Integrand size = 16, antiderivative size = 126 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}-\frac {3 i b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{20 \sqrt [3]{-i d x^3}}+\frac {3 i b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{20 \sqrt [3]{i d x^3}}-\frac {b \sin \left (c+d x^3\right )}{5 x^5} \] Output:
-1/5*a/x^5-3/10*b*d*cos(d*x^3+c)/x^2-3/20*I*b*d^2*exp(I*c)*x*GAMMA(1/3,-I* d*x^3)/(-I*d*x^3)^(1/3)+3/20*I*b*d^2*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/(I*d*x^ 3)^(1/3)-1/5*b*sin(d*x^3+c)/x^5
Time = 0.45 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=\frac {3 b d^2 x^6 \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (-i \cos (c)+\sin (c))+3 b d^2 x^6 \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},i d x^3\right ) (i \cos (c)+\sin (c))-2 \sqrt [3]{d^2 x^6} \left (2 a+3 b d x^3 \cos \left (c+d x^3\right )+2 b \sin \left (c+d x^3\right )\right )}{20 x^5 \sqrt [3]{d^2 x^6}} \] Input:
Integrate[(a + b*Sin[c + d*x^3])/x^6,x]
Output:
(3*b*d^2*x^6*(I*d*x^3)^(1/3)*Gamma[1/3, (-I)*d*x^3]*((-I)*Cos[c] + Sin[c]) + 3*b*d^2*x^6*((-I)*d*x^3)^(1/3)*Gamma[1/3, I*d*x^3]*(I*Cos[c] + Sin[c]) - 2*(d^2*x^6)^(1/3)*(2*a + 3*b*d*x^3*Cos[c + d*x^3] + 2*b*Sin[c + d*x^3])) /(20*x^5*(d^2*x^6)^(1/3))
Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 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.
\(\displaystyle \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {a}{x^6}+\frac {b \sin \left (c+d x^3\right )}{x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a}{5 x^5}-\frac {3 i b e^{i c} d^2 x \Gamma \left (\frac {1}{3},-i d x^3\right )}{20 \sqrt [3]{-i d x^3}}+\frac {3 i b e^{-i c} d^2 x \Gamma \left (\frac {1}{3},i d x^3\right )}{20 \sqrt [3]{i d x^3}}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}\) |
Input:
Int[(a + b*Sin[c + d*x^3])/x^6,x]
Output:
-1/5*a/x^5 - (3*b*d*Cos[c + d*x^3])/(10*x^2) - (((3*I)/20)*b*d^2*E^(I*c)*x *Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3)^(1/3) + (((3*I)/20)*b*d^2*x*Gamma[1/ 3, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(1/3)) - (b*Sin[c + d*x^3])/(5*x^5)
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
\[\int \frac {a +b \sin \left (d \,x^{3}+c \right )}{x^{6}}d x\]
Input:
int((a+b*sin(d*x^3+c))/x^6,x)
Output:
int((a+b*sin(d*x^3+c))/x^6,x)
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=-\frac {6 \, b d x^{3} \cos \left (d x^{3} + c\right ) - 3 \, {\left (b d x^{5} \cos \left (c\right ) - i \, b d x^{5} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 3 \, {\left (b d x^{5} \cos \left (c\right ) + i \, b d x^{5} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) + 4 \, b \sin \left (d x^{3} + c\right ) + 4 \, a}{20 \, x^{5}} \] Input:
integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="fricas")
Output:
-1/20*(6*b*d*x^3*cos(d*x^3 + c) - 3*(b*d*x^5*cos(c) - I*b*d*x^5*sin(c))*(I *d)^(2/3)*gamma(1/3, I*d*x^3) - 3*(b*d*x^5*cos(c) + I*b*d*x^5*sin(c))*(-I* d)^(2/3)*gamma(1/3, -I*d*x^3) + 4*b*sin(d*x^3 + c) + 4*a)/x^5
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=\int \frac {a + b \sin {\left (c + d x^{3} \right )}}{x^{6}}\, dx \] Input:
integrate((a+b*sin(d*x**3+c))/x**6,x)
Output:
Integral((a + b*sin(c + d*x**3))/x**6, x)
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=-\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} b d}{12 \, x^{2}} - \frac {a}{5 \, x^{5}} \] Input:
integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="maxima")
Output:
-1/12*(d*x^3)^(2/3)*(((-I*sqrt(3) - 1)*gamma(-5/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(-5/3, -I*d*x^3))*cos(c) - ((sqrt(3) - I)*gamma(-5/3, I*d*x^3) + (sqrt(3) + I)*gamma(-5/3, -I*d*x^3))*sin(c))*b*d/x^2 - 1/5*a/x^5
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=\int { \frac {b \sin \left (d x^{3} + c\right ) + a}{x^{6}} \,d x } \] Input:
integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="giac")
Output:
integrate((b*sin(d*x^3 + c) + a)/x^6, x)
Timed out. \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=\int \frac {a+b\,\sin \left (d\,x^3+c\right )}{x^6} \,d x \] Input:
int((a + b*sin(c + d*x^3))/x^6,x)
Output:
int((a + b*sin(c + d*x^3))/x^6, x)
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx=\frac {5 \left (\int \frac {\sin \left (d \,x^{3}+c \right )}{x^{6}}d x \right ) b \,x^{5}-a}{5 x^{5}} \] Input:
int((a+b*sin(d*x^3+c))/x^6,x)
Output:
(5*int(sin(c + d*x**3)/x**6,x)*b*x**5 - a)/(5*x**5)