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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 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 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^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 = 0.45, size = 114, normalized size = 0.85 \[ \frac {1}{6} \left (\frac {a^2 d^2 \sin (c+d x)}{x}-\frac {2 a^2 \sin (c+d x)}{x^3}-\frac {a^2 d \cos (c+d x)}{x^2}-a d \cos (c) \left (a d^2-12 b\right ) \text {Ci}(d x)+a d \sin (c) \left (a d^2-12 b\right ) \text {Si}(d x)-\frac {12 a b \sin (c+d x)}{x}-\frac {6 b^2 \cos (c+d x)}{d}\right ) \]

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

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 145, normalized size = 1.08 \[ \frac {2 \, {\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \sin \relax (c) \operatorname {Si}\left (d x\right ) - 2 \, {\left (a^{2} d^{2} x + 6 \, b^{2} x^{3}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{4} - 12 \, a b d^{2}\right )} x^{3} \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) - 2 \, {\left (2 \, a^{2} d - {\left (a^{2} d^{3} - 12 \, a b d\right )} x^{2}\right )} \sin \left (d x + c\right )}{12 \, d x^{3}} \]

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)

________________________________________________________________________________________

giac [C]  time = 0.50, size = 1032, normalized size = 7.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(a^2*d^4*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^4*x^3*real_part(cos_integra
l(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) -
2*a^2*d^4*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^4*x^3*sin_integral(d*x)*tan(1/
2*d*x)^2*tan(1/2*c) - a^2*d^4*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a^2*d^4*x^3*real_part(cos_inte
gral(-d*x))*tan(1/2*d*x)^2 + a^2*d^4*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a^2*d^4*x^3*real_part(cos
_integral(-d*x))*tan(1/2*c)^2 - 12*a*b*d^2*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 12*a
*b*d^2*x^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x^3*imag_part(cos_integral(d*
x))*tan(1/2*c) - 2*a^2*d^4*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a^2*d^4*x^3*sin_integral(d*x)*tan(
1/2*c) - 24*a*b*d^2*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 24*a*b*d^2*x^3*imag_part(cos_
integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 48*a*b*d^2*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - a^2*d
^4*x^3*real_part(cos_integral(d*x)) - a^2*d^4*x^3*real_part(cos_integral(-d*x)) + 12*a*b*d^2*x^3*real_part(cos
_integral(d*x))*tan(1/2*d*x)^2 + 12*a*b*d^2*x^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 4*a^2*d^3*x^2*t
an(1/2*d*x)^2*tan(1/2*c) - 12*a*b*d^2*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 - 12*a*b*d^2*x^3*real_part
(cos_integral(-d*x))*tan(1/2*c)^2 - 4*a^2*d^3*x^2*tan(1/2*d*x)*tan(1/2*c)^2 - 24*a*b*d^2*x^3*imag_part(cos_int
egral(d*x))*tan(1/2*c) + 24*a*b*d^2*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c) - 48*a*b*d^2*x^3*sin_integral
(d*x)*tan(1/2*c) - 2*a^2*d^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - 12*b^2*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*a*b*d
^2*x^3*real_part(cos_integral(d*x)) + 12*a*b*d^2*x^3*real_part(cos_integral(-d*x)) + 4*a^2*d^3*x^2*tan(1/2*d*x
) + 4*a^2*d^3*x^2*tan(1/2*c) + 48*a*b*d*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 48*a*b*d*x^2*tan(1/2*d*x)*tan(1/2*c)^2
+ 2*a^2*d^2*x*tan(1/2*d*x)^2 + 12*b^2*x^3*tan(1/2*d*x)^2 + 8*a^2*d^2*x*tan(1/2*d*x)*tan(1/2*c) + 48*b^2*x^3*t
an(1/2*d*x)*tan(1/2*c) + 2*a^2*d^2*x*tan(1/2*c)^2 + 12*b^2*x^3*tan(1/2*c)^2 - 48*a*b*d*x^2*tan(1/2*d*x) - 48*a
*b*d*x^2*tan(1/2*c) + 8*a^2*d*tan(1/2*d*x)^2*tan(1/2*c) + 8*a^2*d*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d^2*x - 12
*b^2*x^3 - 8*a^2*d*tan(1/2*d*x) - 8*a^2*d*tan(1/2*c))/(d*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*x^3*tan(1/2*d*x)^
2 + d*x^3*tan(1/2*c)^2 + d*x^3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 120, normalized size = 0.90 \[ d^{3} \left (-\frac {b^{2} \cos \left (d x +c \right )}{d^{4}}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d^{2}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )\right ) \]

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)))

________________________________________________________________________________________

maxima [C]  time = 2.28, size = 142, normalized size = 1.06 \[ -\frac {{\left ({\left (a^{2} {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{5} - {\left (12 \, a b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) - a b {\left (12 i \, \Gamma \left (-3, i \, d x\right ) - 12 i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 8 \, a b \sin \left (d x + c\right ) + 2 \, {\left (b^{2} d x^{3} + 2 \, a b d x\right )} \cos \left (d x + c\right )}{2 \, d^{2} x^{3}} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^2+a\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2} \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

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)

________________________________________________________________________________________