Integrand size = 19, antiderivative size = 142 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=-\frac {2 a b \cos (c+d x)}{d}-\frac {a^2 d \cos (c+d x)}{2 x}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {b^2 x^3 \cos (c+d x)}{d}-\frac {1}{2} a^2 d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {6 b^2 \sin (c+d x)}{d^4}-\frac {a^2 \sin (c+d x)}{2 x^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x) \]
-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*s in(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
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\frac {1}{2} \left (-\frac {4 a b \cos (c+d x)}{d}-\frac {a^2 d \cos (c+d x)}{x}+\frac {12 b^2 x \cos (c+d x)}{d^3}-\frac {2 b^2 x^3 \cos (c+d x)}{d}-a^2 d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {12 b^2 \sin (c+d x)}{d^4}-\frac {a^2 \sin (c+d x)}{x^2}+\frac {6 b^2 x^2 \sin (c+d x)}{d^2}-a^2 d^2 \cos (c) \text {Si}(d x)\right ) \]
((-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
Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3820, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx\) |
| \(\Big \downarrow \) 3820 |
| \(\displaystyle \int \left (\frac {a^2 \sin (c+d x)}{x^3}+2 a b \sin (c+d x)+b^2 x^3 \sin (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{2} a^2 d^2 \sin (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {a^2 d \cos (c+d x)}{2 x}-\frac {2 a b \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d}\) |
(-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
3.1.91.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_ )], x_Symbol] :> Int[ExpandIntegrand[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]
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 | \(\frac {8 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \cos \left (d x \right )}{4 \sqrt {\pi }}-\frac {d x \left (-\frac {d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {x d \left (-\frac {5 d^{2} x^{2}}{2}+15\right ) \cos \left (d x \right )}{20 \sqrt {\pi }}-\frac {\left (-\frac {15 d^{2} x^{2}}{2}+15\right ) \sin \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {2 a b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {2 a b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a^{2} \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\) | \(332\) |
-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^4-2 4*cos(d*x+c)*b^2*d*x^3+24*sin(d*x+c)*b^2*x^2)
Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=-\frac {a^{2} d^{6} x^{2} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + a^{2} d^{6} x^{2} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + {\left (2 \, b^{2} d^{3} x^{5} + a^{2} d^{5} x + 4 \, a b d^{3} x^{2} - 12 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, b^{2} d^{2} x^{4} - a^{2} d^{4} - 12 \, b^{2} x^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4} x^{2}} \]
-1/2*(a^2*d^6*x^2*cos_integral(d*x)*sin(c) + a^2*d^6*x^2*cos(c)*sin_integr al(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)
\[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{3}}\, dx \]
Result contains complex when optimal does not.
Time = 2.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\frac {{\left (a^{2} {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 2 \, {\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} - 6 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 6 \, {\left (b^{2} d^{2} x^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \]
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
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 \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\text {Too large to display} \]
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*rea l_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*imag_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))*tan(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_integral(-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)*t an(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_integral(-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*ta n(1/2*c) - 2*a^2*d^6*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*ta...
Timed out. \[ \int \frac {\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx=\int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^3} \,d x \]