3.76 \(\int \frac {\text {Ci}(b x)}{x^3} \, dx\)

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

[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

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Rubi [A]  time = 0.07, 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 = {6504, 12, 3297, 3302} \[ -\frac {1}{4} b^2 \text {CosIntegral}(b x)-\frac {\text {CosIntegral}(b x)}{2 x^2}-\frac {\cos (b x)}{4 x^2}+\frac {b \sin (b x)}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[b*x]/(4*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 3297

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 3302

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 6504

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*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.00 \[ -\frac {1}{4} b^2 \text {Ci}(b x)-\frac {\text {Ci}(b x)}{2 x^2}-\frac {\cos (b x)}{4 x^2}+\frac {b \sin (b x)}{4 x} \]

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)

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fricas [F]  time = 1.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {Ci}\left (b x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos_integral(b*x)/x^3, x)

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giac [A]  time = 0.37, size = 53, normalized size = 1.15 \[ -\frac {b^{2} x^{2} \operatorname {Ci}\left (b x\right ) + b^{2} x^{2} \operatorname {Ci}\left (-b x\right ) - 2 \, b x \sin \left (b x\right ) + 2 \, \cos \left (b x\right )}{8 \, x^{2}} - \frac {\operatorname {Ci}\left (b x\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2*x^2*cos_integral(b*x) + b^2*x^2*cos_integral(-b*x) - 2*b*x*sin(b*x) + 2*cos(b*x))/x^2 - 1/2*cos_inte
gral(b*x)/x^2

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maple [A]  time = 0.01, size = 48, normalized size = 1.04 \[ b^{2} \left (-\frac {\Ci \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{4 b^{2} x^{2}}+\frac {\sin \left (b x \right )}{4 b x}-\frac {\Ci \left (b x \right )}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Ci}\left (b x\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Ci(b*x)/x^3, x)

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

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)

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sympy [B]  time = 2.24, size = 87, normalized size = 1.89 \[ \frac {b^{2} \log {\left (b x \right )}}{4} - \frac {b^{2} \log {\left (b^{2} x^{2} \right )}}{8} - \frac {b^{2} \operatorname {Ci}{\left (b x \right )}}{4} + \frac {b \sin {\left (b x \right )}}{4 x} + \frac {\log {\left (b x \right )}}{2 x^{2}} - \frac {\log {\left (b^{2} x^{2} \right )}}{4 x^{2}} - \frac {\cos {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Ci}{\left (b x \right )}}{2 x^{2}} \]

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)

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