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

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 2637

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

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.44, size = 99, normalized size = 0.87 \[ \frac {1}{2} \left (-\frac {a^2 \sin (c+d x)}{x^2}-\frac {a^2 d \cos (c+d x)}{x}+a \sin (c) \left (4 b-a d^2\right ) \text {Ci}(d x)+a \cos (c) \left (4 b-a d^2\right ) \text {Si}(d x)+\frac {2 b^2 \sin (c+d x)}{d^2}-\frac {2 b^2 x \cos (c+d x)}{d}\right ) \]

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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fricas [A]  time = 0.60, size = 136, normalized size = 1.19 \[ -\frac {2 \, {\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \cos \relax (c) \operatorname {Si}\left (d x\right ) + 2 \, {\left (a^{2} d^{3} x + 2 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) + 2 \, {\left (a^{2} d^{2} - 2 \, b^{2} x^{2}\right )} \sin \left (d x + c\right ) + {\left ({\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{4} - 4 \, a b d^{2}\right )} x^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{4 \, d^{2} x^{2}} \]

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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giac [C]  time = 0.32, size = 1058, normalized size = 9.28 \[ \text {result too large to display} \]

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]

1/4*(a^2*d^4*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^2*imag_part(cos_integral
(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^4*
x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^4*x^2*real_part(cos_integral(-d*x))*tan(1
/2*d*x)^2*tan(1/2*c) - a^2*d^4*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a^2*d^4*x^2*imag_part(cos_int
egral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^4*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 + a^2*d^4*x^2*imag_part(cos_integ
ral(d*x))*tan(1/2*c)^2 - a^2*d^4*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^4*x^2*sin_integral(d
*x)*tan(1/2*c)^2 - 4*a*b*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a*b*d^2*x^2*imag
_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 8*a*b*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2
*c)^2 - 2*a^2*d^4*x^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^4*x^2*real_part(cos_integral(-d*x))*ta
n(1/2*c) + 8*a*b*d^2*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 8*a*b*d^2*x^2*real_part(cos_
integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^3*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*b^2*d*x^3*tan(1/2*d*x)
^2*tan(1/2*c)^2 - a^2*d^4*x^2*imag_part(cos_integral(d*x)) + a^2*d^4*x^2*imag_part(cos_integral(-d*x)) - 2*a^2
*d^4*x^2*sin_integral(d*x) + 4*a*b*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 4*a*b*d^2*x^2*imag_pa
rt(cos_integral(-d*x))*tan(1/2*d*x)^2 + 8*a*b*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 - 4*a*b*d^2*x^2*imag_pa
rt(cos_integral(d*x))*tan(1/2*c)^2 + 4*a*b*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 8*a*b*d^2*x^2*
sin_integral(d*x)*tan(1/2*c)^2 + 2*a^2*d^3*x*tan(1/2*d*x)^2 + 4*b^2*d*x^3*tan(1/2*d*x)^2 + 8*a*b*d^2*x^2*real_
part(cos_integral(d*x))*tan(1/2*c) + 8*a*b*d^2*x^2*real_part(cos_integral(-d*x))*tan(1/2*c) + 8*a^2*d^3*x*tan(
1/2*d*x)*tan(1/2*c) + 16*b^2*d*x^3*tan(1/2*d*x)*tan(1/2*c) + 2*a^2*d^3*x*tan(1/2*c)^2 + 4*b^2*d*x^3*tan(1/2*c)
^2 + 4*a*b*d^2*x^2*imag_part(cos_integral(d*x)) - 4*a*b*d^2*x^2*imag_part(cos_integral(-d*x)) + 8*a*b*d^2*x^2*
sin_integral(d*x) + 4*a^2*d^2*tan(1/2*d*x)^2*tan(1/2*c) - 8*b^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^2*tan(
1/2*d*x)*tan(1/2*c)^2 - 8*b^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d^3*x - 4*b^2*d*x^3 - 4*a^2*d^2*tan(1/2*d*
x) + 8*b^2*x^2*tan(1/2*d*x) - 4*a^2*d^2*tan(1/2*c) + 8*b^2*x^2*tan(1/2*c))/(d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c)^
2 + d^2*x^2*tan(1/2*d*x)^2 + d^2*x^2*tan(1/2*c)^2 + d^2*x^2)

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maple [A]  time = 0.04, size = 124, normalized size = 1.09 \[ d^{2} \left (\frac {\left (1+3 c \right ) b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{4}}+\frac {4 c \,b^{2} \cos \left (d x +c \right )}{d^{4}}+\frac {2 a b \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right )}{d^{2}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )\right ) \]

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 = 2.81, size = 150, normalized size = 1.32 \[ \frac {{\left ({\left (a^{2} {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4} + {\left (a b {\left (-4 i \, \Gamma \left (-2, i \, d x\right ) + 4 i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \relax (c) - 4 \, a b {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{2}\right )} x^{2} - 2 \, {\left (b^{2} d x^{3} + 2 \, a b d x\right )} \cos \left (d x + c\right ) + 2 \, {\left (b^{2} x^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, d^{2} x^{2}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^2+a\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2} \sin {\left (c + d x \right )}}{x^{3}}\, dx \]

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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