3.55 \(\int \frac{(a+b x^2)^2 \sin (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=134 \[ -\frac{1}{6} a^2 d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a^2 d^3 \sin (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{6 x}-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a^2 d \cos (c+d x)}{6 x^2}+2 a b d \cos (c) \text{CosIntegral}(d x)-2 a b d \sin (c) \text{Si}(d x)-\frac{2 a b \sin (c+d x)}{x}-\frac{b^2 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.237871, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3339, 2638, 3297, 3303, 3299, 3302} \[ -\frac{1}{6} a^2 d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a^2 d^3 \sin (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{6 x}-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a^2 d \cos (c+d x)}{6 x^2}+2 a b d \cos (c) \text{CosIntegral}(d x)-2 a b d \sin (c) \text{Si}(d x)-\frac{2 a b \sin (c+d x)}{x}-\frac{b^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sin[c + d*x])/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.417768, size = 114, normalized size = 0.85 \[ \frac{1}{6} \left (\frac{a^2 d^2 \sin (c+d x)}{x}-\frac{2 a^2 \sin (c+d x)}{x^3}-\frac{a^2 d \cos (c+d x)}{x^2}-a d \cos (c) \left (a d^2-12 b\right ) \text{CosIntegral}(d x)+a d \sin (c) \left (a d^2-12 b\right ) \text{Si}(d x)-\frac{12 a b \sin (c+d x)}{x}-\frac{6 b^2 \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sin[c + d*x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.023, size = 120, normalized size = 0.9 \begin{align*}{d}^{3} \left ( -{\frac{{b}^{2}\cos \left ( dx+c \right ) }{{d}^{4}}}+2\,{\frac{ab}{{d}^{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 ) }+{a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) }{3\,{d}^{3}{x}^{3}}}-{\frac{\cos \left ( dx+c \right ) }{6\,{d}^{2}{x}^{2}}}+{\frac{\sin \left ( dx+c \right ) }{6\,dx}}+{\frac{{\it Si} \left ( dx \right ) \sin \left ( c \right ) }{6}}-{\frac{{\it Ci} \left ( dx \right ) \cos \left ( c \right ) }{6}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*sin(d*x+c)/x^4,x)

[Out]

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

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Maxima [C]  time = 11.4995, size = 192, normalized size = 1.43 \begin{align*} -\frac{{\left ({\left (a^{2}{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{5} -{\left (12 \, a b{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) - a b{\left (12 i \, \Gamma \left (-3, i \, d x\right ) - 12 i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3}\right )} x^{3} + 8 \, a b \sin \left (d x + c\right ) + 2 \,{\left (b^{2} d x^{3} + 2 \, a b d x\right )} \cos \left (d x + c\right )}{2 \, d^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*sin(d*x+c)/x^4,x, algorithm="maxima")

[Out]

-1/2*(((a^2*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))*cos(c) + a^2*(-I*gamma(-3, I*d*x) + I*gamma(-3, -I*d*x))*si
n(c))*d^5 - (12*a*b*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))*cos(c) - a*b*(12*I*gamma(-3, I*d*x) - 12*I*gamma(-3
, -I*d*x))*sin(c))*d^3)*x^3 + 8*a*b*sin(d*x + c) + 2*(b^2*d*x^3 + 2*a*b*d*x)*cos(d*x + c))/(d^2*x^3)

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Fricas [A]  time = 1.77639, size = 363, normalized size = 2.71 \begin{align*} \frac{2 \,{\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \sin \left (c\right ) \operatorname{Si}\left (d x\right ) - 2 \,{\left (a^{2} d^{2} x + 6 \, b^{2} x^{3}\right )} \cos \left (d x + c\right ) -{\left ({\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \operatorname{Ci}\left (d x\right ) +{\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \operatorname{Ci}\left (-d x\right )\right )} \cos \left (c\right ) - 2 \,{\left (2 \, a^{2} d -{\left (a^{2} d^{3} - 12 \, a b d\right )} x^{2}\right )} \sin \left (d x + c\right )}{12 \, d x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*sin(d*x+c)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2} \sin{\left (c + d x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*sin(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**2)**2*sin(c + d*x)/x**4, 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^2+a)^2*sin(d*x+c)/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError