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

Optimal. Leaf size=145 \[ a^2 d \cos (c) \text{CosIntegral}(d x)-a^2 d \sin (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{x}+\frac{2 a b \sin (c+d x)}{d^2}-\frac{2 a b x \cos (c+d x)}{d}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{24 b^2 x \sin (c+d x)}{d^4}-\frac{24 b^2 \cos (c+d x)}{d^5}-\frac{b^2 x^4 \cos (c+d x)}{d} \]

[Out]

(-24*b^2*Cos[c + d*x])/d^5 - (2*a*b*x*Cos[c + d*x])/d + (12*b^2*x^2*Cos[c + d*x])/d^3 - (b^2*x^4*Cos[c + d*x])
/d + a^2*d*Cos[c]*CosIntegral[d*x] + (2*a*b*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/x - (24*b^2*x*Sin[c + d*x])
/d^4 + (4*b^2*x^3*Sin[c + d*x])/d^2 - a^2*d*Sin[c]*SinIntegral[d*x]

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Rubi [A]  time = 0.233297, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {3339, 3297, 3303, 3299, 3302, 3296, 2637, 2638} \[ a^2 d \cos (c) \text{CosIntegral}(d x)-a^2 d \sin (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{x}+\frac{2 a b \sin (c+d x)}{d^2}-\frac{2 a b x \cos (c+d x)}{d}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{24 b^2 x \sin (c+d x)}{d^4}-\frac{24 b^2 \cos (c+d x)}{d^5}-\frac{b^2 x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*b^2*Cos[c + d*x])/d^5 - (2*a*b*x*Cos[c + d*x])/d + (12*b^2*x^2*Cos[c + d*x])/d^3 - (b^2*x^4*Cos[c + d*x])
/d + a^2*d*Cos[c]*CosIntegral[d*x] + (2*a*b*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/x - (24*b^2*x*Sin[c + d*x])
/d^4 + (4*b^2*x^3*Sin[c + d*x])/d^2 - a^2*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]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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^2} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{x^2}+2 a b x \sin (c+d x)+b^2 x^4 \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^2} \, dx+(2 a b) \int x \sin (c+d x) \, dx+b^2 \int x^4 \sin (c+d x) \, dx\\ &=-\frac{2 a b x \cos (c+d x)}{d}-\frac{b^2 x^4 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{x}+\frac{(2 a b) \int \cos (c+d x) \, dx}{d}+\frac{\left (4 b^2\right ) \int x^3 \cos (c+d x) \, dx}{d}+\left (a^2 d\right ) \int \frac{\cos (c+d x)}{x} \, dx\\ &=-\frac{2 a b x \cos (c+d x)}{d}-\frac{b^2 x^4 \cos (c+d x)}{d}+\frac{2 a b \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{x}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}-\frac{\left (12 b^2\right ) \int x^2 \sin (c+d x) \, dx}{d^2}+\left (a^2 d \cos (c)\right ) \int \frac{\cos (d x)}{x} \, dx-\left (a^2 d \sin (c)\right ) \int \frac{\sin (d x)}{x} \, dx\\ &=-\frac{2 a b x \cos (c+d x)}{d}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{b^2 x^4 \cos (c+d x)}{d}+a^2 d \cos (c) \text{Ci}(d x)+\frac{2 a b \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{x}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}-a^2 d \sin (c) \text{Si}(d x)-\frac{\left (24 b^2\right ) \int x \cos (c+d x) \, dx}{d^3}\\ &=-\frac{2 a b x \cos (c+d x)}{d}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{b^2 x^4 \cos (c+d x)}{d}+a^2 d \cos (c) \text{Ci}(d x)+\frac{2 a b \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{x}-\frac{24 b^2 x \sin (c+d x)}{d^4}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}-a^2 d \sin (c) \text{Si}(d x)+\frac{\left (24 b^2\right ) \int \sin (c+d x) \, dx}{d^4}\\ &=-\frac{24 b^2 \cos (c+d x)}{d^5}-\frac{2 a b x \cos (c+d x)}{d}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{b^2 x^4 \cos (c+d x)}{d}+a^2 d \cos (c) \text{Ci}(d x)+\frac{2 a b \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{x}-\frac{24 b^2 x \sin (c+d x)}{d^4}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}-a^2 d \sin (c) \text{Si}(d x)\\ \end{align*}

Mathematica [A]  time = 0.367577, size = 145, normalized size = 1. \[ a^2 d \cos (c) \text{CosIntegral}(d x)-a^2 d \sin (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{x}+\frac{2 a b \sin (c+d x)}{d^2}-\frac{2 a b x \cos (c+d x)}{d}+\frac{4 b^2 x^3 \sin (c+d x)}{d^2}+\frac{12 b^2 x^2 \cos (c+d x)}{d^3}-\frac{24 b^2 x \sin (c+d x)}{d^4}-\frac{24 b^2 \cos (c+d x)}{d^5}-\frac{b^2 x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*b^2*Cos[c + d*x])/d^5 - (2*a*b*x*Cos[c + d*x])/d + (12*b^2*x^2*Cos[c + d*x])/d^3 - (b^2*x^4*Cos[c + d*x])
/d + a^2*d*Cos[c]*CosIntegral[d*x] + (2*a*b*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/x - (24*b^2*x*Sin[c + d*x])
/d^4 + (4*b^2*x^3*Sin[c + d*x])/d^2 - a^2*d*Sin[c]*SinIntegral[d*x]

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Maple [B]  time = 0.031, size = 365, normalized size = 2.5 \begin{align*} d \left ({\frac{ \left ( 5\,{c}^{4}+4\,{c}^{3}+3\,{c}^{2}+2\,c+1 \right ){b}^{2} \left ( - \left ( dx+c \right ) ^{4}\cos \left ( dx+c \right ) +4\, \left ( dx+c \right ) ^{3}\sin \left ( dx+c \right ) +12\, \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) -24\,\cos \left ( dx+c \right ) -24\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{6}}}-6\,{\frac{c{b}^{2} \left ( 4\,{c}^{3}+3\,{c}^{2}+2\,c+1 \right ) \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}}}+15\,{\frac{ \left ( 3\,{c}^{2}+2\,c+1 \right ){c}^{2}{b}^{2} \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}}}+2\,{\frac{ \left ( 1+2\,c \right ) ab \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{3}}}-20\,{\frac{{b}^{2}{c}^{3} \left ( 1+2\,c \right ) \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{6}}}+6\,{\frac{cab\cos \left ( dx+c \right ) }{{d}^{3}}}-15\,{\frac{{b}^{2}{c}^{4}\cos \left ( dx+c \right ) }{{d}^{6}}}+{a}^{2} \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)^2*sin(d*x+c)/x^2,x)

[Out]

d*((5*c^4+4*c^3+3*c^2+2*c+1)/d^6*b^2*(-(d*x+c)^4*cos(d*x+c)+4*(d*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos(d*x+c)-24*
cos(d*x+c)-24*(d*x+c)*sin(d*x+c))-6*c*b^2*(4*c^3+3*c^2+2*c+1)/d^6*(-(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))+15*(3*c^2+2*c+1)/d^6*c^2*b^2*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c
)*sin(d*x+c))+2*(1+2*c)/d^3*a*b*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-20*b^2*c^3*(1+2*c)/d^6*(sin(d*x+c)-(d*x+c)*cos
(d*x+c))+6*c/d^3*a*b*cos(d*x+c)-15*c^4/d^6*b^2*cos(d*x+c)+a^2*(-sin(d*x+c)/x/d-Si(d*x)*sin(c)+Ci(d*x)*cos(c)))

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Maxima [C]  time = 41.2719, size = 174, normalized size = 1.2 \begin{align*} \frac{{\left (a^{2}{\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (-i \, \Gamma \left (-1, i \, d x\right ) + i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 2 \,{\left (b^{2} d^{4} x^{4} + 2 \, a b d^{4} x - 12 \, b^{2} d^{2} x^{2} + 24 \, b^{2}\right )} \cos \left (d x + c\right ) + 4 \,{\left (2 \, b^{2} d^{3} x^{3} + a b d^{3} - 12 \, b^{2} d x\right )} \sin \left (d x + c\right )}{2 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*(gamma(-1, I*d*x) + gamma(-1, -I*d*x))*cos(c) + a^2*(-I*gamma(-1, I*d*x) + I*gamma(-1, -I*d*x))*sin(
c))*d^6 - 2*(b^2*d^4*x^4 + 2*a*b*d^4*x - 12*b^2*d^2*x^2 + 24*b^2)*cos(d*x + c) + 4*(2*b^2*d^3*x^3 + a*b*d^3 -
12*b^2*d*x)*sin(d*x + c))/d^5

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Fricas [A]  time = 1.75666, size = 365, normalized size = 2.52 \begin{align*} -\frac{2 \, a^{2} d^{6} x \sin \left (c\right ) \operatorname{Si}\left (d x\right ) + 2 \,{\left (b^{2} d^{4} x^{5} + 2 \, a b d^{4} x^{2} - 12 \, b^{2} d^{2} x^{3} + 24 \, b^{2} x\right )} \cos \left (d x + c\right ) -{\left (a^{2} d^{6} x \operatorname{Ci}\left (d x\right ) + a^{2} d^{6} x \operatorname{Ci}\left (-d x\right )\right )} \cos \left (c\right ) - 2 \,{\left (4 \, b^{2} d^{3} x^{4} - a^{2} d^{5} + 2 \, a b d^{3} x - 24 \, b^{2} d x^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{5} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*d^6*x*sin(c)*sin_integral(d*x) + 2*(b^2*d^4*x^5 + 2*a*b*d^4*x^2 - 12*b^2*d^2*x^3 + 24*b^2*x)*cos(d
*x + c) - (a^2*d^6*x*cos_integral(d*x) + a^2*d^6*x*cos_integral(-d*x))*cos(c) - 2*(4*b^2*d^3*x^4 - a^2*d^5 + 2
*a*b*d^3*x - 24*b^2*d*x^2)*sin(d*x + c))/(d^5*x)

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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^{2}}\, 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**2,x)

[Out]

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

[Out]

Exception raised: TypeError