Integrand size = 12, antiderivative size = 42 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {\cos \left (a+b x^2\right )}{2 x^2}-\frac {1}{2} b \operatorname {CosIntegral}\left (b x^2\right ) \sin (a)-\frac {1}{2} b \cos (a) \text {Si}\left (b x^2\right ) \] Output:
-1/2*cos(b*x^2+a)/x^2-1/2*b*Ci(b*x^2)*sin(a)-1/2*b*cos(a)*Si(b*x^2)
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {\cos \left (a+b x^2\right )+b x^2 \operatorname {CosIntegral}\left (b x^2\right ) \sin (a)+b x^2 \cos (a) \text {Si}\left (b x^2\right )}{2 x^2} \] Input:
Integrate[Cos[a + b*x^2]/x^3,x]
Output:
-1/2*(Cos[a + b*x^2] + b*x^2*CosIntegral[b*x^2]*Sin[a] + b*x^2*Cos[a]*SinI ntegral[b*x^2])/x^2
Time = 0.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3861, 3042, 3778, 25, 3042, 3784, 3042, 3780, 3783}
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.
\(\displaystyle \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 3861 |
\(\displaystyle \frac {1}{2} \int \frac {\cos \left (b x^2+a\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sin \left (b x^2+a+\frac {\pi }{2}\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{2} \left (b \int -\frac {\sin \left (b x^2+a\right )}{x^2}dx^2-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-b \int \frac {\sin \left (b x^2+a\right )}{x^2}dx^2-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (-b \int \frac {\sin \left (b x^2+a\right )}{x^2}dx^2-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {1}{2} \left (-b \left (\sin (a) \int \frac {\cos \left (b x^2\right )}{x^2}dx^2+\cos (a) \int \frac {\sin \left (b x^2\right )}{x^2}dx^2\right )-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (-b \left (\sin (a) \int \frac {\sin \left (b x^2+\frac {\pi }{2}\right )}{x^2}dx^2+\cos (a) \int \frac {\sin \left (b x^2\right )}{x^2}dx^2\right )-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {1}{2} \left (-b \left (\sin (a) \int \frac {\sin \left (b x^2+\frac {\pi }{2}\right )}{x^2}dx^2+\cos (a) \text {Si}\left (b x^2\right )\right )-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {1}{2} \left (-b \left (\sin (a) \operatorname {CosIntegral}\left (b x^2\right )+\cos (a) \text {Si}\left (b x^2\right )\right )-\frac {\cos \left (a+b x^2\right )}{x^2}\right )\) |
Input:
Int[Cos[a + b*x^2]/x^3,x]
Output:
(-(Cos[a + b*x^2]/x^2) - b*(CosIntegral[b*x^2]*Sin[a] + Cos[a]*SinIntegral [b*x^2]))/2
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* f)/d] Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] && NeQ[d*e - c*f, 0]
Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
Time = 0.68 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {\cos \left (b \,x^{2}+a \right )}{2 x^{2}}-b \left (\frac {\cos \left (a \right ) \operatorname {Si}\left (b \,x^{2}\right )}{2}+\frac {\operatorname {Ci}\left (b \,x^{2}\right ) \sin \left (a \right )}{2}\right )\) | \(39\) |
risch | \(\frac {{\mathrm e}^{-i a} \pi \,\operatorname {csgn}\left (b \,x^{2}\right ) b}{4}-\frac {{\mathrm e}^{-i a} \operatorname {Si}\left (b \,x^{2}\right ) b}{2}+\frac {i \operatorname {expIntegral}_{1}\left (-i b \,x^{2}\right ) {\mathrm e}^{-i a} b}{4}-\frac {i {\mathrm e}^{i a} b \,\operatorname {expIntegral}_{1}\left (-i b \,x^{2}\right )}{4}-\frac {\cos \left (b \,x^{2}+a \right )}{2 x^{2}}\) | \(80\) |
meijerg | \(\frac {\cos \left (a \right ) \sqrt {\pi }\, \sqrt {b^{2}}\, \left (-\frac {4 b^{2} \cos \left (x^{2} \sqrt {b^{2}}\right )}{x^{2} \left (b^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x^{2} \sqrt {b^{2}}\right )}{\sqrt {\pi }}\right )}{8}-\frac {\sin \left (a \right ) \sqrt {\pi }\, b \left (\frac {4 \gamma -4+8 \ln \left (x \right )+4 \ln \left (b \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {b \,x^{2}}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, b \,x^{2}}+\frac {4 \,\operatorname {Ci}\left (b \,x^{2}\right )}{\sqrt {\pi }}\right )}{8}\) | \(141\) |
Input:
int(cos(b*x^2+a)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*cos(b*x^2+a)/x^2-b*(1/2*cos(a)*Si(b*x^2)+1/2*Ci(b*x^2)*sin(a))
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {b x^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) + b x^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + \cos \left (b x^{2} + a\right )}{2 \, x^{2}} \] Input:
integrate(cos(b*x^2+a)/x^3,x, algorithm="fricas")
Output:
-1/2*(b*x^2*cos_integral(b*x^2)*sin(a) + b*x^2*cos(a)*sin_integral(b*x^2) + cos(b*x^2 + a))/x^2
\[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cos {\left (a + b x^{2} \right )}}{x^{3}}\, dx \] Input:
integrate(cos(b*x**2+a)/x**3,x)
Output:
Integral(cos(a + b*x**2)/x**3, x)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4} \, {\left ({\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} b \] Input:
integrate(cos(b*x^2+a)/x^3,x, algorithm="maxima")
Output:
-1/4*((I*gamma(-1, I*b*x^2) - I*gamma(-1, -I*b*x^2))*cos(a) + (gamma(-1, I *b*x^2) + gamma(-1, -I*b*x^2))*sin(a))*b
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (36) = 72\).
Time = 0.39 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {{\left (b x^{2} + a\right )} b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) - a b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) + {\left (b x^{2} + a\right )} b^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) - a b^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + b^{2} \cos \left (b x^{2} + a\right )}{2 \, b^{2} x^{2}} \] Input:
integrate(cos(b*x^2+a)/x^3,x, algorithm="giac")
Output:
-1/2*((b*x^2 + a)*b^2*cos_integral(b*x^2)*sin(a) - a*b^2*cos_integral(b*x^ 2)*sin(a) + (b*x^2 + a)*b^2*cos(a)*sin_integral(b*x^2) - a*b^2*cos(a)*sin_ integral(b*x^2) + b^2*cos(b*x^2 + a))/(b^2*x^2)
Timed out. \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cos \left (b\,x^2+a\right )}{x^3} \,d x \] Input:
int(cos(a + b*x^2)/x^3,x)
Output:
int(cos(a + b*x^2)/x^3, x)
\[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=\frac {2 \left (\int \frac {\cos \left (b \,x^{2}+a \right )}{x^{3}}d x \right ) x^{2}+2 \left (\int \frac {1}{x^{3}}d x \right ) x^{2}+1}{2 x^{2}} \] Input:
int(cos(b*x^2+a)/x^3,x)
Output:
(2*int(cos(a + b*x**2)/x**3,x)*x**2 + 2*int(1/x**3,x)*x**2 + 1)/(2*x**2)