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

Optimal. Leaf size=142 \[ -\frac {1}{2} a^2 d^2 \sin (c) \text {Ci}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {a^2 d \cos (c+d x)}{2 x}-\frac {2 a b \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3339, 2638, 3297, 3303, 3299, 3302, 3296, 2637} \[ -\frac {1}{2} a^2 d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {a^2 d \cos (c+d x)}{2 x}-\frac {2 a b \cos (c+d x)}{d}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {b^2 x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.41, size = 138, normalized size = 0.97 \[ \frac {1}{2} \left (-a^2 d^2 \sin (c) \text {Ci}(d x)-a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x^2}-\frac {a^2 d \cos (c+d x)}{x}-\frac {4 a b \cos (c+d x)}{d}-\frac {12 b^2 \sin (c+d x)}{d^4}+\frac {12 b^2 x \cos (c+d x)}{d^3}+\frac {6 b^2 x^2 \sin (c+d x)}{d^2}-\frac {2 b^2 x^3 \cos (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*b*Cos[c + d*x])/d - (a^2*d*Cos[c + d*x])/x + (12*b^2*x*Cos[c + d*x])/d^3 - (2*b^2*x^3*Cos[c + d*x])/d -
a^2*d^2*CosIntegral[d*x]*Sin[c] - (12*b^2*Sin[c + d*x])/d^4 - (a^2*Sin[c + d*x])/x^2 + (6*b^2*x^2*Sin[c + d*x
])/d^2 - a^2*d^2*Cos[c]*SinIntegral[d*x])/2

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fricas [A]  time = 0.69, size = 142, normalized size = 1.00 \[ -\frac {2 \, a^{2} d^{6} x^{2} \cos \relax (c) \operatorname {Si}\left (d x\right ) + 2 \, {\left (2 \, b^{2} d^{3} x^{5} + a^{2} d^{5} x + 4 \, a b d^{3} x^{2} - 12 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) - 2 \, {\left (6 \, b^{2} d^{2} x^{4} - a^{2} d^{4} - 12 \, b^{2} x^{2}\right )} \sin \left (d x + c\right ) + {\left (a^{2} d^{6} x^{2} \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} x^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{4 \, d^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 1.15, size = 2171, normalized size = 15.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*d^6*x^2*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)^2 - a^2*d^6*x^2
*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)^2 + 2*a^2*d^6*x^2*sin_integral
(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^6*x^2*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^2*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) + 4*b^2*d^3*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^2*ima
g_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^2*imag_part(cos_integral(-d*x))*ta
n(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 - 2*a^2*d^6*x^2*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 +
a^2*d^6*x^2*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^2*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*x^2*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/
2*c)^2 + a^2*d^6*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^6*x^2*imag_part(cos_inte
gral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^6*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*b^2*
d^3*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 - 2*a^2*d^6*x^2*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^2*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^2
*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^6*x^2*real_part(cos_integral(-d*x))*tan(1/2*
d*x)^2*tan(1/2*c) + 4*b^2*d^3*x^5*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 4*b^2*d^3*x^5*tan(1/2*d*x)^2*tan(1/2*c
)^2 - 2*a^2*d^5*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^2*imag_part(cos_integral(d*x)
)*tan(1/2*d*x + 1/2*c)^2 + a^2*d^6*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2 - 2*a^2*d^6*x^2*si
n_integral(d*x)*tan(1/2*d*x + 1/2*c)^2 - a^2*d^6*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a^2*d^6*x^2
*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^6*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 + a^2*d^6*x^2*i
mag_part(cos_integral(d*x))*tan(1/2*c)^2 - a^2*d^6*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^6*
x^2*sin_integral(d*x)*tan(1/2*c)^2 + 24*b^2*d^2*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*b*d
^3*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*b^2*d^3*x^5*tan(1/2*d*x + 1/2*c)^2 - 4*b^2*d^3*x
^5*tan(1/2*d*x)^2 + 2*a^2*d^5*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 - 2*a^2*d^6*x^2*real_part(cos_integral(d
*x))*tan(1/2*c) - 2*a^2*d^6*x^2*real_part(cos_integral(-d*x))*tan(1/2*c) + 8*a^2*d^5*x*tan(1/2*d*x + 1/2*c)^2*
tan(1/2*d*x)*tan(1/2*c) - 4*b^2*d^3*x^5*tan(1/2*c)^2 + 2*a^2*d^5*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2
*d^5*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*b^2*d*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^2*imag_part(cos_integral(d*x)) + a^2*d^6*x^2*imag_part(cos_integral(-d*x)) - 2*a^2*d^6*x^2*sin_integral(d*x
) + 24*b^2*d^2*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2 + 8*a*b*d^3*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 +
4*a^2*d^4*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c) + 24*b^2*d^2*x^4*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2
+ 8*a*b*d^3*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 4*a^2*d^4*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)*tan(1/2*c
)^2 - 8*a*b*d^3*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*b^2*d^3*x^5 - 2*a^2*d^5*x*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d
^5*x*tan(1/2*d*x)^2 - 24*b^2*d*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + 8*a^2*d^5*x*tan(1/2*d*x)*tan(1/2*c)
+ 2*a^2*d^5*x*tan(1/2*c)^2 - 24*b^2*d*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 24*b^2*d*x^3*tan(1/2*d*x)^2*t
an(1/2*c)^2 + 24*b^2*d^2*x^4*tan(1/2*d*x + 1/2*c) + 8*a*b*d^3*x^2*tan(1/2*d*x + 1/2*c)^2 - 4*a^2*d^4*tan(1/2*d
*x + 1/2*c)^2*tan(1/2*d*x) - 8*a*b*d^3*x^2*tan(1/2*d*x)^2 - 4*a^2*d^4*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 4*a^
2*d^4*tan(1/2*d*x)^2*tan(1/2*c) - 8*a*b*d^3*x^2*tan(1/2*c)^2 + 4*a^2*d^4*tan(1/2*d*x)*tan(1/2*c)^2 - 48*b^2*x^
2*tan(1/2*d*x + 1/2*c)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^5*x - 24*b^2*d*x^3*tan(1/2*d*x + 1/2*c)^2 + 24*b^
2*d*x^3*tan(1/2*d*x)^2 + 24*b^2*d*x^3*tan(1/2*c)^2 - 8*a*b*d^3*x^2 - 4*a^2*d^4*tan(1/2*d*x) - 48*b^2*x^2*tan(1
/2*d*x + 1/2*c)*tan(1/2*d*x)^2 - 4*a^2*d^4*tan(1/2*c) - 48*b^2*x^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 24*b^2*
d*x^3 - 48*b^2*x^2*tan(1/2*d*x + 1/2*c))/(d^4*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^4*x^2
*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + d^4*x^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + d^4*x^2*tan(1/2*d*x)^2*
tan(1/2*c)^2 + d^4*x^2*tan(1/2*d*x + 1/2*c)^2 + d^4*x^2*tan(1/2*d*x)^2 + d^4*x^2*tan(1/2*c)^2 + d^4*x^2)

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maple [A]  time = 0.05, size = 251, normalized size = 1.77 \[ d^{2} \left (\frac {20 c^{3} b^{2} \cos \left (d x +c \right )}{d^{6}}-\frac {2 a b \cos \left (d x +c \right )}{d^{3}}+\frac {\left (10 c^{3}+6 c^{2}+3 c +1\right ) b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )+\frac {15 \left (1+3 c \right ) c^{2} b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}-\frac {6 b^{2} c \left (6 c^{2}+3 c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(20*c^3/d^6*b^2*cos(d*x+c)-2*a*b*cos(d*x+c)/d^3+(10*c^3+6*c^2+3*c+1)/d^6*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))+a^2*(-1/2*sin(d*x+c)/x^2/d^2-1/2*cos(d*x+c)/x/d-1/2*Si(d*x
)*cos(c)-1/2*Ci(d*x)*sin(c))+15*(1+3*c)/d^6*c^2*b^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-6*b^2*c*(6*c^2+3*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)))

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maxima [C]  time = 3.07, size = 110, normalized size = 0.77 \[ \frac {{\left (a^{2} {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} - 2 \, {\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} - 6 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 6 \, {\left (b^{2} d^{2} x^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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