3.93 \(\int \frac {(a+b x^3)^2 \sin (c+d x)}{x^5} \, dx\)

Optimal. Leaf size=167 \[ \frac {1}{24} a^2 d^4 \sin (c) \text {Ci}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^3 \cos (c+d x)}{24 x}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}+2 a b d \cos (c) \text {Ci}(d x)-2 a b d \sin (c) \text {Si}(d x)-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {b^2 x \cos (c+d x)}{d} \]

[Out]

2*a*b*d*Ci(d*x)*cos(c)-1/12*a^2*d*cos(d*x+c)/x^3+1/24*a^2*d^3*cos(d*x+c)/x-b^2*x*cos(d*x+c)/d+1/24*a^2*d^4*cos
(c)*Si(d*x)+1/24*a^2*d^4*Ci(d*x)*sin(c)-2*a*b*d*Si(d*x)*sin(c)+b^2*sin(d*x+c)/d^2-1/4*a^2*sin(d*x+c)/x^4+1/24*
a^2*d^2*sin(d*x+c)/x^2-2*a*b*sin(d*x+c)/x

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Rubi [A]  time = 0.28, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ \frac {1}{24} a^2 d^4 \sin (c) \text {CosIntegral}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}+2 a b d \cos (c) \text {CosIntegral}(d x)-2 a b d \sin (c) \text {Si}(d x)-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {b^2 x \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*Sin[c + d*x])/x^5,x]

[Out]

-(a^2*d*Cos[c + d*x])/(12*x^3) + (a^2*d^3*Cos[c + d*x])/(24*x) - (b^2*x*Cos[c + d*x])/d + 2*a*b*d*Cos[c]*CosIn
tegral[d*x] + (a^2*d^4*CosIntegral[d*x]*Sin[c])/24 + (b^2*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/(4*x^4) + (a^
2*d^2*Sin[c + d*x])/(24*x^2) - (2*a*b*Sin[c + d*x])/x + (a^2*d^4*Cos[c]*SinIntegral[d*x])/24 - 2*a*b*d*Sin[c]*
SinIntegral[d*x]

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^3\right )^2 \sin (c+d x)}{x^5} \, dx &=\int \left (\frac {a^2 \sin (c+d x)}{x^5}+\frac {2 a b \sin (c+d x)}{x^2}+b^2 x \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x^5} \, dx+(2 a b) \int \frac {\sin (c+d x)}{x^2} \, 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)}{4 x^4}-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \int \cos (c+d x) \, dx}{d}+\frac {1}{4} \left (a^2 d\right ) \int \frac {\cos (c+d x)}{x^4} \, dx+(2 a b d) \int \frac {\cos (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {b^2 x \cos (c+d x)}{d}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{x}-\frac {1}{12} \left (a^2 d^2\right ) \int \frac {\sin (c+d x)}{x^3} \, 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 {a^2 d \cos (c+d x)}{12 x^3}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \sin (c) \text {Si}(d x)-\frac {1}{24} \left (a^2 d^3\right ) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \sin (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \sin (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx+\frac {1}{24} \left (a^2 d^4 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {1}{24} a^2 d^4 \text {Ci}(d x) \sin (c)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)-2 a b d \sin (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 148, normalized size = 0.89 \[ \frac {1}{24} \left (\frac {a^2 d^3 \cos (c+d x)}{x}+\frac {a^2 d^2 \sin (c+d x)}{x^2}-\frac {6 a^2 \sin (c+d x)}{x^4}-\frac {2 a^2 d \cos (c+d x)}{x^3}+a d \text {Ci}(d x) \left (a d^3 \sin (c)+48 b \cos (c)\right )+a d \text {Si}(d x) \left (a d^3 \cos (c)-48 b \sin (c)\right )-\frac {48 a b \sin (c+d x)}{x}+\frac {24 b^2 \sin (c+d x)}{d^2}-\frac {24 b^2 x \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*Sin[c + d*x])/x^5,x]

[Out]

((-2*a^2*d*Cos[c + d*x])/x^3 + (a^2*d^3*Cos[c + d*x])/x - (24*b^2*x*Cos[c + d*x])/d + a*d*CosIntegral[d*x]*(48
*b*Cos[c] + a*d^3*Sin[c]) + (24*b^2*Sin[c + d*x])/d^2 - (6*a^2*Sin[c + d*x])/x^4 + (a^2*d^2*Sin[c + d*x])/x^2
- (48*a*b*Sin[c + d*x])/x + a*d*(a*d^3*Cos[c] - 48*b*Sin[c])*SinIntegral[d*x])/24

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fricas [A]  time = 0.75, size = 186, normalized size = 1.11 \[ \frac {2 \, {\left (a^{2} d^{5} x^{3} - 24 \, b^{2} d x^{5} - 2 \, a^{2} d^{3} x\right )} \cos \left (d x + c\right ) + 2 \, {\left (a^{2} d^{6} x^{4} \operatorname {Si}\left (d x\right ) + 24 \, a b d^{3} x^{4} \operatorname {Ci}\left (d x\right ) + 24 \, a b d^{3} x^{4} \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) + 2 \, {\left (a^{2} d^{4} x^{2} - 48 \, a b d^{2} x^{3} + 24 \, b^{2} x^{4} - 6 \, a^{2} d^{2}\right )} \sin \left (d x + c\right ) + {\left (a^{2} d^{6} x^{4} \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} x^{4} \operatorname {Ci}\left (-d x\right ) - 96 \, a b d^{3} x^{4} \operatorname {Si}\left (d x\right )\right )} \sin \relax (c)}{48 \, d^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^5,x, algorithm="fricas")

[Out]

1/48*(2*(a^2*d^5*x^3 - 24*b^2*d*x^5 - 2*a^2*d^3*x)*cos(d*x + c) + 2*(a^2*d^6*x^4*sin_integral(d*x) + 24*a*b*d^
3*x^4*cos_integral(d*x) + 24*a*b*d^3*x^4*cos_integral(-d*x))*cos(c) + 2*(a^2*d^4*x^2 - 48*a*b*d^2*x^3 + 24*b^2
*x^4 - 6*a^2*d^2)*sin(d*x + c) + (a^2*d^6*x^4*cos_integral(d*x) + a^2*d^6*x^4*cos_integral(-d*x) - 96*a*b*d^3*
x^4*sin_integral(d*x))*sin(c))/(d^2*x^4)

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giac [C]  time = 1.33, size = 1255, normalized size = 7.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^5,x, algorithm="giac")

[Out]

-1/48*(a^2*d^6*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^6*x^4*imag_part(cos_integr
al(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^6*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^
6*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^6*x^4*real_part(cos_integral(-d*x))*tan
(1/2*d*x)^2*tan(1/2*c) - a^2*d^6*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a^2*d^6*x^4*imag_part(cos_i
ntegral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^6*x^4*sin_integral(d*x)*tan(1/2*d*x)^2 + a^2*d^6*x^4*imag_part(cos_int
egral(d*x))*tan(1/2*c)^2 - a^2*d^6*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^6*x^4*sin_integral
(d*x)*tan(1/2*c)^2 - 2*a^2*d^6*x^4*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^6*x^4*real_part(cos_integ
ral(-d*x))*tan(1/2*c) - 2*a^2*d^5*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 48*a*b*d^3*x^4*real_part(cos_integral(d*x)
)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 48*a*b*d^3*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2
*d^6*x^4*imag_part(cos_integral(d*x)) + a^2*d^6*x^4*imag_part(cos_integral(-d*x)) - 2*a^2*d^6*x^4*sin_integral
(d*x) + 96*a*b*d^3*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 96*a*b*d^3*x^4*imag_part(cos_i
ntegral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 192*a*b*d^3*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2
*d^5*x^3*tan(1/2*d*x)^2 - 48*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 48*a*b*d^3*x^4*real_par
t(cos_integral(-d*x))*tan(1/2*d*x)^2 + 8*a^2*d^5*x^3*tan(1/2*d*x)*tan(1/2*c) + 2*a^2*d^5*x^3*tan(1/2*c)^2 + 48
*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*c)^2 + 48*a*b*d^3*x^4*real_part(cos_integral(-d*x))*tan(1/2*
c)^2 + 48*b^2*d*x^5*tan(1/2*d*x)^2*tan(1/2*c)^2 + 96*a*b*d^3*x^4*imag_part(cos_integral(d*x))*tan(1/2*c) - 96*
a*b*d^3*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c) + 192*a*b*d^3*x^4*sin_integral(d*x)*tan(1/2*c) + 4*a^2*d^
4*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^4*x^2*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d^5*x^3 - 48*a*b*d^3*x^4*rea
l_part(cos_integral(d*x)) - 48*a*b*d^3*x^4*real_part(cos_integral(-d*x)) - 48*b^2*d*x^5*tan(1/2*d*x)^2 - 192*b
^2*d*x^5*tan(1/2*d*x)*tan(1/2*c) - 192*a*b*d^2*x^3*tan(1/2*d*x)^2*tan(1/2*c) - 48*b^2*d*x^5*tan(1/2*c)^2 - 192
*a*b*d^2*x^3*tan(1/2*d*x)*tan(1/2*c)^2 + 4*a^2*d^3*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*a^2*d^4*x^2*tan(1/2*d*x)
- 4*a^2*d^4*x^2*tan(1/2*c) + 96*b^2*x^4*tan(1/2*d*x)^2*tan(1/2*c) + 96*b^2*x^4*tan(1/2*d*x)*tan(1/2*c)^2 + 48*
b^2*d*x^5 + 192*a*b*d^2*x^3*tan(1/2*d*x) - 4*a^2*d^3*x*tan(1/2*d*x)^2 + 192*a*b*d^2*x^3*tan(1/2*c) - 16*a^2*d^
3*x*tan(1/2*d*x)*tan(1/2*c) - 4*a^2*d^3*x*tan(1/2*c)^2 - 96*b^2*x^4*tan(1/2*d*x) - 96*b^2*x^4*tan(1/2*c) - 24*
a^2*d^2*tan(1/2*d*x)^2*tan(1/2*c) - 24*a^2*d^2*tan(1/2*d*x)*tan(1/2*c)^2 + 4*a^2*d^3*x + 24*a^2*d^2*tan(1/2*d*
x) + 24*a^2*d^2*tan(1/2*c))/(d^2*x^4*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^2*x^4*tan(1/2*d*x)^2 + d^2*x^4*tan(1/2*c)
^2 + d^2*x^4)

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maple [A]  time = 0.05, size = 167, normalized size = 1.00 \[ d^{4} \left (\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d^{3}}+\frac {6 c \,b^{2} \cos \left (d x +c \right )}{d^{6}}+\frac {\left (1+5 c \right ) b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{4 x^{4} d^{4}}-\frac {\cos \left (d x +c \right )}{12 x^{3} d^{3}}+\frac {\sin \left (d x +c \right )}{24 x^{2} d^{2}}+\frac {\cos \left (d x +c \right )}{24 x d}+\frac {\Si \left (d x \right ) \cos \relax (c )}{24}+\frac {\Ci \left (d x \right ) \sin \relax (c )}{24}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*sin(d*x+c)/x^5,x)

[Out]

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

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maxima [C]  time = 16.43, size = 166, normalized size = 0.99 \[ \frac {{\left ({\left (a^{2} {\left (-i \, \Gamma \left (-4, i \, d x\right ) + i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - a^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{7} - {\left (48 \, a b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - a b {\left (48 i \, \Gamma \left (-4, i \, d x\right ) - 48 i \, \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4}\right )} x^{4} - 2 \, {\left (b^{2} d^{2} x^{5} + 2 \, a b d^{2} x^{2} - 12 \, a b\right )} \cos \left (d x + c\right ) + 2 \, {\left (b^{2} d x^{4} - 4 \, a b d x\right )} \sin \left (d x + c\right )}{2 \, d^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^5,x, algorithm="maxima")

[Out]

1/2*(((a^2*(-I*gamma(-4, I*d*x) + I*gamma(-4, -I*d*x))*cos(c) - a^2*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*sin
(c))*d^7 - (48*a*b*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*cos(c) - a*b*(48*I*gamma(-4, I*d*x) - 48*I*gamma(-4,
-I*d*x))*sin(c))*d^4)*x^4 - 2*(b^2*d^2*x^5 + 2*a*b*d^2*x^2 - 12*a*b)*cos(d*x + c) + 2*(b^2*d*x^4 - 4*a*b*d*x)
*sin(d*x + c))/(d^3*x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3)^2)/x^5,x)

[Out]

int((sin(c + d*x)*(a + b*x^3)^2)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*sin(d*x+c)/x**5,x)

[Out]

Integral((a + b*x**3)**2*sin(c + d*x)/x**5, x)

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