\(\int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx\) [75]

Optimal result

Integrand size = 8, antiderivative size = 26 \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=-\frac {\cos (b x)}{x}-\frac {\operatorname {CosIntegral}(b x)}{x}-b \text {Si}(b x) \]

[Out]

-Ci(b*x)/x-cos(b*x)/x-b*Si(b*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6639, 12, 3378, 3380} \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=-\frac {\operatorname {CosIntegral}(b x)}{x}-b \text {Si}(b x)-\frac {\cos (b x)}{x} \]

[In]

Int[CosIntegral[b*x]/x^2,x]

[Out]

-(Cos[b*x]/x) - CosIntegral[b*x]/x - b*SinIntegral[b*x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 6639

Int[CosIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(CosIntegr
al[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Cos[a + b*x]/(a + b*x)), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {CosIntegral}(b x)}{x}+b \int \frac {\cos (b x)}{b x^2} \, dx \\ & = -\frac {\operatorname {CosIntegral}(b x)}{x}+\int \frac {\cos (b x)}{x^2} \, dx \\ & = -\frac {\cos (b x)}{x}-\frac {\operatorname {CosIntegral}(b x)}{x}-b \int \frac {\sin (b x)}{x} \, dx \\ & = -\frac {\cos (b x)}{x}-\frac {\operatorname {CosIntegral}(b x)}{x}-b \text {Si}(b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=-\frac {\cos (b x)}{x}-\frac {\operatorname {CosIntegral}(b x)}{x}-b \text {Si}(b x) \]

[In]

Integrate[CosIntegral[b*x]/x^2,x]

[Out]

-(Cos[b*x]/x) - CosIntegral[b*x]/x - b*SinIntegral[b*x]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23

method	result	size
parts	\(-\frac {\operatorname {Ci}\left (b x \right )}{x}+b \left (-\frac {\cos \left (b x \right )}{b x}-\operatorname {Si}\left (b x \right )\right )\)	\(32\)
derivativedivides	\(b \left (-\frac {\operatorname {Ci}\left (b x \right )}{b x}-\frac {\cos \left (b x \right )}{b x}-\operatorname {Si}\left (b x \right )\right )\)	\(34\)
default	\(b \left (-\frac {\operatorname {Ci}\left (b x \right )}{b x}-\frac {\cos \left (b x \right )}{b x}-\operatorname {Si}\left (b x \right )\right )\)	\(34\)
meijerg	\(\frac {b \sqrt {\pi }\, \left (-\frac {2 b x \operatorname {hypergeom}\left (\left [\frac {1}{2}, 1, 1\right ], \left [\frac {3}{2}, \frac {3}{2}, 2, 2\right ], -\frac {b^{2} x^{2}}{4}\right )}{\sqrt {\pi }}-\frac {4 \left (2+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right )}{\sqrt {\pi }\, x b}\right )}{8}\)	\(57\)

[In]

int(Ci(b*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-Ci(b*x)/x+b*(-cos(b*x)/b/x-Si(b*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=\frac {b x \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \operatorname {C}\left (b x\right )}{2 \, x} \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="fricas")

[Out]

1/2*(b*x*cos_integral(1/2*pi*b^2*x^2) - 2*fresnel_cos(b*x))/x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.62 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=- \frac {b^{2} x {{}_{3}F_{4}\left (\begin {matrix} \frac {1}{2}, 1, 1 \\ \frac {3}{2}, \frac {3}{2}, 2, 2 \end {matrix}\middle | {- \frac {b^{2} x^{2}}{4}} \right )}}{4} - \frac {\log {\left (b^{2} x^{2} \right )}}{2 x} - \frac {1}{x} - \frac {\gamma }{x} \]

[In]

integrate(Ci(b*x)/x**2,x)

[Out]

-b**2*x*hyper((1/2, 1, 1), (3/2, 3/2, 2, 2), -b**2*x**2/4)/4 - log(b**2*x**2)/(2*x) - 1/x - EulerGamma/x

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=\frac {1}{4} \, b {\left ({\rm Ei}\left (\frac {1}{2} i \, \pi b^{2} x^{2}\right ) + {\rm Ei}\left (-\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} - \frac {\operatorname {C}\left (b x\right )}{x} \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="maxima")

[Out]

1/4*b*(Ei(1/2*I*pi*b^2*x^2) + Ei(-1/2*I*pi*b^2*x^2)) - fresnel_cos(b*x)/x

Giac [F]

\[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {CosIntegral}(b x)}{x^2} \, dx=-b\,\mathrm {sinint}\left (b\,x\right )-\frac {\mathrm {cosint}\left (b\,x\right )}{x}-\frac {\cos \left (b\,x\right )}{x} \]

[In]

int(cosint(b*x)/x^2,x)

[Out]

- b*sinint(b*x) - cosint(b*x)/x - cos(b*x)/x