Integrand size = 17, antiderivative size = 149 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=-\frac {a d \cos (c+d x)}{12 x^3}-\frac {b d \cos (c+d x)}{2 x}+\frac {a d^3 \cos (c+d x)}{24 x}-\frac {1}{2} b d^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {1}{24} a d^4 \operatorname {CosIntegral}(d x) \sin (c)-\frac {a \sin (c+d x)}{4 x^4}-\frac {b \sin (c+d x)}{2 x^2}+\frac {a d^2 \sin (c+d x)}{24 x^2}-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} a d^4 \cos (c) \text {Si}(d x) \] Output:
-1/12*a*d*cos(d*x+c)/x^3-1/2*b*d*cos(d*x+c)/x+1/24*a*d^3*cos(d*x+c)/x-1/2* b*d^2*Ci(d*x)*sin(c)+1/24*a*d^4*Ci(d*x)*sin(c)-1/4*a*sin(d*x+c)/x^4-1/2*b* sin(d*x+c)/x^2+1/24*a*d^2*sin(d*x+c)/x^2-1/2*b*d^2*cos(c)*Si(d*x)+1/24*a*d ^4*cos(c)*Si(d*x)
Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\frac {-2 a d x \cos (c+d x)-12 b d x^3 \cos (c+d x)+a d^3 x^3 \cos (c+d x)+d^2 \left (-12 b+a d^2\right ) x^4 \operatorname {CosIntegral}(d x) \sin (c)-6 a \sin (c+d x)-12 b x^2 \sin (c+d x)+a d^2 x^2 \sin (c+d x)+d^2 \left (-12 b+a d^2\right ) x^4 \cos (c) \text {Si}(d x)}{24 x^4} \] Input:
Integrate[((a + b*x^2)*Sin[c + d*x])/x^5,x]
Output:
(-2*a*d*x*Cos[c + d*x] - 12*b*d*x^3*Cos[c + d*x] + a*d^3*x^3*Cos[c + d*x] + d^2*(-12*b + a*d^2)*x^4*CosIntegral[d*x]*Sin[c] - 6*a*Sin[c + d*x] - 12* b*x^2*Sin[c + d*x] + a*d^2*x^2*Sin[c + d*x] + d^2*(-12*b + a*d^2)*x^4*Cos[ c]*SinIntegral[d*x])/(24*x^4)
Time = 0.47 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx\) |
| \(\Big \downarrow \) 3820 |
| \(\displaystyle \int \left (\frac {a \sin (c+d x)}{x^5}+\frac {b \sin (c+d x)}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{24} a d^4 \sin (c) \operatorname {CosIntegral}(d x)+\frac {1}{24} a d^4 \cos (c) \text {Si}(d x)+\frac {a d^3 \cos (c+d x)}{24 x}+\frac {a d^2 \sin (c+d x)}{24 x^2}-\frac {a \sin (c+d x)}{4 x^4}-\frac {a d \cos (c+d x)}{12 x^3}-\frac {1}{2} b d^2 \sin (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{2 x^2}-\frac {b d \cos (c+d x)}{2 x}\) |
Input:
Int[((a + b*x^2)*Sin[c + d*x])/x^5,x]
Output:
-1/12*(a*d*Cos[c + d*x])/x^3 - (b*d*Cos[c + d*x])/(2*x) + (a*d^3*Cos[c + d *x])/(24*x) - (b*d^2*CosIntegral[d*x]*Sin[c])/2 + (a*d^4*CosIntegral[d*x]* Sin[c])/24 - (a*Sin[c + d*x])/(4*x^4) - (b*Sin[c + d*x])/(2*x^2) + (a*d^2* Sin[c + d*x])/(24*x^2) - (b*d^2*Cos[c]*SinIntegral[d*x])/2 + (a*d^4*Cos[c] *SinIntegral[d*x])/24
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]
Time = 1.01 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(d^{4} \left (a \left (-\frac {\sin \left (d x +c \right )}{4 d^{4} x^{4}}-\frac {\cos \left (d x +c \right )}{12 d^{3} x^{3}}+\frac {\sin \left (d x +c \right )}{24 d^{2} x^{2}}+\frac {\cos \left (d x +c \right )}{24 d x}+\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{24}+\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{24}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )}{d^{2}}\right )\) | \(131\) |
| default | \(d^{4} \left (a \left (-\frac {\sin \left (d x +c \right )}{4 d^{4} x^{4}}-\frac {\cos \left (d x +c \right )}{12 d^{3} x^{3}}+\frac {\sin \left (d x +c \right )}{24 d^{2} x^{2}}+\frac {\cos \left (d x +c \right )}{24 d x}+\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{24}+\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{24}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )}{d^{2}}\right )\) | \(131\) |
| risch | \(-\frac {i \operatorname {expIntegral}_{1}\left (i d x \right ) \cos \left (c \right ) a \,d^{4}}{48}+\frac {i \cos \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) a \,d^{4}}{48}+\frac {i \operatorname {expIntegral}_{1}\left (i d x \right ) \cos \left (c \right ) b \,d^{2}}{4}-\frac {i \cos \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) b \,d^{2}}{4}-\frac {\operatorname {expIntegral}_{1}\left (i d x \right ) \sin \left (c \right ) a \,d^{4}}{48}-\frac {\sin \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) a \,d^{4}}{48}+\frac {\operatorname {expIntegral}_{1}\left (i d x \right ) \sin \left (c \right ) b \,d^{2}}{4}+\frac {\sin \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) b \,d^{2}}{4}-\frac {i \left (2 i a \,d^{9} x^{7}-24 i b \,d^{7} x^{7}-4 i a \,d^{7} x^{5}\right ) \cos \left (d x +c \right )}{48 d^{6} x^{8}}-\frac {\left (-2 a \,d^{8} x^{6}+24 b \,d^{6} x^{6}+12 a \,d^{6} x^{4}\right ) \sin \left (d x +c \right )}{48 d^{6} x^{8}}\) | \(214\) |
| meijerg | \(\frac {d^{2} b \sin \left (c \right ) \sqrt {\pi }\, \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 x^{2} d^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, x d}-\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {d^{2} b \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{x^{2} d^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \sin \left (c \right ) \sqrt {\pi }\, d^{4} \left (-\frac {8}{\sqrt {\pi }\, x^{4} d^{4}}+\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {\frac {4 \gamma }{3}-\frac {25}{9}+\frac {4 \ln \left (x \right )}{3}+\frac {2 \ln \left (d^{2}\right )}{3}}{\sqrt {\pi }}+\frac {\frac {25}{9} x^{4} d^{4}-8 x^{2} d^{2}+8}{\sqrt {\pi }\, x^{4} d^{4}}-\frac {4 \gamma }{3 \sqrt {\pi }}-\frac {4 \ln \left (2\right )}{3 \sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{3 \sqrt {\pi }}-\frac {8 \left (-\frac {15 x^{2} d^{2}}{2}+45\right ) \cos \left (d x \right )}{45 \sqrt {\pi }\, x^{4} d^{4}}+\frac {8 \left (-\frac {15 x^{2} d^{2}}{2}+15\right ) \sin \left (d x \right )}{45 \sqrt {\pi }\, x^{3} d^{3}}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{3 \sqrt {\pi }}\right )}{32}+\frac {a \cos \left (c \right ) \sqrt {\pi }\, d^{4} \left (-\frac {8 \left (-\frac {x^{2} d^{2}}{2}+1\right ) \cos \left (d x \right )}{3 d^{3} x^{3} \sqrt {\pi }}-\frac {8 \left (-\frac {x^{2} d^{2}}{2}+3\right ) \sin \left (d x \right )}{3 d^{4} x^{4} \sqrt {\pi }}+\frac {4 \,\operatorname {Si}\left (d x \right )}{3 \sqrt {\pi }}\right )}{32}\) | \(411\) |
Input:
int((b*x^2+a)*sin(d*x+c)/x^5,x,method=_RETURNVERBOSE)
Output:
d^4*(a*(-1/4*sin(d*x+c)/d^4/x^4-1/12*cos(d*x+c)/d^3/x^3+1/24*sin(d*x+c)/d^ 2/x^2+1/24*cos(d*x+c)/d/x+1/24*Si(d*x)*cos(c)+1/24*Ci(d*x)*sin(c))+1/d^2*b *(-1/2*sin(d*x+c)/d^2/x^2-1/2*cos(d*x+c)/d/x-1/2*Si(d*x)*cos(c)-1/2*Ci(d*x )*sin(c)))
Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\frac {{\left (a d^{4} - 12 \, b d^{2}\right )} x^{4} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + {\left (a d^{4} - 12 \, b d^{2}\right )} x^{4} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + {\left ({\left (a d^{3} - 12 \, b d\right )} x^{3} - 2 \, a d x\right )} \cos \left (d x + c\right ) + {\left ({\left (a d^{2} - 12 \, b\right )} x^{2} - 6 \, a\right )} \sin \left (d x + c\right )}{24 \, x^{4}} \] Input:
integrate((b*x^2+a)*sin(d*x+c)/x^5,x, algorithm="fricas")
Output:
1/24*((a*d^4 - 12*b*d^2)*x^4*cos_integral(d*x)*sin(c) + (a*d^4 - 12*b*d^2) *x^4*cos(c)*sin_integral(d*x) + ((a*d^3 - 12*b*d)*x^3 - 2*a*d*x)*cos(d*x + c) + ((a*d^2 - 12*b)*x^2 - 6*a)*sin(d*x + c))/x^4
\[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\int \frac {\left (a + b x^{2}\right ) \sin {\left (c + d x \right )}}{x^{5}}\, dx \] Input:
integrate((b*x**2+a)*sin(d*x+c)/x**5,x)
Output:
Integral((a + b*x**2)*sin(c + d*x)/x**5, x)
Result contains complex when optimal does not.
Time = 0.70 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=-\frac {{\left ({\left (a {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + a {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 12 \, {\left (b {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4}\right )} x^{4} + 2 \, b d x \cos \left (d x + c\right ) + 6 \, b \sin \left (d x + c\right )}{2 \, d^{2} x^{4}} \] Input:
integrate((b*x^2+a)*sin(d*x+c)/x^5,x, algorithm="maxima")
Output:
-1/2*(((a*(I*gamma(-4, I*d*x) - I*gamma(-4, -I*d*x))*cos(c) + a*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*sin(c))*d^6 - 12*(b*(I*gamma(-4, I*d*x) - I*g amma(-4, -I*d*x))*cos(c) + b*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*sin(c) )*d^4)*x^4 + 2*b*d*x*cos(d*x + c) + 6*b*sin(d*x + c))/(d^2*x^4)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 1086, normalized size of antiderivative = 7.29 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\text {Too large to display} \] Input:
integrate((b*x^2+a)*sin(d*x+c)/x^5,x, algorithm="giac")
Output:
-1/48*(a*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2* a*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^4*x^4*real _part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^4*x^4*real_part (cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - a*d^4*x^4*imag_part(cos_i ntegral(d*x))*tan(1/2*d*x)^2 + a*d^4*x^4*imag_part(cos_integral(-d*x))*tan (1/2*d*x)^2 - 2*a*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2 + a*d^4*x^4*ima g_part(cos_integral(d*x))*tan(1/2*c)^2 - a*d^4*x^4*imag_part(cos_integral( -d*x))*tan(1/2*c)^2 + 2*a*d^4*x^4*sin_integral(d*x)*tan(1/2*c)^2 - 12*b*d^ 2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*b*d^2* x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*b*d^2*x ^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^4*x^4*real_part(c os_integral(d*x))*tan(1/2*c) - 2*a*d^4*x^4*real_part(cos_integral(-d*x))*t an(1/2*c) + 24*b*d^2*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1 /2*c) + 24*b*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2* c) - 2*a*d^3*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^4*x^4*imag_part(cos_int egral(d*x)) + a*d^4*x^4*imag_part(cos_integral(-d*x)) - 2*a*d^4*x^4*sin_in tegral(d*x) + 12*b*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 1 2*b*d^2*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*b*d^2*x^4*si n_integral(d*x)*tan(1/2*d*x)^2 - 12*b*d^2*x^4*imag_part(cos_integral(d*...
Timed out. \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^5} \,d x \] Input:
int((sin(c + d*x)*(a + b*x^2))/x^5,x)
Output:
int((sin(c + d*x)*(a + b*x^2))/x^5, x)
\[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^5} \, dx=\frac {-12 \cos \left (d x +c \right ) b x -6 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{4}+x^{4}}d x \right ) a \,d^{2} x^{4}+72 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{4}+x^{4}}d x \right ) b \,x^{4}-3 \sin \left (d x +c \right ) a d -a \,d^{2} x +12 b x}{12 d \,x^{4}} \] Input:
int((b*x^2+a)*sin(d*x+c)/x^5,x)
Output:
( - 12*cos(c + d*x)*b*x - 6*int(tan((c + d*x)/2)**2/(tan((c + d*x)/2)**2*x **4 + x**4),x)*a*d**2*x**4 + 72*int(tan((c + d*x)/2)**2/(tan((c + d*x)/2)* *2*x**4 + x**4),x)*b*x**4 - 3*sin(c + d*x)*a*d - a*d**2*x + 12*b*x)/(12*d* x**4)