.18 \begin{align*} -\frac{2 \, a d^{3} x \sin \left (c\right ) \operatorname{Si}\left (d x\right ) + 2 \, b d x^{2} \cos \left (d x + c\right ) -{\left (a d^{3} x \operatorname{Ci}\left (d x\right ) + a d^{3} x \operatorname{Ci}\left (-d x\right )\right )} \cos \left (c\right ) + 2 \,{\left (a d^{2} - b x\right )} \sin \left (d x + c\right )}{2 \, d^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^2,x, algorithm="fricas")

[Out]

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

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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^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*sin(d*x+c)/x**2,x)

[Out]

Integral((a + b*x**3)*sin(c + d*x)/x**2, 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^2,x, algorithm="giac")

[Out]

Exception raised: TypeError