\(\int \frac {(a+b x^3)^2 \sin (c+d x)}{x^4} \, dx\) [92]

Optimal result

Integrand size = 19, antiderivative size = 151 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=\frac {2 b^2 \cos (c+d x)}{d^3}-\frac {a^2 d \cos (c+d x)}{6 x^2}-\frac {b^2 x^2 \cos (c+d x)}{d}-\frac {1}{6} a^2 d^3 \cos (c) \operatorname {CosIntegral}(d x)+2 a b \operatorname {CosIntegral}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{3 x^3}+\frac {a^2 d^2 \sin (c+d x)}{6 x}+\frac {2 b^2 x \sin (c+d x)}{d^2}+2 a b \cos (c) \text {Si}(d x)+\frac {1}{6} a^2 d^3 \sin (c) \text {Si}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3420, 3378, 3384, 3380, 3383, 3377, 2718} \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=-\frac {1}{6} a^2 d^3 \cos (c) \operatorname {CosIntegral}(d x)+\frac {1}{6} a^2 d^3 \sin (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{6 x}-\frac {a^2 \sin (c+d x)}{3 x^3}-\frac {a^2 d \cos (c+d x)}{6 x^2}+2 a b \sin (c) \operatorname {CosIntegral}(d x)+2 a b \cos (c) \text {Si}(d x)+\frac {2 b^2 \cos (c+d x)}{d^3}+\frac {2 b^2 x \sin (c+d x)}{d^2}-\frac {b^2 x^2 \cos (c+d x)}{d} \]

[In]

Int[((a + b*x^3)^2*Sin[c + d*x])/x^4,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=\frac {1}{6} \left (\frac {12 b^2 \cos (c+d x)}{d^3}-\frac {a^2 d \cos (c+d x)}{x^2}-\frac {6 b^2 x^2 \cos (c+d x)}{d}-a \operatorname {CosIntegral}(d x) \left (a d^3 \cos (c)-12 b \sin (c)\right )-\frac {2 a^2 \sin (c+d x)}{x^3}+\frac {a^2 d^2 \sin (c+d x)}{x}+\frac {12 b^2 x \sin (c+d x)}{d^2}+a \left (12 b \cos (c)+a d^3 \sin (c)\right ) \text {Si}(d x)\right ) \]

[In]

Integrate[((a + b*x^3)^2*Sin[c + d*x])/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(d^{3} \left (\frac {2 a b \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )}{d^{3}}-\frac {15 b^{2} c^{2} \cos \left (d x +c \right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{3 d^{3} x^{3}}-\frac {\cos \left (d x +c \right )}{6 d^{2} x^{2}}+\frac {\sin \left (d x +c \right )}{6 d x}+\frac {\operatorname {Si}\left (d x \right ) \sin \left (c \right )}{6}-\frac {\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{6}\right )+\frac {\left (10 c^{2}+4 c +1\right ) b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}-\frac {6 c \,b^{2} \left (4 c +1\right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}\right )\)	\(196\)
default	\(d^{3} \left (\frac {2 a b \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )}{d^{3}}-\frac {15 b^{2} c^{2} \cos \left (d x +c \right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{3 d^{3} x^{3}}-\frac {\cos \left (d x +c \right )}{6 d^{2} x^{2}}+\frac {\sin \left (d x +c \right )}{6 d x}+\frac {\operatorname {Si}\left (d x \right ) \sin \left (c \right )}{6}-\frac {\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{6}\right )+\frac {\left (10 c^{2}+4 c +1\right ) b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{6}}-\frac {6 c \,b^{2} \left (4 c +1\right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{6}}\right )\)	\(196\)
risch	\(-\frac {\pi \,\operatorname {csgn}\left (d x \right ) \sin \left (c \right ) a^{2} d^{6} x^{3}-2 \,\operatorname {Si}\left (d x \right ) \sin \left (c \right ) a^{2} d^{6} x^{3}+i \pi \,\operatorname {csgn}\left (d x \right ) \cos \left (c \right ) a^{2} d^{6} x^{3}-2 i \operatorname {Si}\left (d x \right ) \cos \left (c \right ) a^{2} d^{6} x^{3}+12 \pi \,\operatorname {csgn}\left (d x \right ) \cos \left (c \right ) a b \,d^{3} x^{3}-2 \,\operatorname {Ei}_{1}\left (-i d x \right ) \cos \left (c \right ) a^{2} d^{6} x^{3}-24 \,\operatorname {Si}\left (d x \right ) \cos \left (c \right ) a b \,d^{3} x^{3}+24 i \operatorname {Si}\left (d x \right ) \sin \left (c \right ) a b \,d^{3} x^{3}-12 i \pi \,\operatorname {csgn}\left (d x \right ) \sin \left (c \right ) a b \,d^{3} x^{3}-2 \sin \left (d x +c \right ) a^{2} d^{5} x^{2}+12 \cos \left (d x +c \right ) b^{2} d^{2} x^{5}+24 \,\operatorname {Ei}_{1}\left (-i d x \right ) \sin \left (c \right ) a b \,d^{3} x^{3}-24 \sin \left (d x +c \right ) b^{2} d \,x^{4}+2 \cos \left (d x +c \right ) a^{2} d^{4} x +4 \sin \left (d x +c \right ) a^{2} d^{3}-24 \cos \left (d x +c \right ) b^{2} x^{3}}{12 d^{3} x^{3}}\)	\(280\)
meijerg	\(\frac {4 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {4 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {d^{2} x^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+a b \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )+2 a b \cos \left (c \right ) \operatorname {Si}\left (d x \right )+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) d^{4} \left (-\frac {8 \left (-d^{2} x^{2}+2\right ) d^{2} \cos \left (x \sqrt {d^{2}}\right )}{3 x^{3} \left (d^{2}\right )^{\frac {5}{2}} \sqrt {\pi }}+\frac {8 \sin \left (x \sqrt {d^{2}}\right )}{3 d^{2} x^{2} \sqrt {\pi }}+\frac {8 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{3 \sqrt {\pi }}\right )}{16 \sqrt {d^{2}}}+\frac {a^{2} \sqrt {\pi }\, \cos \left (c \right ) d^{3} \left (-\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {4 \left (2 \gamma -\frac {11}{3}+2 \ln \left (x \right )+2 \ln \left (d \right )\right )}{3 \sqrt {\pi }}+\frac {-\frac {44 d^{2} x^{2}}{9}+8}{d^{2} x^{2} \sqrt {\pi }}+\frac {8 \gamma }{3 \sqrt {\pi }}+\frac {8 \ln \left (2\right )}{3 \sqrt {\pi }}+\frac {8 \ln \left (\frac {d x}{2}\right )}{3 \sqrt {\pi }}-\frac {8 \cos \left (d x \right )}{3 \sqrt {\pi }\, d^{2} x^{2}}-\frac {16 \left (-\frac {5 d^{2} x^{2}}{2}+5\right ) \sin \left (d x \right )}{15 \sqrt {\pi }\, d^{3} x^{3}}-\frac {8 \,\operatorname {Ci}\left (d x \right )}{3 \sqrt {\pi }}\right )}{16}\)	\(403\)

[In]

int((b*x^3+a)^2*sin(d*x+c)/x^4,x,method=_RETURNVERBOSE)

[Out]

d^3*(2/d^3*a*b*(Si(d*x)*cos(c)+Ci(d*x)*sin(c))-15/d^6*b^2*c^2*cos(d*x+c)+a^2*(-1/3*sin(d*x+c)/d^3/x^3-1/6*cos(
d*x+c)/d^2/x^2+1/6*sin(d*x+c)/d/x+1/6*Si(d*x)*sin(c)-1/6*Ci(d*x)*cos(c))+(10*c^2+4*c+1)/d^6*b^2*(-(d*x+c)^2*co
s(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))-6*c*b^2*(4*c+1)/d^6*(sin(d*x+c)-cos(d*x+c)*(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=-\frac {{\left (6 \, b^{2} d^{2} x^{5} + a^{2} d^{4} x - 12 \, b^{2} x^{3}\right )} \cos \left (d x + c\right ) + {\left (a^{2} d^{6} x^{3} \operatorname {Ci}\left (d x\right ) - 12 \, a b d^{3} x^{3} \operatorname {Si}\left (d x\right )\right )} \cos \left (c\right ) - {\left (a^{2} d^{5} x^{2} + 12 \, b^{2} d x^{4} - 2 \, a^{2} d^{3}\right )} \sin \left (d x + c\right ) - {\left (a^{2} d^{6} x^{3} \operatorname {Si}\left (d x\right ) + 12 \, a b d^{3} x^{3} \operatorname {Ci}\left (d x\right )\right )} \sin \left (c\right )}{6 \, d^{3} x^{3}} \]

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=\int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

[In]

integrate((b*x**3+a)**2*sin(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**3)**2*sin(c + d*x)/x**4, x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.11 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=-\frac {{\left ({\left (a^{2} {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) - a^{2} {\left (i \, \Gamma \left (-3, i \, d x\right ) - i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} + 12 \, {\left (a b {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) - a b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3}\right )} x^{3} + 2 \, {\left (b^{2} d^{2} x^{5} + 2 \, a b d^{2} x^{2} - 2 \, b^{2} x^{3} - 4 \, a b\right )} \cos \left (d x + c\right ) - 4 \, {\left (b^{2} d x^{4} - a b d x\right )} \sin \left (d x + c\right )}{2 \, d^{3} x^{3}} \]

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^4,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

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

Time = 0.34 (sec) , antiderivative size = 1181, normalized size of antiderivative = 7.82 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=\text {Too large to display} \]

[In]

integrate((b*x^3+a)^2*sin(d*x+c)/x^4,x, algorithm="giac")

[Out]

1/12*(a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^6*x^3*real_part(cos_integra
l(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^6*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) -
2*a^2*d^6*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^6*x^3*sin_integral(d*x)*tan(1/
2*d*x)^2*tan(1/2*c) - a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a^2*d^6*x^3*real_part(cos_inte
gral(-d*x))*tan(1/2*d*x)^2 + a^2*d^6*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a^2*d^6*x^3*real_part(cos
_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^6*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^6*x^3*imag_par
t(cos_integral(-d*x))*tan(1/2*c) + 4*a^2*d^6*x^3*sin_integral(d*x)*tan(1/2*c) - 12*b^2*d^2*x^5*tan(1/2*d*x)^2*
tan(1/2*c)^2 - 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*a*b*d^3*x^3*imag_p
art(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*
c)^2 - a^2*d^6*x^3*real_part(cos_integral(d*x)) - a^2*d^6*x^3*real_part(cos_integral(-d*x)) - 4*a^2*d^5*x^2*ta
n(1/2*d*x)^2*tan(1/2*c) + 24*a*b*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 24*a*b*d^3*x
^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d^5*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 12*b^2*
d^2*x^5*tan(1/2*d*x)^2 + 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 12*a*b*d^3*x^3*imag_part
(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2 + 48*b^2*d^2*x^5*tan(1/2
*d*x)*tan(1/2*c) + 12*b^2*d^2*x^5*tan(1/2*c)^2 - 12*a*b*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 12
*a*b*d^3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 24*a*b*d^3*x^3*sin_integral(d*x)*tan(1/2*c)^2 - 2*a^
2*d^4*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a^2*d^5*x^2*tan(1/2*d*x) + 4*a^2*d^5*x^2*tan(1/2*c) + 24*a*b*d^3*x^3*r
eal_part(cos_integral(d*x))*tan(1/2*c) + 24*a*b*d^3*x^3*real_part(cos_integral(-d*x))*tan(1/2*c) - 48*b^2*d*x^
4*tan(1/2*d*x)^2*tan(1/2*c) - 48*b^2*d*x^4*tan(1/2*d*x)*tan(1/2*c)^2 - 12*b^2*d^2*x^5 + 12*a*b*d^3*x^3*imag_pa
rt(cos_integral(d*x)) - 12*a*b*d^3*x^3*imag_part(cos_integral(-d*x)) + 24*a*b*d^3*x^3*sin_integral(d*x) + 2*a^
2*d^4*x*tan(1/2*d*x)^2 + 8*a^2*d^4*x*tan(1/2*d*x)*tan(1/2*c) + 2*a^2*d^4*x*tan(1/2*c)^2 + 24*b^2*x^3*tan(1/2*d
*x)^2*tan(1/2*c)^2 + 48*b^2*d*x^4*tan(1/2*d*x) + 48*b^2*d*x^4*tan(1/2*c) + 8*a^2*d^3*tan(1/2*d*x)^2*tan(1/2*c)
+ 8*a^2*d^3*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d^4*x - 24*b^2*x^3*tan(1/2*d*x)^2 - 96*b^2*x^3*tan(1/2*d*x)*tan
(1/2*c) - 24*b^2*x^3*tan(1/2*c)^2 - 8*a^2*d^3*tan(1/2*d*x) - 8*a^2*d^3*tan(1/2*c) + 24*b^2*x^3)/(d^3*x^3*tan(1
/2*d*x)^2*tan(1/2*c)^2 + d^3*x^3*tan(1/2*d*x)^2 + d^3*x^3*tan(1/2*c)^2 + d^3*x^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^4} \, dx=\int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^4} \,d x \]

[In]

int((sin(c + d*x)*(a + b*x^3)^2)/x^4,x)

[Out]

int((sin(c + d*x)*(a + b*x^3)^2)/x^4, x)