\(\int \frac {\cos (a+b x^2)}{x^3} \, dx\) [7]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {\cos \left (a+b x^2\right )}{2 x^2}-\frac {1}{2} b \operatorname {CosIntegral}\left (b x^2\right ) \sin (a)-\frac {1}{2} b \cos (a) \text {Si}\left (b x^2\right ) \]

[Out]

-1/2*cos(b*x^2+a)/x^2-1/2*b*cos(a)*Si(b*x^2)-1/2*b*Ci(b*x^2)*sin(a)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3461, 3378, 3384, 3380, 3383} \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{2} b \sin (a) \operatorname {CosIntegral}\left (b x^2\right )-\frac {1}{2} b \cos (a) \text {Si}\left (b x^2\right )-\frac {\cos \left (a+b x^2\right )}{2 x^2} \]

[In]

Int[Cos[a + b*x^2]/x^3,x]

[Out]

-1/2*Cos[a + b*x^2]/x^2 - (b*CosIntegral[b*x^2]*Sin[a])/2 - (b*Cos[a]*SinIntegral[b*x^2])/2

Rule 3378

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] - Dist[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]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[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]

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\cos (a+b x)}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {\cos \left (a+b x^2\right )}{2 x^2}-\frac {1}{2} b \text {Subst}\left (\int \frac {\sin (a+b x)}{x} \, dx,x,x^2\right ) \\ & = -\frac {\cos \left (a+b x^2\right )}{2 x^2}-\frac {1}{2} (b \cos (a)) \text {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,x^2\right )-\frac {1}{2} (b \sin (a)) \text {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,x^2\right ) \\ & = -\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 ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {\cos \left (a+b x^2\right )+b x^2 \operatorname {CosIntegral}\left (b x^2\right ) \sin (a)+b x^2 \cos (a) \text {Si}\left (b x^2\right )}{2 x^2} \]

[In]

Integrate[Cos[a + b*x^2]/x^3,x]

[Out]

-1/2*(Cos[a + b*x^2] + b*x^2*CosIntegral[b*x^2]*Sin[a] + b*x^2*Cos[a]*SinIntegral[b*x^2])/x^2

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {\cos \left (b \,x^{2}+a \right )}{2 x^{2}}-b \left (\frac {\cos \left (a \right ) \operatorname {Si}\left (b \,x^{2}\right )}{2}+\frac {\sin \left (a \right ) \operatorname {Ci}\left (b \,x^{2}\right )}{2}\right )\)	\(39\)
risch	\(\frac {{\mathrm e}^{-i a} \pi \,\operatorname {csgn}\left (b \,x^{2}\right ) b}{4}-\frac {{\mathrm e}^{-i a} \operatorname {Si}\left (b \,x^{2}\right ) b}{2}+\frac {i \operatorname {Ei}_{1}\left (-i b \,x^{2}\right ) {\mathrm e}^{-i a} b}{4}-\frac {i {\mathrm e}^{i a} b \,\operatorname {Ei}_{1}\left (-i b \,x^{2}\right )}{4}-\frac {\cos \left (b \,x^{2}+a \right )}{2 x^{2}}\)	\(80\)
meijerg	\(\frac {\cos \left (a \right ) \sqrt {\pi }\, \sqrt {b^{2}}\, \left (-\frac {4 b^{2} \cos \left (x^{2} \sqrt {b^{2}}\right )}{x^{2} \left (b^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x^{2} \sqrt {b^{2}}\right )}{\sqrt {\pi }}\right )}{8}-\frac {\sin \left (a \right ) \sqrt {\pi }\, b \left (\frac {4 \gamma -4+8 \ln \left (x \right )+4 \ln \left (b \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {b \,x^{2}}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{2} b}+\frac {4 \,\operatorname {Ci}\left (b \,x^{2}\right )}{\sqrt {\pi }}\right )}{8}\)	\(141\)

[In]

int(cos(b*x^2+a)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*cos(b*x^2+a)/x^2-b*(1/2*cos(a)*Si(b*x^2)+1/2*sin(a)*Ci(b*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {b x^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) + b x^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + \cos \left (b x^{2} + a\right )}{2 \, x^{2}} \]

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="fricas")

[Out]

-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

Sympy [F]

\[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cos {\left (a + b x^{2} \right )}}{x^{3}}\, dx \]

[In]

integrate(cos(b*x**2+a)/x**3,x)

[Out]

Integral(cos(a + b*x**2)/x**3, x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4} \, {\left ({\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} b \]

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="maxima")

[Out]

-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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (36) = 72\).

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=-\frac {{\left (b x^{2} + a\right )} b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) - a b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) + {\left (b x^{2} + a\right )} b^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) - a b^{2} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + b^{2} \cos \left (b x^{2} + a\right )}{2 \, b^{2} x^{2}} \]

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="giac")

[Out]

-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)*s
in_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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cos \left (b\,x^2+a\right )}{x^3} \,d x \]

[In]

int(cos(a + b*x^2)/x^3,x)

[Out]

int(cos(a + b*x^2)/x^3, x)