\(\int \frac {(a+b x^3)^2 \sin (c+d x)}{x^3} \, dx\) [91]

Optimal result

Integrand size = 19, antiderivative size = 142 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=-\frac {2 a b \cos (c+d x)}{d}-\frac {a^2 d \cos (c+d x)}{2 x}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {b^2 x^3 \cos (c+d x)}{d}-\frac {1}{2} a^2 d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {6 b^2 \sin (c+d x)}{d^4}-\frac {a^2 \sin (c+d x)}{2 x^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x) \]

[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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3420, 2718, 3378, 3384, 3380, 3383, 3377, 2717} \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=-\frac {1}{2} a^2 d^2 \sin (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {a^2 d \cos (c+d x)}{2 x}-\frac {2 a b \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d} \]

[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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 3378

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 3380

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 3383

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 3384

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 3420

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*} \text {integral}& = \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 \operatorname {CosIntegral}(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*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\frac {1}{2} \left (-\frac {4 a b \cos (c+d x)}{d}-\frac {a^2 d \cos (c+d x)}{x}+\frac {12 b^2 x \cos (c+d x)}{d^3}-\frac {2 b^2 x^3 \cos (c+d x)}{d}-a^2 d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {12 b^2 \sin (c+d x)}{d^4}-\frac {a^2 \sin (c+d x)}{x^2}+\frac {6 b^2 x^2 \sin (c+d x)}{d^2}-a^2 d^2 \cos (c) \text {Si}(d x)\right ) \]

[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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.50 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.48

method	result	size
risch	\(-\frac {-\pi \,\operatorname {csgn}\left (d x \right ) \cos \left (c \right ) a^{2} d^{6} x^{2}+2 \,\operatorname {Si}\left (d x \right ) \cos \left (c \right ) a^{2} d^{6} x^{2}+i \pi \,\operatorname {csgn}\left (d x \right ) \sin \left (c \right ) a^{2} d^{6} x^{2}-2 i \operatorname {Si}\left (d x \right ) \sin \left (c \right ) a^{2} d^{6} x^{2}-2 \,\operatorname {Ei}_{1}\left (-i d x \right ) \sin \left (c \right ) a^{2} d^{6} x^{2}+4 \cos \left (d x +c \right ) b^{2} d^{3} x^{5}-12 \sin \left (d x +c \right ) b^{2} d^{2} x^{4}+2 \cos \left (d x +c \right ) a^{2} d^{5} x +8 \cos \left (d x +c \right ) a b \,d^{3} x^{2}+2 \sin \left (d x +c \right ) a^{2} d^{4}-24 \cos \left (d x +c \right ) b^{2} d \,x^{3}+24 \sin \left (d x +c \right ) b^{2} x^{2}}{4 x^{2} d^{4}}\)	\(210\)
derivativedivides	\(d^{2} \left (\frac {20 b^{2} c^{3} \cos \left (d x +c \right )}{d^{6}}-\frac {2 a b \cos \left (d x +c \right )}{d^{3}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {\left (10 c^{3}+6 c^{2}+3 c +1\right ) b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}-\frac {6 c \,b^{2} \left (6 c^{2}+3 c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}+\frac {15 \left (3 c +1\right ) c^{2} b^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}\right )\)	\(251\)
default	\(d^{2} \left (\frac {20 b^{2} c^{3} \cos \left (d x +c \right )}{d^{6}}-\frac {2 a b \cos \left (d x +c \right )}{d^{3}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {\left (10 c^{3}+6 c^{2}+3 c +1\right ) b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}-\frac {6 c \,b^{2} \left (6 c^{2}+3 c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}+\frac {15 \left (3 c +1\right ) c^{2} b^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}\right )\)	\(251\)
meijerg	\(\frac {8 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \cos \left (d x \right )}{4 \sqrt {\pi }}-\frac {d x \left (-\frac {d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {x d \left (-\frac {5 d^{2} x^{2}}{2}+15\right ) \cos \left (d x \right )}{20 \sqrt {\pi }}-\frac {\left (-\frac {15 d^{2} x^{2}}{2}+15\right ) \sin \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {2 a b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {2 a b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a^{2} \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(332\)

[In]

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

[Out]

-1/4/x^2/d^4*(-Pi*csgn(d*x)*cos(c)*a^2*d^6*x^2+2*Si(d*x)*cos(c)*a^2*d^6*x^2+I*Pi*csgn(d*x)*sin(c)*a^2*d^6*x^2-
2*I*Si(d*x)*sin(c)*a^2*d^6*x^2-2*Ei(1,-I*d*x)*sin(c)*a^2*d^6*x^2+4*cos(d*x+c)*b^2*d^3*x^5-12*sin(d*x+c)*b^2*d^
2*x^4+2*cos(d*x+c)*a^2*d^5*x+8*cos(d*x+c)*a*b*d^3*x^2+2*sin(d*x+c)*a^2*d^4-24*cos(d*x+c)*b^2*d*x^3+24*sin(d*x+
c)*b^2*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=-\frac {a^{2} d^{6} x^{2} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + a^{2} d^{6} x^{2} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + {\left (2 \, b^{2} d^{3} x^{5} + a^{2} d^{5} x + 4 \, a b d^{3} x^{2} - 12 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, b^{2} d^{2} x^{4} - a^{2} d^{4} - 12 \, b^{2} x^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4} x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{3}}\, dx \]

[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)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\frac {{\left (a^{2} {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 2 \, {\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} - 6 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 6 \, {\left (b^{2} d^{2} x^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \]

[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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.32 (sec) , antiderivative size = 2171, normalized size of antiderivative = 15.29 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\text {Too large to display} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^3} \,d x \]

[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)