3.92 \(\int \frac {(a+b x^3)^2 \sin (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=151 \[ -\frac {1}{6} a^2 d^3 \cos (c) \text {Ci}(d x)+\frac {1}{6} a^2 d^3 \sin (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{6 x}-\frac {a^2 \sin (c+d x)}{3 x^3}-\frac {a^2 d \cos (c+d x)}{6 x^2}+2 a b \sin (c) \text {Ci}(d x)+2 a b \cos (c) \text {Si}(d x)+\frac {2 b^2 \cos (c+d x)}{d^3}+\frac {2 b^2 x \sin (c+d x)}{d^2}-\frac {b^2 x^2 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 14, 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, 2638} \[ -\frac {1}{6} a^2 d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{6} a^2 d^3 \sin (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{6 x}-\frac {a^2 \sin (c+d x)}{3 x^3}-\frac {a^2 d \cos (c+d x)}{6 x^2}+2 a b \sin (c) \text {CosIntegral}(d x)+2 a b \cos (c) \text {Si}(d x)+\frac {2 b^2 x \sin (c+d x)}{d^2}+\frac {2 b^2 \cos (c+d x)}{d^3}-\frac {b^2 x^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*Sin[c + d*x])/x^4,x]

[Out]

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

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Mathematica [A]  time = 0.63, size = 135, normalized size = 0.89 \[ \frac {1}{6} \left (\frac {a^2 d^2 \sin (c+d x)}{x}-\frac {2 a^2 \sin (c+d x)}{x^3}-\frac {a^2 d \cos (c+d x)}{x^2}-a \text {Ci}(d x) \left (a d^3 \cos (c)-12 b \sin (c)\right )+a \text {Si}(d x) \left (a d^3 \sin (c)+12 b \cos (c)\right )+\frac {12 b^2 \cos (c+d x)}{d^3}+\frac {12 b^2 x \sin (c+d x)}{d^2}-\frac {6 b^2 x^2 \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*Sin[c + d*x])/x^4,x]

[Out]

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

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fricas [A]  time = 0.84, size = 176, normalized size = 1.17 \[ -\frac {2 \, {\left (6 \, b^{2} d^{2} x^{5} + a^{2} d^{4} x - 12 \, b^{2} x^{3}\right )} \cos \left (d x + c\right ) + {\left (a^{2} d^{6} x^{3} \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} x^{3} \operatorname {Ci}\left (-d x\right ) - 24 \, a b d^{3} x^{3} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) - 2 \, {\left (a^{2} d^{5} x^{2} + 12 \, b^{2} d x^{4} - 2 \, a^{2} d^{3}\right )} \sin \left (d x + c\right ) - 2 \, {\left (a^{2} d^{6} x^{3} \operatorname {Si}\left (d x\right ) + 6 \, a b d^{3} x^{3} \operatorname {Ci}\left (d x\right ) + 6 \, a b d^{3} x^{3} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{12 \, d^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^4,x, algorithm="fricas")

[Out]

-1/12*(2*(6*b^2*d^2*x^5 + a^2*d^4*x - 12*b^2*x^3)*cos(d*x + c) + (a^2*d^6*x^3*cos_integral(d*x) + a^2*d^6*x^3*
cos_integral(-d*x) - 24*a*b*d^3*x^3*sin_integral(d*x))*cos(c) - 2*(a^2*d^5*x^2 + 12*b^2*d*x^4 - 2*a^2*d^3)*sin
(d*x + c) - 2*(a^2*d^6*x^3*sin_integral(d*x) + 6*a*b*d^3*x^3*cos_integral(d*x) + 6*a*b*d^3*x^3*cos_integral(-d
*x))*sin(c))/(d^3*x^3)

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giac [C]  time = 0.95, size = 1181, normalized size = 7.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^4,x, algorithm="giac")

[Out]

1/12*(a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^6*x^3*real_part(cos_integra
l(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^6*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) -
2*a^2*d^6*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^6*x^3*sin_integral(d*x)*tan(1/
2*d*x)^2*tan(1/2*c) - a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a^2*d^6*x^3*real_part(cos_inte
gral(-d*x))*tan(1/2*d*x)^2 + a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a^2*d^6*x^3*real_part(cos
_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^6*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^6*x^3*imag_par
t(cos_integral(-d*x))*tan(1/2*c) + 4*a^2*d^6*x^3*sin_integral(d*x)*tan(1/2*c) - 12*b^2*d^2*x^5*tan(1/2*d*x)^2*
tan(1/2*c)^2 - 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*a*b*d^3*x^3*imag_p
art(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*
c)^2 - a^2*d^6*x^3*real_part(cos_integral(d*x)) - a^2*d^6*x^3*real_part(cos_integral(-d*x)) - 4*a^2*d^5*x^2*ta
n(1/2*d*x)^2*tan(1/2*c) + 24*a*b*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 24*a*b*d^3*x
^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d^5*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 12*b^2*
d^2*x^5*tan(1/2*d*x)^2 + 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 12*a*b*d^3*x^3*imag_part
(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2 + 48*b^2*d^2*x^5*tan(1/2
*d*x)*tan(1/2*c) + 12*b^2*d^2*x^5*tan(1/2*c)^2 - 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 12
*a*b*d^3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*c)^2 - 2*a^
2*d^4*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a^2*d^5*x^2*tan(1/2*d*x) + 4*a^2*d^5*x^2*tan(1/2*c) + 24*a*b*d^3*x^3*r
eal_part(cos_integral(d*x))*tan(1/2*c) + 24*a*b*d^3*x^3*real_part(cos_integral(-d*x))*tan(1/2*c) - 48*b^2*d*x^
4*tan(1/2*d*x)^2*tan(1/2*c) - 48*b^2*d*x^4*tan(1/2*d*x)*tan(1/2*c)^2 - 12*b^2*d^2*x^5 + 12*a*b*d^3*x^3*imag_pa
rt(cos_integral(d*x)) - 12*a*b*d^3*x^3*imag_part(cos_integral(-d*x)) + 24*a*b*d^3*x^3*sin_integral(d*x) + 2*a^
2*d^4*x*tan(1/2*d*x)^2 + 8*a^2*d^4*x*tan(1/2*d*x)*tan(1/2*c) + 2*a^2*d^4*x*tan(1/2*c)^2 + 24*b^2*x^3*tan(1/2*d
*x)^2*tan(1/2*c)^2 + 48*b^2*d*x^4*tan(1/2*d*x) + 48*b^2*d*x^4*tan(1/2*c) + 8*a^2*d^3*tan(1/2*d*x)^2*tan(1/2*c)
+ 8*a^2*d^3*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d^4*x - 24*b^2*x^3*tan(1/2*d*x)^2 - 96*b^2*x^3*tan(1/2*d*x)*tan
(1/2*c) - 24*b^2*x^3*tan(1/2*c)^2 - 8*a^2*d^3*tan(1/2*d*x) - 8*a^2*d^3*tan(1/2*c) + 24*b^2*x^3)/(d^3*x^3*tan(1
/2*d*x)^2*tan(1/2*c)^2 + d^3*x^3*tan(1/2*d*x)^2 + d^3*x^3*tan(1/2*c)^2 + d^3*x^3)

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maple [A]  time = 0.06, size = 196, normalized size = 1.30 \[ d^{3} \left (\frac {2 a b \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right )}{d^{3}}-\frac {15 c^{2} b^{2} \cos \left (d x +c \right )}{d^{6}}+\frac {\left (10 c^{2}+4 c +1\right ) b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )-\frac {6 b^{2} c \left (1+4 c \right ) \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*sin(d*x+c)/x^4,x)

[Out]

d^3*(2/d^3*a*b*(Si(d*x)*cos(c)+Ci(d*x)*sin(c))-15*c^2/d^6*b^2*cos(d*x+c)+(10*c^2+4*c+1)/d^6*b^2*(-(d*x+c)^2*co
s(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+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))-6*b^2*c*(1+4*c)/d^6*(sin(d*x+c)-(d*x+c)*cos(d*x+c)))

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maxima [C]  time = 10.00, size = 173, normalized size = 1.15 \[ -\frac {{\left ({\left (a^{2} {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) - a^{2} {\left (i \, \Gamma \left (-3, i \, d x\right ) - i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} - {\left (a b {\left (12 i \, \Gamma \left (-3, i \, d x\right ) - 12 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + 12 \, a b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 2 \, {\left (b^{2} d^{2} x^{5} + 2 \, a b d^{2} x^{2} - 2 \, b^{2} x^{3} - 4 \, a b\right )} \cos \left (d x + c\right ) - 4 \, {\left (b^{2} d x^{4} - a b d x\right )} \sin \left (d x + c\right )}{2 \, d^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+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))*sin
(c))*d^6 - (a*b*(12*I*gamma(-3, I*d*x) - 12*I*gamma(-3, -I*d*x))*cos(c) + 12*a*b*(gamma(-3, I*d*x) + gamma(-3,
-I*d*x))*sin(c))*d^3)*x^3 + 2*(b^2*d^2*x^5 + 2*a*b*d^2*x^2 - 2*b^2*x^3 - 4*a*b)*cos(d*x + c) - 4*(b^2*d*x^4 -
a*b*d*x)*sin(d*x + c))/(d^3*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3)^2)/x^4,x)

[Out]

int((sin(c + d*x)*(a + b*x^3)^2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*sin(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**3)**2*sin(c + d*x)/x**4, x)

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