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

Optimal. Leaf size=145 \[ a^2 d \cos (c) \text {Ci}(d x)-a^2 d \sin (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x}+\frac {2 a b \sin (c+d x)}{d^2}-\frac {2 a b x \cos (c+d x)}{d}-\frac {24 b^2 \cos (c+d x)}{d^5}-\frac {24 b^2 x \sin (c+d x)}{d^4}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}-\frac {b^2 x^4 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3339, 3297, 3303, 3299, 3302, 3296, 2637, 2638} \[ a^2 d \cos (c) \text {CosIntegral}(d x)-a^2 d \sin (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x}+\frac {2 a b \sin (c+d x)}{d^2}-\frac {2 a b x \cos (c+d x)}{d}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}-\frac {24 b^2 x \sin (c+d x)}{d^4}-\frac {24 b^2 \cos (c+d x)}{d^5}-\frac {b^2 x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.39, size = 145, normalized size = 1.00 \[ a^2 d \cos (c) \text {Ci}(d x)-a^2 d \sin (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x}+\frac {2 a b \sin (c+d x)}{d^2}-\frac {2 a b x \cos (c+d x)}{d}-\frac {24 b^2 \cos (c+d x)}{d^5}-\frac {24 b^2 x \sin (c+d x)}{d^4}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}-\frac {b^2 x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*b^2*Cos[c + d*x])/d^5 - (2*a*b*x*Cos[c + d*x])/d + (12*b^2*x^2*Cos[c + d*x])/d^3 - (b^2*x^4*Cos[c + d*x])
/d + a^2*d*Cos[c]*CosIntegral[d*x] + (2*a*b*Sin[c + d*x])/d^2 - (a^2*Sin[c + d*x])/x - (24*b^2*x*Sin[c + d*x])
/d^4 + (4*b^2*x^3*Sin[c + d*x])/d^2 - a^2*d*Sin[c]*SinIntegral[d*x]

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fricas [A]  time = 0.77, size = 145, normalized size = 1.00 \[ -\frac {2 \, a^{2} d^{6} x \sin \relax (c) \operatorname {Si}\left (d x\right ) + 2 \, {\left (b^{2} d^{4} x^{5} + 2 \, a b d^{4} x^{2} - 12 \, b^{2} d^{2} x^{3} + 24 \, b^{2} x\right )} \cos \left (d x + c\right ) - {\left (a^{2} d^{6} x \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} x \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) - 2 \, {\left (4 \, b^{2} d^{3} x^{4} - a^{2} d^{5} + 2 \, a b d^{3} x - 24 \, b^{2} d x^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 1.34, size = 2038, normalized size = 14.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*d^4*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^6*x*real_part(cos_integral(d*x))
*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^6*x*real_part(cos_integral(-d*x))*tan(1/2*d*x + 1/
2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*b^2*d^4*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 - 2*a^2*d^6*x*imag_pa
rt(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d^6*x*imag_part(cos_integral(-d
*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d^6*x*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*t
an(1/2*d*x)^2*tan(1/2*c) + 2*b^2*d^4*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*b^2*d^4*x^5*tan(1/2*d*x)^2*ta
n(1/2*c)^2 + a^2*d^6*x*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + a^2*d^6*x*real_par
t(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 - a^2*d^6*x*real_part(cos_integral(d*x))*tan(1/2*d
*x + 1/2*c)^2*tan(1/2*c)^2 - a^2*d^6*x*real_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - a^2
*d^6*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^6*x*real_part(cos_integral(-d*x))*tan(
1/2*d*x)^2*tan(1/2*c)^2 + 16*b^2*d^3*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a*b*d^4*x^2*tan(
1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*b^2*d^4*x^5*tan(1/2*d*x + 1/2*c)^2 - 2*b^2*d^4*x^5*tan(1/2*
d*x)^2 - 2*a^2*d^6*x*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*a^2*d^6*x*imag_part(co
s_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) - 4*a^2*d^6*x*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan
(1/2*c) - 2*a^2*d^6*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d^6*x*imag_part(cos_integ
ral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d^6*x*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - 2*b^2*d^4*x^5
*tan(1/2*c)^2 - 24*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^6*x*real_part(cos_in
tegral(d*x))*tan(1/2*d*x + 1/2*c)^2 + a^2*d^6*x*real_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2 + a^2*d^6
*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a^2*d^6*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 16*b
^2*d^3*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2 + 4*a*b*d^4*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + 4*a^2*d
^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c) - a^2*d^6*x*real_part(cos_integral(d*x))*tan(1/2*c)^2 - a^
2*d^6*x*real_part(cos_integral(-d*x))*tan(1/2*c)^2 + 16*b^2*d^3*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 4*a*b*
d^4*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 4*a^2*d^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)*tan(1/2*c)^2 - 4*a
*b*d^4*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*b^2*d^4*x^5 - 24*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2
- 2*a^2*d^6*x*imag_part(cos_integral(d*x))*tan(1/2*c) + 2*a^2*d^6*x*imag_part(cos_integral(-d*x))*tan(1/2*c) -
4*a^2*d^6*x*sin_integral(d*x)*tan(1/2*c) - 24*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 24*b^2*d^2*x^
3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*b*d^3*x*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^6*x*real_
part(cos_integral(d*x)) + a^2*d^6*x*real_part(cos_integral(-d*x)) + 16*b^2*d^3*x^4*tan(1/2*d*x + 1/2*c) + 4*a*
b*d^4*x^2*tan(1/2*d*x + 1/2*c)^2 - 4*a^2*d^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x) - 4*a*b*d^4*x^2*tan(1/2*d*x)^
2 - 4*a^2*d^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 4*a^2*d^5*tan(1/2*d*x)^2*tan(1/2*c) - 4*a*b*d^4*x^2*tan(1/2*
c)^2 + 4*a^2*d^5*tan(1/2*d*x)*tan(1/2*c)^2 - 96*b^2*d*x^2*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2
4*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*d^2*x^3*tan(1/2*d*x)^2 + 8*a*b*d^3*x*tan(1/2*d*x + 1/2*c)*tan(1/
2*d*x)^2 + 24*b^2*d^2*x^3*tan(1/2*c)^2 + 8*a*b*d^3*x*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 48*b^2*x*tan(1/2*d*x
+ 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*a*b*d^4*x^2 - 4*a^2*d^5*tan(1/2*d*x) - 96*b^2*d*x^2*tan(1/2*d*x + 1
/2*c)*tan(1/2*d*x)^2 - 4*a^2*d^5*tan(1/2*c) - 96*b^2*d*x^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 24*b^2*d^2*x^3
+ 8*a*b*d^3*x*tan(1/2*d*x + 1/2*c) + 48*b^2*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + 48*b^2*x*tan(1/2*d*x + 1
/2*c)^2*tan(1/2*c)^2 - 48*b^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - 96*b^2*d*x^2*tan(1/2*d*x + 1/2*c) + 48*b^2*x*tan
(1/2*d*x + 1/2*c)^2 - 48*b^2*x*tan(1/2*d*x)^2 - 48*b^2*x*tan(1/2*c)^2 - 48*b^2*x)/(d^5*x*tan(1/2*d*x + 1/2*c)^
2*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^5*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + d^5*x*tan(1/2*d*x + 1/2*c)^2*tan
(1/2*c)^2 + d^5*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^5*x*tan(1/2*d*x + 1/2*c)^2 + d^5*x*tan(1/2*d*x)^2 + d^5*x*ta
n(1/2*c)^2 + d^5*x)

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maple [B]  time = 0.06, size = 365, normalized size = 2.52 \[ d \left (\frac {\left (5 c^{4}+4 c^{3}+3 c^{2}+2 c +1\right ) b^{2} \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )-\frac {20 b^{2} c^{3} \left (1+2 c \right ) \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+\frac {2 \left (1+2 c \right ) a b \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}-\frac {6 b^{2} c \left (4 c^{3}+3 c^{2}+2 c +1\right ) \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+\frac {6 c a b \cos \left (d x +c \right )}{d^{3}}-\frac {15 c^{4} b^{2} \cos \left (d x +c \right )}{d^{6}}+\frac {15 \left (3 c^{2}+2 c +1\right ) c^{2} b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 14.76, size = 129, normalized size = 0.89 \[ \frac {{\left (a^{2} {\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (-i \, \Gamma \left (-1, i \, d x\right ) + i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} - 2 \, {\left (b^{2} d^{4} x^{4} + 2 \, a b d^{4} x - 12 \, b^{2} d^{2} x^{2} + 24 \, b^{2}\right )} \cos \left (d x + c\right ) + 4 \, {\left (2 \, b^{2} d^{3} x^{3} + a b d^{3} - 12 \, b^{2} d x\right )} \sin \left (d x + c\right )}{2 \, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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