3.1.76 \(\int \frac {\text {CosIntegral}(b x)}{x^3} \, dx\) [76]

Optimal. Leaf size=46 \[ -\frac {\cos (b x)}{4 x^2}-\frac {1}{4} b^2 \text {CosIntegral}(b x)-\frac {\text {CosIntegral}(b x)}{2 x^2}+\frac {b \sin (b x)}{4 x} \]

[Out]

-1/4*b^2*Ci(b*x)-1/2*Ci(b*x)/x^2-1/4*cos(b*x)/x^2+1/4*b*sin(b*x)/x

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6639, 12, 3378, 3383} \begin {gather*} -\frac {1}{4} b^2 \text {CosIntegral}(b x)-\frac {\text {CosIntegral}(b x)}{2 x^2}-\frac {\cos (b x)}{4 x^2}+\frac {b \sin (b x)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[CosIntegral[b*x]/x^3,x]

[Out]

-1/4*Cos[b*x]/x^2 - (b^2*CosIntegral[b*x])/4 - CosIntegral[b*x]/(2*x^2) + (b*Sin[b*x])/(4*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 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 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*} \int \frac {\text {Ci}(b x)}{x^3} \, dx &=-\frac {\text {Ci}(b x)}{2 x^2}+\frac {1}{2} b \int \frac {\cos (b x)}{b x^3} \, dx\\ &=-\frac {\text {Ci}(b x)}{2 x^2}+\frac {1}{2} \int \frac {\cos (b x)}{x^3} \, dx\\ &=-\frac {\cos (b x)}{4 x^2}-\frac {\text {Ci}(b x)}{2 x^2}-\frac {1}{4} b \int \frac {\sin (b x)}{x^2} \, dx\\ &=-\frac {\cos (b x)}{4 x^2}-\frac {\text {Ci}(b x)}{2 x^2}+\frac {b \sin (b x)}{4 x}-\frac {1}{4} b^2 \int \frac {\cos (b x)}{x} \, dx\\ &=-\frac {\cos (b x)}{4 x^2}-\frac {1}{4} b^2 \text {Ci}(b x)-\frac {\text {Ci}(b x)}{2 x^2}+\frac {b \sin (b x)}{4 x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 46, normalized size = 1.00 \begin {gather*} -\frac {\cos (b x)}{4 x^2}-\frac {1}{4} b^2 \text {CosIntegral}(b x)-\frac {\text {CosIntegral}(b x)}{2 x^2}+\frac {b \sin (b x)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[CosIntegral[b*x]/x^3,x]

[Out]

-1/4*Cos[b*x]/x^2 - (b^2*CosIntegral[b*x])/4 - CosIntegral[b*x]/(2*x^2) + (b*Sin[b*x])/(4*x)

________________________________________________________________________________________

Maple [A]
time = 0.22, size = 48, normalized size = 1.04

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\cosineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{4 b^{2} x^{2}}+\frac {\sin \left (b x \right )}{4 b x}-\frac {\cosineIntegral \left (b x \right )}{4}\right )\)	\(48\)
default	\(b^{2} \left (-\frac {\cosineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{4 b^{2} x^{2}}+\frac {\sin \left (b x \right )}{4 b x}-\frac {\cosineIntegral \left (b x \right )}{4}\right )\)	\(48\)
meijerg	\(\frac {\sqrt {\pi }\, b^{2} \left (\frac {-8 b^{2} x^{2}+4}{\sqrt {\pi }\, b^{2} x^{2}}+\frac {4 \left (3 b^{2} x^{2}+6\right ) \gamma }{3 \sqrt {\pi }\, b^{2} x^{2}}+\frac {4 \left (3 b^{2} x^{2}+6\right ) \ln \left (2\right )}{3 \sqrt {\pi }\, b^{2} x^{2}}+\frac {4 \left (3 b^{2} x^{2}+6\right ) \ln \left (\frac {b x}{2}\right )}{3 \sqrt {\pi }\, b^{2} x^{2}}-\frac {4 \cos \left (b x \right )}{\sqrt {\pi }\, b^{2} x^{2}}+\frac {4 \sin \left (b x \right )}{\sqrt {\pi }\, b x}-\frac {4 \left (3 b^{2} x^{2}+6\right ) \cosineIntegral \left (b x \right )}{3 \sqrt {\pi }\, b^{2} x^{2}}-\frac {4 \left (1+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right )}{\sqrt {\pi }\, x^{2} b^{2}}-\frac {2 \left (2 \gamma -4+2 \ln \left (x \right )+2 \ln \left (b \right )\right )}{\sqrt {\pi }}\right )}{16}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(Ci(b*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

b^2*(-1/2*Ci(b*x)/b^2/x^2-1/4*cos(b*x)/b^2/x^2+1/4*sin(b*x)/b/x-1/4*Ci(b*x))

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.53, size = 61, normalized size = 1.33 \begin {gather*} -\frac {\sqrt {\frac {1}{2}} \sqrt {\pi x^{2}} {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{2}}{16 \, x} - \frac {\operatorname {C}\left (b x\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="maxima")

[Out]

-1/16*sqrt(1/2)*sqrt(pi*x^2)*((I + 1)*sqrt(2)*gamma(-1/2, 1/2*I*pi*b^2*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -1/2
*I*pi*b^2*x^2))*b^2/x - 1/2*fresnel_cos(b*x)/x^2

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 42, normalized size = 0.91 \begin {gather*} -\frac {\pi \sqrt {b^{2}} b x^{2} \operatorname {S}\left (\sqrt {b^{2}} x\right ) + b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \operatorname {C}\left (b x\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="fricas")

[Out]

-1/2*(pi*sqrt(b^2)*b*x^2*fresnel_sin(sqrt(b^2)*x) + b*x*cos(1/2*pi*b^2*x^2) + fresnel_cos(b*x))/x^2

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (39) = 78\).
time = 1.50, size = 87, normalized size = 1.89 \begin {gather*} \frac {b^{2} \log {\left (b x \right )}}{4} - \frac {b^{2} \log {\left (b^{2} x^{2} \right )}}{8} - \frac {b^{2} \operatorname {Ci}{\left (b x \right )}}{4} + \frac {b \sin {\left (b x \right )}}{4 x} + \frac {\log {\left (b x \right )}}{2 x^{2}} - \frac {\log {\left (b^{2} x^{2} \right )}}{4 x^{2}} - \frac {\cos {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Ci}{\left (b x \right )}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Ci(b*x)/x**3,x)

[Out]

b**2*log(b*x)/4 - b**2*log(b**2*x**2)/8 - b**2*Ci(b*x)/4 + b*sin(b*x)/(4*x) + log(b*x)/(2*x**2) - log(b**2*x**
2)/(4*x**2) - cos(b*x)/(4*x**2) - Ci(b*x)/(2*x**2)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^3, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\frac {\frac {\cos \left (b\,x\right )}{2}-\frac {b\,x\,\sin \left (b\,x\right )}{2}}{2\,x^2}-\frac {b^2\,\mathrm {cosint}\left (b\,x\right )}{4}-\frac {\mathrm {cosint}\left (b\,x\right )}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosint(b*x)/x^3,x)

[Out]

- (cos(b*x)/2 - (b*x*sin(b*x))/2)/(2*x^2) - (b^2*cosint(b*x))/4 - cosint(b*x)/(2*x^2)

________________________________________________________________________________________