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

Optimal. Leaf size=114 \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}+2 a b \sin (c) \text{CosIntegral}(d x)+2 a b \cos (c) \text{Si}(d x)+\frac{b^2 \sin (c+d x)}{d^2}-\frac{b^2 x \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.20306, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3339, 3297, 3303, 3299, 3302, 3296, 2637} \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}+2 a b \sin (c) \text{CosIntegral}(d x)+2 a b \cos (c) \text{Si}(d x)+\frac{b^2 \sin (c+d x)}{d^2}-\frac{b^2 x \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*d*Cos[c + d*x])/(2*x) - (b^2*x*Cos[c + d*x])/d + 2*a*b*CosIntegral[d*x]*Sin[c] - (a^2*d^2*CosIntegral[d*
x]*Sin[c])/2 + (b^2*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/(2*x^2) + 2*a*b*Cos[c]*SinIntegral[d*x] - (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 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^2\right )^2 \sin (c+d x)}{x^3} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{x^3}+\frac{2 a b \sin (c+d x)}{x}+b^2 x \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^3} \, dx+(2 a b) \int \frac{\sin (c+d x)}{x} \, dx+b^2 \int x \sin (c+d x) \, dx\\ &=-\frac{b^2 x \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{b^2 \int \cos (c+d x) \, dx}{d}+\frac{1}{2} \left (a^2 d\right ) \int \frac{\cos (c+d x)}{x^2} \, dx+(2 a b \cos (c)) \int \frac{\sin (d x)}{x} \, dx+(2 a b \sin (c)) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{b^2 x \cos (c+d x)}{d}+2 a b \text{Ci}(d x) \sin (c)+\frac{b^2 \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{2 x^2}+2 a b \cos (c) \text{Si}(d x)-\frac{1}{2} \left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{b^2 x \cos (c+d x)}{d}+2 a b \text{Ci}(d x) \sin (c)+\frac{b^2 \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{2 x^2}+2 a b \cos (c) \text{Si}(d x)-\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{a^2 d \cos (c+d x)}{2 x}-\frac{b^2 x \cos (c+d x)}{d}+2 a b \text{Ci}(d x) \sin (c)-\frac{1}{2} a^2 d^2 \text{Ci}(d x) \sin (c)+\frac{b^2 \sin (c+d x)}{d^2}-\frac{a^2 \sin (c+d x)}{2 x^2}+2 a b \cos (c) \text{Si}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)\\ \end{align*}

Mathematica [A]  time = 0.413665, size = 99, normalized size = 0.87 \[ \frac{1}{2} \left (-\frac{a^2 \sin (c+d x)}{x^2}-\frac{a^2 d \cos (c+d x)}{x}+a \sin (c) \left (4 b-a d^2\right ) \text{CosIntegral}(d x)+a \cos (c) \left (4 b-a d^2\right ) \text{Si}(d x)+\frac{2 b^2 \sin (c+d x)}{d^2}-\frac{2 b^2 x \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 124, normalized size = 1.1 \begin{align*}{d}^{2} \left ({\frac{ \left ( 1+3\,c \right ){b}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{4}}}+4\,{\frac{c{b}^{2}\cos \left ( dx+c \right ) }{{d}^{4}}}+2\,{\frac{ab \left ({\it Si} \left ( dx \right ) \cos \left ( c \right ) +{\it Ci} \left ( dx \right ) \sin \left ( c \right ) \right ) }{{d}^{2}}}+{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^2+a)^2*sin(d*x+c)/x^3,x)

[Out]

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

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Maxima [C]  time = 16.2023, size = 203, normalized size = 1.78 \begin{align*} \frac{{\left ({\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^{4} +{\left (a b{\left (-4 i \, \Gamma \left (-2, i \, d x\right ) + 4 i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - 4 \, a b{\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2}\right )} x^{2} - 2 \,{\left (b^{2} d x^{3} + 2 \, a b d x\right )} \cos \left (d x + c\right ) + 2 \,{\left (b^{2} x^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.86032, size = 344, normalized size = 3.02 \begin{align*} -\frac{2 \,{\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \cos \left (c\right ) \operatorname{Si}\left (d x\right ) + 2 \,{\left (a^{2} d^{3} x + 2 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) + 2 \,{\left (a^{2} d^{2} - 2 \, b^{2} x^{2}\right )} \sin \left (d x + c\right ) +{\left ({\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \operatorname{Ci}\left (d x\right ) +{\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \operatorname{Ci}\left (-d x\right )\right )} \sin \left (c\right )}{4 \, d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

[Out]

Exception raised: TypeError