Integrand size = 16, antiderivative size = 130 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {3 b d \cos \left (c+d x^3\right )}{4 x}-\frac {3 i b d^2 e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{8 \left (-i d x^3\right )^{2/3}}+\frac {3 i b d^2 e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{8 \left (i d x^3\right )^{2/3}}-\frac {b \sin \left (c+d x^3\right )}{4 x^4} \] Output:
-1/4*a/x^4-3/4*b*d*cos(d*x^3+c)/x-3/8*I*b*d^2*exp(I*c)*x^2*GAMMA(2/3,-I*d* x^3)/(-I*d*x^3)^(2/3)+3/8*I*b*d^2*x^2*GAMMA(2/3,I*d*x^3)/exp(I*c)/(I*d*x^3 )^(2/3)-1/4*b*sin(d*x^3+c)/x^4
Time = 0.50 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\frac {3 b d^2 x^6 \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (-i \cos (c)+\sin (c))+3 b d^2 x^6 \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (i \cos (c)+\sin (c))-2 \left (d^2 x^6\right )^{2/3} \left (a+3 b d x^3 \cos \left (c+d x^3\right )+b \sin \left (c+d x^3\right )\right )}{8 x^4 \left (d^2 x^6\right )^{2/3}} \] Input:
Integrate[(a + b*Sin[c + d*x^3])/x^5,x]
Output:
(3*b*d^2*x^6*(I*d*x^3)^(2/3)*Gamma[2/3, (-I)*d*x^3]*((-I)*Cos[c] + Sin[c]) + 3*b*d^2*x^6*((-I)*d*x^3)^(2/3)*Gamma[2/3, I*d*x^3]*(I*Cos[c] + Sin[c]) - 2*(d^2*x^6)^(2/3)*(a + 3*b*d*x^3*Cos[c + d*x^3] + b*Sin[c + d*x^3]))/(8* x^4*(d^2*x^6)^(2/3))
Time = 0.28 (sec) , antiderivative size = 130, 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.
\(\displaystyle \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {a}{x^5}+\frac {b \sin \left (c+d x^3\right )}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a}{4 x^4}-\frac {3 i b e^{i c} d^2 x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{8 \left (-i d x^3\right )^{2/3}}+\frac {3 i b e^{-i c} d^2 x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{8 \left (i d x^3\right )^{2/3}}-\frac {3 b d \cos \left (c+d x^3\right )}{4 x}-\frac {b \sin \left (c+d x^3\right )}{4 x^4}\) |
Input:
Int[(a + b*Sin[c + d*x^3])/x^5,x]
Output:
-1/4*a/x^4 - (3*b*d*Cos[c + d*x^3])/(4*x) - (((3*I)/8)*b*d^2*E^(I*c)*x^2*G amma[2/3, (-I)*d*x^3])/((-I)*d*x^3)^(2/3) + (((3*I)/8)*b*d^2*x^2*Gamma[2/3 , I*d*x^3])/(E^(I*c)*(I*d*x^3)^(2/3)) - (b*Sin[c + d*x^3])/(4*x^4)
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]]
\[\int \frac {a +b \sin \left (d \,x^{3}+c \right )}{x^{5}}d x\]
Input:
int((a+b*sin(d*x^3+c))/x^5,x)
Output:
int((a+b*sin(d*x^3+c))/x^5,x)
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=-\frac {6 \, b d x^{3} \cos \left (d x^{3} + c\right ) - 3 \, {\left (b d x^{4} \cos \left (c\right ) - i \, b d x^{4} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 3 \, {\left (b d x^{4} \cos \left (c\right ) + i \, b d x^{4} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) + 2 \, b \sin \left (d x^{3} + c\right ) + 2 \, a}{8 \, x^{4}} \] Input:
integrate((a+b*sin(d*x^3+c))/x^5,x, algorithm="fricas")
Output:
-1/8*(6*b*d*x^3*cos(d*x^3 + c) - 3*(b*d*x^4*cos(c) - I*b*d*x^4*sin(c))*(I* d)^(1/3)*gamma(2/3, I*d*x^3) - 3*(b*d*x^4*cos(c) + I*b*d*x^4*sin(c))*(-I*d )^(1/3)*gamma(2/3, -I*d*x^3) + 2*b*sin(d*x^3 + c) + 2*a)/x^4
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\int \frac {a + b \sin {\left (c + d x^{3} \right )}}{x^{5}}\, dx \] Input:
integrate((a+b*sin(d*x**3+c))/x**5,x)
Output:
Integral((a + b*sin(c + d*x**3))/x**5, x)
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.70 \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} b d}{12 \, x} - \frac {a}{4 \, x^{4}} \] Input:
integrate((a+b*sin(d*x^3+c))/x^5,x, algorithm="maxima")
Output:
1/12*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(-4/3, I*d*x^3) + (sqrt(3) - I)*ga mma(-4/3, -I*d*x^3))*cos(c) - ((I*sqrt(3) - 1)*gamma(-4/3, I*d*x^3) + (-I* sqrt(3) - 1)*gamma(-4/3, -I*d*x^3))*sin(c))*b*d/x - 1/4*a/x^4
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\int { \frac {b \sin \left (d x^{3} + c\right ) + a}{x^{5}} \,d x } \] Input:
integrate((a+b*sin(d*x^3+c))/x^5,x, algorithm="giac")
Output:
integrate((b*sin(d*x^3 + c) + a)/x^5, x)
Timed out. \[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\int \frac {a+b\,\sin \left (d\,x^3+c\right )}{x^5} \,d x \] Input:
int((a + b*sin(c + d*x^3))/x^5,x)
Output:
int((a + b*sin(c + d*x^3))/x^5, x)
\[ \int \frac {a+b \sin \left (c+d x^3\right )}{x^5} \, dx=\frac {4 \left (\int \frac {\sin \left (d \,x^{3}+c \right )}{x^{5}}d x \right ) b \,x^{4}-a}{4 x^{4}} \] Input:
int((a+b*sin(d*x^3+c))/x^5,x)
Output:
(4*int(sin(c + d*x**3)/x**5,x)*b*x**4 - a)/(4*x**4)