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

Optimal. Leaf size=161 \[ a^2 \sin (c) \text {Ci}(d x)+a^2 \cos (c) \text {Si}(d x)+\frac {4 a b \cos (c+d x)}{d^3}+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {120 b^2 \sin (c+d x)}{d^6}-\frac {120 b^2 x \cos (c+d x)}{d^5}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}-\frac {b^2 x^5 \cos (c+d x)}{d} \]

[Out]

4*a*b*cos(d*x+c)/d^3-120*b^2*x*cos(d*x+c)/d^5-2*a*b*x^2*cos(d*x+c)/d+20*b^2*x^3*cos(d*x+c)/d^3-b^2*x^5*cos(d*x
+c)/d+a^2*cos(c)*Si(d*x)+a^2*Ci(d*x)*sin(c)+120*b^2*sin(d*x+c)/d^6+4*a*b*x*sin(d*x+c)/d^2-60*b^2*x^2*sin(d*x+c
)/d^4+5*b^2*x^4*sin(d*x+c)/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 161, 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, 3303, 3299, 3302, 3296, 2638, 2637} \[ a^2 \sin (c) \text {CosIntegral}(d x)+a^2 \cos (c) \text {Si}(d x)+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 a b \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}+\frac {120 b^2 \sin (c+d x)}{d^6}-\frac {120 b^2 x \cos (c+d x)}{d^5}-\frac {b^2 x^5 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*Sin[c + d*x])/x,x]

[Out]

(4*a*b*Cos[c + d*x])/d^3 - (120*b^2*x*Cos[c + d*x])/d^5 - (2*a*b*x^2*Cos[c + d*x])/d + (20*b^2*x^3*Cos[c + d*x
])/d^3 - (b^2*x^5*Cos[c + d*x])/d + a^2*CosIntegral[d*x]*Sin[c] + (120*b^2*Sin[c + d*x])/d^6 + (4*a*b*x*Sin[c
+ d*x])/d^2 - (60*b^2*x^2*Sin[c + d*x])/d^4 + (5*b^2*x^4*Sin[c + d*x])/d^2 + a^2*Cos[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 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 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} \, dx &=\int \left (\frac {a^2 \sin (c+d x)}{x}+2 a b x^2 \sin (c+d x)+b^2 x^5 \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x} \, dx+(2 a b) \int x^2 \sin (c+d x) \, dx+b^2 \int x^5 \sin (c+d x) \, dx\\ &=-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {(4 a b) \int x \cos (c+d x) \, dx}{d}+\frac {\left (5 b^2\right ) \int x^4 \cos (c+d x) \, dx}{d}+\left (a^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx+\left (a^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)-\frac {(4 a b) \int \sin (c+d x) \, dx}{d^2}-\frac {\left (20 b^2\right ) \int x^3 \sin (c+d x) \, dx}{d^2}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {b^2 x^5 \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)-\frac {\left (60 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d^3}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {b^2 x^5 \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)+\frac {\left (120 b^2\right ) \int x \sin (c+d x) \, dx}{d^4}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {120 b^2 x \cos (c+d x)}{d^5}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {b^2 x^5 \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)+\frac {\left (120 b^2\right ) \int \cos (c+d x) \, dx}{d^5}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {120 b^2 x \cos (c+d x)}{d^5}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {b^2 x^5 \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {120 b^2 \sin (c+d x)}{d^6}+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.53, size = 108, normalized size = 0.67 \[ a^2 \sin (c) \text {Ci}(d x)+a^2 \cos (c) \text {Si}(d x)+\frac {b \left (4 a d^4 x+5 b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \sin (c+d x)}{d^6}-\frac {b \left (2 a d^2 \left (d^2 x^2-2\right )+b x \left (d^4 x^4-20 d^2 x^2+120\right )\right ) \cos (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(2*a*d^2*(-2 + d^2*x^2) + b*x*(120 - 20*d^2*x^2 + d^4*x^4))*Cos[c + d*x])/d^5) + a^2*CosIntegral[d*x]*Sin
[c] + (b*(4*a*d^4*x + 5*b*(24 - 12*d^2*x^2 + d^4*x^4))*Sin[c + d*x])/d^6 + a^2*Cos[c]*SinIntegral[d*x]

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 145, normalized size = 0.90 \[ \frac {2 \, a^{2} d^{6} \cos \relax (c) \operatorname {Si}\left (d x\right ) - 2 \, {\left (b^{2} d^{5} x^{5} + 2 \, a b d^{5} x^{2} - 20 \, b^{2} d^{3} x^{3} - 4 \, a b d^{3} + 120 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 2 \, {\left (5 \, b^{2} d^{4} x^{4} + 4 \, a b d^{4} x - 60 \, b^{2} d^{2} x^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right ) + {\left (a^{2} d^{6} \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{2 \, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*d^6*cos(c)*sin_integral(d*x) - 2*(b^2*d^5*x^5 + 2*a*b*d^5*x^2 - 20*b^2*d^3*x^3 - 4*a*b*d^3 + 120*b^
2*d*x)*cos(d*x + c) + 2*(5*b^2*d^4*x^4 + 4*a*b*d^4*x - 60*b^2*d^2*x^2 + 120*b^2)*sin(d*x + c) + (a^2*d^6*cos_i
ntegral(d*x) + a^2*d^6*cos_integral(-d*x))*sin(c))/d^6

________________________________________________________________________________________

giac [C]  time = 0.36, size = 921, normalized size = 5.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*d^5*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 2*b^2*d^5*x^5*tan(1/2*d*x + 1/2*c)^2 - 2*b^2*d^5*x^5*
tan(1/2*c)^2 + 20*b^2*d^4*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 4*a*b*d^5*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2
*c)^2 - a^2*d^6*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + a^2*d^6*imag_part(cos_integ
ral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2*d^6*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)
^2 - 2*b^2*d^5*x^5 + 2*a^2*d^6*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*a^2*d^6*real
_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) - 40*b^2*d^3*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)
^2 + 20*b^2*d^4*x^4*tan(1/2*d*x + 1/2*c) + 4*a*b*d^5*x^2*tan(1/2*d*x + 1/2*c)^2 + a^2*d^6*imag_part(cos_integr
al(d*x))*tan(1/2*d*x + 1/2*c)^2 - a^2*d^6*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^6*sin
_integral(d*x)*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*d^5*x^2*tan(1/2*c)^2 - a^2*d^6*imag_part(cos_integral(d*x))*tan(
1/2*c)^2 + a^2*d^6*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a^2*d^6*sin_integral(d*x)*tan(1/2*c)^2 - 40*
b^2*d^3*x^3*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^6*real_part(cos_integral(d*x))*tan(1/2*c) + 2*a^2*d^6*real_part(c
os_integral(-d*x))*tan(1/2*c) + 40*b^2*d^3*x^3*tan(1/2*c)^2 + 16*a*b*d^4*x*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 -
4*a*b*d^5*x^2 + a^2*d^6*imag_part(cos_integral(d*x)) - a^2*d^6*imag_part(cos_integral(-d*x)) + 2*a^2*d^6*sin_
integral(d*x) - 240*b^2*d^2*x^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 - 8*a*b*d^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c
)^2 + 40*b^2*d^3*x^3 + 16*a*b*d^4*x*tan(1/2*d*x + 1/2*c) + 240*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2
40*b^2*d^2*x^2*tan(1/2*d*x + 1/2*c) - 8*a*b*d^3*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*d^3*tan(1/2*c)^2 + 240*b^2*d*x*
tan(1/2*d*x + 1/2*c)^2 - 240*b^2*d*x*tan(1/2*c)^2 + 8*a*b*d^3 + 480*b^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 - 24
0*b^2*d*x + 480*b^2*tan(1/2*d*x + 1/2*c))/(d^6*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + d^6*tan(1/2*d*x + 1/2*c)^
2 + d^6*tan(1/2*c)^2 + d^6)

________________________________________________________________________________________

maple [B]  time = 0.04, size = 487, normalized size = 3.02 \[ \frac {\left (c^{5}+c^{4}+c^{3}+c^{2}+c +1\right ) b^{2} \left (-\left (d x +c \right )^{5} \cos \left (d x +c \right )+5 \left (d x +c \right )^{4} \sin \left (d x +c \right )+20 \left (d x +c \right )^{3} \cos \left (d x +c \right )-60 \left (d x +c \right )^{2} \sin \left (d x +c \right )+120 \sin \left (d x +c \right )-120 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}-\frac {6 b^{2} c \left (c^{4}+c^{3}+c^{2}+c +1\right ) \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}+\frac {15 \left (c^{3}+c^{2}+c +1\right ) c^{2} b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+\frac {2 \left (c^{2}+c +1\right ) a b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{3}}-\frac {20 b^{2} c^{3} \left (c^{2}+c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}-\frac {6 c a b \left (1+c \right ) \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}+\frac {15 \left (1+c \right ) b^{2} c^{4} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}-\frac {6 c^{2} a b \cos \left (d x +c \right )}{d^{3}}+\frac {6 c^{5} b^{2} \cos \left (d x +c \right )}{d^{6}}+a^{2} \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^5+c^4+c^3+c^2+c+1)/d^6*b^2*(-(d*x+c)^5*cos(d*x+c)+5*(d*x+c)^4*sin(d*x+c)+20*(d*x+c)^3*cos(d*x+c)-60*(d*x+c)
^2*sin(d*x+c)+120*sin(d*x+c)-120*(d*x+c)*cos(d*x+c))-6*b^2*c*(c^4+c^3+c^2+c+1)/d^6*(-(d*x+c)^4*cos(d*x+c)+4*(d
*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos(d*x+c)-24*cos(d*x+c)-24*(d*x+c)*sin(d*x+c))+15*(c^3+c^2+c+1)/d^6*c^2*b^2*(
-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))+2*(c^2+c+1)/d^3*a*b*(-(d*x+c)^
2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))-20*b^2*c^3*(c^2+c+1)/d^6*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2
*(d*x+c)*sin(d*x+c))-6*c*a*b*(1+c)/d^3*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+15*(1+c)/d^6*b^2*c^4*(sin(d*x+c)-(d*x+c
)*cos(d*x+c))-6*c^2/d^3*a*b*cos(d*x+c)+6*c^5/d^6*b^2*cos(d*x+c)+a^2*(Si(d*x)*cos(c)+Ci(d*x)*sin(c))

________________________________________________________________________________________

maxima [C]  time = 12.02, size = 147, normalized size = 0.91 \[ \frac {{\left (a^{2} {\left (-i \, {\rm Ei}\left (i \, d x\right ) + i \, {\rm Ei}\left (-i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left ({\rm Ei}\left (i \, d x\right ) + {\rm Ei}\left (-i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} - 2 \, {\left (b^{2} d^{5} x^{5} + 2 \, a b d^{5} x^{2} - 20 \, b^{2} d^{3} x^{3} - 4 \, a b d^{3} + 120 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 2 \, {\left (5 \, b^{2} d^{4} x^{4} + 4 \, a b d^{4} x - 60 \, b^{2} d^{2} x^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*(-I*Ei(I*d*x) + I*Ei(-I*d*x))*cos(c) + a^2*(Ei(I*d*x) + Ei(-I*d*x))*sin(c))*d^6 - 2*(b^2*d^5*x^5 + 2
*a*b*d^5*x^2 - 20*b^2*d^3*x^3 - 4*a*b*d^3 + 120*b^2*d*x)*cos(d*x + c) + 2*(5*b^2*d^4*x^4 + 4*a*b*d^4*x - 60*b^
2*d^2*x^2 + 120*b^2)*sin(d*x + c))/d^6

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3)^2)/x,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 10.64, size = 211, normalized size = 1.31 \[ a^{2} \sin {\relax (c )} \operatorname {Ci}{\left (d x \right )} + a^{2} \cos {\relax (c )} \operatorname {Si}{\left (d x \right )} + 2 a b x^{2} \left (\begin {cases} - \cos {\relax (c )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - 4 a b \left (\begin {cases} - \frac {x^{2} \cos {\relax (c )}}{2} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {x \sin {\left (c + d x \right )}}{d} + \frac {\cos {\left (c + d x \right )}}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{2} \cos {\relax (c )}}{2} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) + b^{2} x^{5} \left (\begin {cases} - \cos {\relax (c )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - 5 b^{2} \left (\begin {cases} - \frac {x^{5} \cos {\relax (c )}}{5} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {x^{4} \sin {\left (c + d x \right )}}{d} + \frac {4 x^{3} \cos {\left (c + d x \right )}}{d^{2}} - \frac {12 x^{2} \sin {\left (c + d x \right )}}{d^{3}} - \frac {24 x \cos {\left (c + d x \right )}}{d^{4}} + \frac {24 \sin {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\frac {x^{5} \cos {\relax (c )}}{5} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*sin(c)*Ci(d*x) + a**2*cos(c)*Si(d*x) + 2*a*b*x**2*Piecewise((-cos(c), Eq(d, 0)), (-cos(c + d*x)/d, True))
- 4*a*b*Piecewise((-x**2*cos(c)/2, Eq(d, 0)), (-Piecewise((x*sin(c + d*x)/d + cos(c + d*x)/d**2, Ne(d, 0)), (
x**2*cos(c)/2, True))/d, True)) + b**2*x**5*Piecewise((-cos(c), Eq(d, 0)), (-cos(c + d*x)/d, True)) - 5*b**2*P
iecewise((-x**5*cos(c)/5, Eq(d, 0)), (-Piecewise((x**4*sin(c + d*x)/d + 4*x**3*cos(c + d*x)/d**2 - 12*x**2*sin
(c + d*x)/d**3 - 24*x*cos(c + d*x)/d**4 + 24*sin(c + d*x)/d**5, Ne(d, 0)), (x**5*cos(c)/5, True))/d, True))

________________________________________________________________________________________