\(\int \frac {(a+b x^2) \sin (c+d x)}{x^4} \, dx\) [47]

Optimal result

Integrand size = 17, antiderivative size = 106 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=-\frac {a d \cos (c+d x)}{6 x^2}+b d \cos (c) \operatorname {CosIntegral}(d x)-\frac {1}{6} a d^3 \cos (c) \operatorname {CosIntegral}(d x)-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{x}+\frac {a d^2 \sin (c+d x)}{6 x}-b d \sin (c) \text {Si}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3420, 3378, 3384, 3380, 3383} \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=-\frac {1}{6} a d^3 \cos (c) \operatorname {CosIntegral}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}+b d \cos (c) \operatorname {CosIntegral}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x} \]

[In]

Int[((a + b*x^2)*Sin[c + d*x])/x^4,x]

[Out]

-1/6*(a*d*Cos[c + d*x])/x^2 + b*d*Cos[c]*CosIntegral[d*x] - (a*d^3*Cos[c]*CosIntegral[d*x])/6 - (a*Sin[c + d*x
])/(3*x^3) - (b*Sin[c + d*x])/x + (a*d^2*Sin[c + d*x])/(6*x) - b*d*Sin[c]*SinIntegral[d*x] + (a*d^3*Sin[c]*Sin
Integral[d*x])/6

Rule 3378

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 3380

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 3383

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 3384

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 3420

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=\frac {-a d x \cos (c+d x)+d \left (6 b-a d^2\right ) x^3 \cos (c) \operatorname {CosIntegral}(d x)-2 a \sin (c+d x)-6 b x^2 \sin (c+d x)+a d^2 x^2 \sin (c+d x)+d \left (-6 b+a d^2\right ) x^3 \sin (c) \text {Si}(d x)}{6 x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(d^{3} \left (a \left (-\frac {\sin \left (d x +c \right )}{3 d^{3} x^{3}}-\frac {\cos \left (d x +c \right )}{6 d^{2} x^{2}}+\frac {\sin \left (d x +c \right )}{6 d x}+\frac {\operatorname {Si}\left (d x \right ) \sin \left (c \right )}{6}-\frac {\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{6}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )}{d^{2}}\right )\)	\(102\)
default	\(d^{3} \left (a \left (-\frac {\sin \left (d x +c \right )}{3 d^{3} x^{3}}-\frac {\cos \left (d x +c \right )}{6 d^{2} x^{2}}+\frac {\sin \left (d x +c \right )}{6 d x}+\frac {\operatorname {Si}\left (d x \right ) \sin \left (c \right )}{6}-\frac {\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{6}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )}{d^{2}}\right )\)	\(102\)
risch	\(\frac {\cos \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) a \,d^{3}}{12}+\frac {\cos \left (c \right ) \operatorname {Ei}_{1}\left (-i d x \right ) a \,d^{3}}{12}-\frac {\cos \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) b d}{2}-\frac {\cos \left (c \right ) \operatorname {Ei}_{1}\left (-i d x \right ) b d}{2}-\frac {i \sin \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) a \,d^{3}}{12}+\frac {i \sin \left (c \right ) \operatorname {Ei}_{1}\left (-i d x \right ) a \,d^{3}}{12}+\frac {i \sin \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) b d}{2}-\frac {i \sin \left (c \right ) \operatorname {Ei}_{1}\left (-i d x \right ) b d}{2}-\frac {a d \cos \left (d x +c \right )}{6 x^{2}}+\frac {i \left (-2 i a \,d^{7} x^{5}+12 i b \,d^{5} x^{5}+4 i a \,d^{5} x^{3}\right ) \sin \left (d x +c \right )}{12 d^{5} x^{6}}\)	\(177\)
meijerg	\(\frac {d^{2} b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {d b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sqrt {\pi }\, \sin \left (c \right ) d^{4} \left (-\frac {8 \left (-d^{2} x^{2}+2\right ) d^{2} \cos \left (x \sqrt {d^{2}}\right )}{3 x^{3} \left (d^{2}\right )^{\frac {5}{2}} \sqrt {\pi }}+\frac {8 \sin \left (x \sqrt {d^{2}}\right )}{3 d^{2} x^{2} \sqrt {\pi }}+\frac {8 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{3 \sqrt {\pi }}\right )}{16 \sqrt {d^{2}}}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) d^{3} \left (-\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {4 \left (2 \gamma -\frac {11}{3}+2 \ln \left (x \right )+2 \ln \left (d \right )\right )}{3 \sqrt {\pi }}+\frac {-\frac {44 d^{2} x^{2}}{9}+8}{d^{2} x^{2} \sqrt {\pi }}+\frac {8 \gamma }{3 \sqrt {\pi }}+\frac {8 \ln \left (2\right )}{3 \sqrt {\pi }}+\frac {8 \ln \left (\frac {d x}{2}\right )}{3 \sqrt {\pi }}-\frac {8 \cos \left (d x \right )}{3 \sqrt {\pi }\, d^{2} x^{2}}-\frac {16 \left (-\frac {5 d^{2} x^{2}}{2}+5\right ) \sin \left (d x \right )}{15 \sqrt {\pi }\, d^{3} x^{3}}-\frac {8 \,\operatorname {Ci}\left (d x \right )}{3 \sqrt {\pi }}\right )}{16}\)	\(353\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=-\frac {{\left (a d^{3} - 6 \, b d\right )} x^{3} \cos \left (c\right ) \operatorname {Ci}\left (d x\right ) - {\left (a d^{3} - 6 \, b d\right )} x^{3} \sin \left (c\right ) \operatorname {Si}\left (d x\right ) + a d x \cos \left (d x + c\right ) - {\left ({\left (a d^{2} - 6 \, b\right )} x^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{6 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=\int \frac {\left (a + b x^{2}\right ) \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.72 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=-\frac {{\left ({\left (a {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + a {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{5} - 6 \, {\left (b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + b {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3}\right )} x^{3} + 2 \, b d x \cos \left (d x + c\right ) + 4 \, b \sin \left (d x + c\right )}{2 \, d^{2} x^{3}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.31 (sec) , antiderivative size = 834, normalized size of antiderivative = 7.87 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/12*(a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*d^3*x^3*real_part(cos_integral(-d
*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^
3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*t
an(1/2*c) - a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a*d^3*x^3*real_part(cos_integral(-d*x))*ta
n(1/2*d*x)^2 + a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a*d^3*x^3*real_part(cos_integral(-d*x))*t
an(1/2*c)^2 - 6*b*d*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 6*b*d*x^3*real_part(cos_int
egral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^3*x^3*i
mag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a*d^3*x^3*sin_integral(d*x)*tan(1/2*c) - 12*b*d*x^3*imag_part(cos_
integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 12*b*d*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)
- 24*b*d*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - a*d^3*x^3*real_part(cos_integral(d*x)) - a*d^3*x^3*
real_part(cos_integral(-d*x)) + 6*b*d*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 6*b*d*x^3*real_part(co
s_integral(-d*x))*tan(1/2*d*x)^2 - 4*a*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 6*b*d*x^3*real_part(cos_integral(d*
x))*tan(1/2*c)^2 - 6*b*d*x^3*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 4*a*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^
2 - 12*b*d*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) + 12*b*d*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c) -
24*b*d*x^3*sin_integral(d*x)*tan(1/2*c) - 2*a*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 6*b*d*x^3*real_part(cos_integ
ral(d*x)) + 6*b*d*x^3*real_part(cos_integral(-d*x)) + 4*a*d^2*x^2*tan(1/2*d*x) + 4*a*d^2*x^2*tan(1/2*c) + 24*b
*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 24*b*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*a*d*x*tan(1/2*d*x)^2 + 8*a*d*x*tan(1/2
*d*x)*tan(1/2*c) + 2*a*d*x*tan(1/2*c)^2 - 24*b*x^2*tan(1/2*d*x) - 24*b*x^2*tan(1/2*c) + 8*a*tan(1/2*d*x)^2*tan
(1/2*c) + 8*a*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a*d*x - 8*a*tan(1/2*d*x) - 8*a*tan(1/2*c))/(x^3*tan(1/2*d*x)^2*tan
(1/2*c)^2 + x^3*tan(1/2*d*x)^2 + x^3*tan(1/2*c)^2 + x^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^4} \, dx=\int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^4} \,d x \]

[In]

int((sin(c + d*x)*(a + b*x^2))/x^4,x)

[Out]

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