Integrand size = 16, antiderivative size = 74 \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {1}{4} b d^2 \operatorname {CosIntegral}\left (d x^2\right ) \sin (c)-\frac {b \sin \left (c+d x^2\right )}{4 x^4}-\frac {1}{4} b d^2 \cos (c) \text {Si}\left (d x^2\right ) \] Output:
-1/4*a/x^4-1/4*b*d*cos(d*x^2+c)/x^2-1/4*b*d^2*Ci(d*x^2)*sin(c)-1/4*b*sin(d *x^2+c)/x^4-1/4*b*d^2*cos(c)*Si(d*x^2)
Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {b \cos \left (d x^2\right ) \left (d x^2 \cos (c)+\sin (c)\right )}{4 x^4}+\frac {b \left (-\cos (c)+d x^2 \sin (c)\right ) \sin \left (d x^2\right )}{4 x^4}-\frac {1}{4} b d^2 \left (\operatorname {CosIntegral}\left (d x^2\right ) \sin (c)+\cos (c) \text {Si}\left (d x^2\right )\right ) \] Input:
Integrate[(a + b*Sin[c + d*x^2])/x^5,x]
Output:
-1/4*a/x^4 - (b*Cos[d*x^2]*(d*x^2*Cos[c] + Sin[c]))/(4*x^4) + (b*(-Cos[c] + d*x^2*Sin[c])*Sin[d*x^2])/(4*x^4) - (b*d^2*(CosIntegral[d*x^2]*Sin[c] + Cos[c]*SinIntegral[d*x^2]))/4
Time = 0.29 (sec) , antiderivative size = 74, 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.
\(\displaystyle \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {a}{x^5}+\frac {b \sin \left (c+d x^2\right )}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a}{4 x^4}-\frac {1}{4} b d^2 \sin (c) \operatorname {CosIntegral}\left (d x^2\right )-\frac {1}{4} b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {b \sin \left (c+d x^2\right )}{4 x^4}\) |
Input:
Int[(a + b*Sin[c + d*x^2])/x^5,x]
Output:
-1/4*a/x^4 - (b*d*Cos[c + d*x^2])/(4*x^2) - (b*d^2*CosIntegral[d*x^2]*Sin[ c])/4 - (b*Sin[c + d*x^2])/(4*x^4) - (b*d^2*Cos[c]*SinIntegral[d*x^2])/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]]
Time = 0.72 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {a}{4 x^{4}}+b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{4 x^{4}}+\frac {d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{2 x^{2}}-d \left (\frac {\cos \left (c \right ) \operatorname {Si}\left (d \,x^{2}\right )}{2}+\frac {\sin \left (c \right ) \operatorname {Ci}\left (d \,x^{2}\right )}{2}\right )\right )}{2}\right )\) | \(65\) |
parts | \(-\frac {a}{4 x^{4}}+b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{4 x^{4}}+\frac {d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{2 x^{2}}-d \left (\frac {\cos \left (c \right ) \operatorname {Si}\left (d \,x^{2}\right )}{2}+\frac {\sin \left (c \right ) \operatorname {Ci}\left (d \,x^{2}\right )}{2}\right )\right )}{2}\right )\) | \(65\) |
risch | \(-\frac {-\pi \,\operatorname {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-i c} b \,d^{2} x^{4}-i {\mathrm e}^{-i c} \operatorname {expIntegral}_{1}\left (-i d \,x^{2}\right ) b \,d^{2} x^{4}+i b \,d^{2} \operatorname {expIntegral}_{1}\left (-i d \,x^{2}\right ) {\mathrm e}^{i c} x^{4}+2 \,\operatorname {Si}\left (d \,x^{2}\right ) {\mathrm e}^{-i c} b \,d^{2} x^{4}+2 x^{2} b d \cos \left (d \,x^{2}+c \right )+2 b \sin \left (d \,x^{2}+c \right )+2 a}{8 x^{4}}\) | \(125\) |
Input:
int((a+b*sin(d*x^2+c))/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*a/x^4+b*(-1/4/x^4*sin(d*x^2+c)+1/2*d*(-1/2/x^2*cos(d*x^2+c)-d*(1/2*co s(c)*Si(d*x^2)+1/2*sin(c)*Ci(d*x^2))))
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=-\frac {b d^{2} x^{4} \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) + b d^{2} x^{4} \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + b d x^{2} \cos \left (d x^{2} + c\right ) + b \sin \left (d x^{2} + c\right ) + a}{4 \, x^{4}} \] Input:
integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="fricas")
Output:
-1/4*(b*d^2*x^4*cos_integral(d*x^2)*sin(c) + b*d^2*x^4*cos(c)*sin_integral (d*x^2) + b*d*x^2*cos(d*x^2 + c) + b*sin(d*x^2 + c) + a)/x^4
\[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=\int \frac {a + b \sin {\left (c + d x^{2} \right )}}{x^{5}}\, dx \] Input:
integrate((a+b*sin(d*x**2+c))/x**5,x)
Output:
Integral((a + b*sin(c + d*x**2))/x**5, x)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=\frac {1}{4} \, {\left ({\left (i \, \Gamma \left (-2, i \, d x^{2}\right ) - i \, \Gamma \left (-2, -i \, d x^{2}\right )\right )} \cos \left (c\right ) + {\left (\Gamma \left (-2, i \, d x^{2}\right ) + \Gamma \left (-2, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} b d^{2} - \frac {a}{4 \, x^{4}} \] Input:
integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="maxima")
Output:
1/4*((I*gamma(-2, I*d*x^2) - I*gamma(-2, -I*d*x^2))*cos(c) + (gamma(-2, I* d*x^2) + gamma(-2, -I*d*x^2))*sin(c))*b*d^2 - 1/4*a/x^4
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (64) = 128\).
Time = 0.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.76 \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=-\frac {{\left (d x^{2} + c\right )}^{2} b d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) - 2 \, {\left (d x^{2} + c\right )} b c d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) + b c^{2} d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) + {\left (d x^{2} + c\right )}^{2} b d^{3} \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) - 2 \, {\left (d x^{2} + c\right )} b c d^{3} \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + b c^{2} d^{3} \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + {\left (d x^{2} + c\right )} b d^{3} \cos \left (d x^{2} + c\right ) - b c d^{3} \cos \left (d x^{2} + c\right ) + b d^{3} \sin \left (d x^{2} + c\right ) + a d^{3}}{4 \, {\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c + c^{2}\right )} d} \] Input:
integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="giac")
Output:
-1/4*((d*x^2 + c)^2*b*d^3*cos_integral(d*x^2)*sin(c) - 2*(d*x^2 + c)*b*c*d ^3*cos_integral(d*x^2)*sin(c) + b*c^2*d^3*cos_integral(d*x^2)*sin(c) + (d* x^2 + c)^2*b*d^3*cos(c)*sin_integral(d*x^2) - 2*(d*x^2 + c)*b*c*d^3*cos(c) *sin_integral(d*x^2) + b*c^2*d^3*cos(c)*sin_integral(d*x^2) + (d*x^2 + c)* b*d^3*cos(d*x^2 + c) - b*c*d^3*cos(d*x^2 + c) + b*d^3*sin(d*x^2 + c) + a*d ^3)/(((d*x^2 + c)^2 - 2*(d*x^2 + c)*c + c^2)*d)
Timed out. \[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=\int \frac {a+b\,\sin \left (d\,x^2+c\right )}{x^5} \,d x \] Input:
int((a + b*sin(c + d*x^2))/x^5,x)
Output:
int((a + b*sin(c + d*x^2))/x^5, x)
\[ \int \frac {a+b \sin \left (c+d x^2\right )}{x^5} \, dx=\frac {-\cos \left (d \,x^{2}+c \right ) b d \,x^{2}-2 \left (\int \frac {\sin \left (d \,x^{2}+c \right )}{x}d x \right ) b \,d^{2} x^{4}-\sin \left (d \,x^{2}+c \right ) b -a}{4 x^{4}} \] Input:
int((a+b*sin(d*x^2+c))/x^5,x)
Output:
( - cos(c + d*x**2)*b*d*x**2 - 2*int(sin(c + d*x**2)/x,x)*b*d**2*x**4 - si n(c + d*x**2)*b - a)/(4*x**4)