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

Optimal. Leaf size=142 \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{2 a b \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.218588, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {3339, 2638, 3297, 3303, 3299, 3302, 3296, 2637} \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{2 a b \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 )^2 \sin (c+d x)}{x^3} \, dx &=\int \left (2 a b \sin (c+d x)+\frac{a^2 \sin (c+d x)}{x^3}+b^2 x^3 \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^3} \, dx+(2 a b) \int \sin (c+d x) \, dx+b^2 \int x^3 \sin (c+d x) \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{\left (3 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d}+\frac{1}{2} \left (a^2 d\right ) \int \frac{\cos (c+d x)}{x^2} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{\left (6 b^2\right ) \int x \sin (c+d x) \, dx}{d^2}-\frac{1}{2} \left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{\left (6 b^2\right ) \int \cos (c+d x) \, dx}{d^3}-\frac{1}{2} \left (a^2 d^2 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx-\frac{1}{2} \left (a^2 d^2 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{1}{2} a^2 d^2 \text{Ci}(d x) \sin (c)-\frac{6 b^2 \sin (c+d x)}{d^4}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)\\ \end{align*}

Mathematica [A]  time = 0.382464, size = 138, normalized size = 0.97 \[ \frac{1}{2} \left (-a^2 d^2 \sin (c) \text{CosIntegral}(d x)-a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{x^2}-\frac{a^2 d \cos (c+d x)}{x}-\frac{4 a b \cos (c+d x)}{d}+\frac{6 b^2 x^2 \sin (c+d x)}{d^2}-\frac{12 b^2 \sin (c+d x)}{d^4}+\frac{12 b^2 x \cos (c+d x)}{d^3}-\frac{2 b^2 x^3 \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*b*Cos[c + d*x])/d - (a^2*d*Cos[c + d*x])/x + (12*b^2*x*Cos[c + d*x])/d^3 - (2*b^2*x^3*Cos[c + d*x])/d -
a^2*d^2*CosIntegral[d*x]*Sin[c] - (12*b^2*Sin[c + d*x])/d^4 - (a^2*Sin[c + d*x])/x^2 + (6*b^2*x^2*Sin[c + d*x
])/d^2 - a^2*d^2*Cos[c]*SinIntegral[d*x])/2

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 251, normalized size = 1.8 \begin{align*}{d}^{2} \left ({\frac{ \left ( 10\,{c}^{3}+6\,{c}^{2}+3\,c+1 \right ){b}^{2} \left ( - \left ( dx+c \right ) ^{3}\cos \left ( dx+c \right ) +3\, \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) -6\,\sin \left ( dx+c \right ) +6\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{6}}}-6\,{\frac{c{b}^{2} \left ( 6\,{c}^{2}+3\,c+1 \right ) \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{6}}}+15\,{\frac{ \left ( 1+3\,c \right ){c}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{6}}}-2\,{\frac{ab\cos \left ( dx+c \right ) }{{d}^{3}}}+20\,{\frac{{b}^{2}{c}^{3}\cos \left ( dx+c \right ) }{{d}^{6}}}+{a}^{2} \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)^2*sin(d*x+c)/x^3,x)

[Out]

d^2*((10*c^3+6*c^2+3*c+1)/d^6*b^2*(-(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))-6*c*b^2*(6*c^2+3*c+1)/d^6*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+15*(1+3*c)/d^6*c^2*b^
2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-2*a*b*cos(d*x+c)/d^3+20*c^3/d^6*b^2*cos(d*x+c)+a^2*(-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)))

________________________________________________________________________________________

Maxima [C]  time = 12.6194, size = 149, normalized size = 1.05 \begin{align*} \frac{{\left (a^{2}{\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 2 \,{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} - 6 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 6 \,{\left (b^{2} d^{2} x^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*(I*gamma(-2, I*d*x) - I*gamma(-2, -I*d*x))*cos(c) + a^2*(gamma(-2, I*d*x) + gamma(-2, -I*d*x))*sin(c
))*d^6 - 2*(b^2*d^3*x^3 + 2*a*b*d^3 - 6*b^2*d*x)*cos(d*x + c) + 6*(b^2*d^2*x^2 - 2*b^2)*sin(d*x + c))/d^4

________________________________________________________________________________________

Fricas [A]  time = 1.9102, size = 355, normalized size = 2.5 \begin{align*} -\frac{2 \, a^{2} d^{6} x^{2} \cos \left (c\right ) \operatorname{Si}\left (d x\right ) + 2 \,{\left (2 \, b^{2} d^{3} x^{5} + a^{2} d^{5} x + 4 \, a b d^{3} x^{2} - 12 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) - 2 \,{\left (6 \, b^{2} d^{2} x^{4} - a^{2} d^{4} - 12 \, b^{2} x^{2}\right )} \sin \left (d x + c\right ) +{\left (a^{2} d^{6} x^{2} \operatorname{Ci}\left (d x\right ) + a^{2} d^{6} x^{2} \operatorname{Ci}\left (-d x\right )\right )} \sin \left (c\right )}{4 \, d^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^2*d^6*x^2*cos(c)*sin_integral(d*x) + 2*(2*b^2*d^3*x^5 + a^2*d^5*x + 4*a*b*d^3*x^2 - 12*b^2*d*x^3)*co
s(d*x + c) - 2*(6*b^2*d^2*x^4 - a^2*d^4 - 12*b^2*x^2)*sin(d*x + c) + (a^2*d^6*x^2*cos_integral(d*x) + a^2*d^6*
x^2*cos_integral(-d*x))*sin(c))/(d^4*x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{2} \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)**2*sin(d*x+c)/x**3,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError