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

Optimal. Leaf size=46 \[ -\frac {\cosh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Chi}(b x)-\frac {\text {Chi}(b x)}{2 x^2}-\frac {b \sinh (b x)}{4 x} \]

[Out]

1/4*b^2*Chi(b*x)-1/2*Chi(b*x)/x^2-1/4*cosh(b*x)/x^2-1/4*b*sinh(b*x)/x

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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 = {6668, 12, 3378, 3382} \begin {gather*} \frac {1}{4} b^2 \text {Chi}(b x)-\frac {\text {Chi}(b x)}{2 x^2}-\frac {\cosh (b x)}{4 x^2}-\frac {b \sinh (b x)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Cosh[b*x]/x^2 + (b^2*CoshIntegral[b*x])/4 - CoshIntegral[b*x]/(2*x^2) - (b*Sinh[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 3382

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

Rule 6668

Int[CoshIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(CoshInte
gral[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Cosh[a + b*x]/(a + b*x)), x], x] /
; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\text {Chi}(b x)}{x^3} \, dx &=-\frac {\text {Chi}(b x)}{2 x^2}+\frac {1}{2} b \int \frac {\cosh (b x)}{b x^3} \, dx\\ &=-\frac {\text {Chi}(b x)}{2 x^2}+\frac {1}{2} \int \frac {\cosh (b x)}{x^3} \, dx\\ &=-\frac {\cosh (b x)}{4 x^2}-\frac {\text {Chi}(b x)}{2 x^2}+\frac {1}{4} b \int \frac {\sinh (b x)}{x^2} \, dx\\ &=-\frac {\cosh (b x)}{4 x^2}-\frac {\text {Chi}(b x)}{2 x^2}-\frac {b \sinh (b x)}{4 x}+\frac {1}{4} b^2 \int \frac {\cosh (b x)}{x} \, dx\\ &=-\frac {\cosh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Chi}(b x)-\frac {\text {Chi}(b x)}{2 x^2}-\frac {b \sinh (b x)}{4 x}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Cosh[b*x]/x^2 + (b^2*CoshIntegral[b*x])/4 - CoshIntegral[b*x]/(2*x^2) - (b*Sinh[b*x])/(4*x)

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Maple [A]
time = 0.21, size = 48, normalized size = 1.04

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\hyperbolicCosineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b x}+\frac {\hyperbolicCosineIntegral \left (b x \right )}{4}\right )\)	\(48\)
default	\(b^{2} \left (-\frac {\hyperbolicCosineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b x}+\frac {\hyperbolicCosineIntegral \left (b x \right )}{4}\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(-1/2*Chi(b*x)/b^2/x^2-1/4/b^2/x^2*cosh(b*x)-1/4*sinh(b*x)/b/x+1/4*Chi(b*x))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x)/x^3, x)

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Fricas [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(Chi(b*x)/x^3,x, algorithm="fricas")

[Out]

integral(cosh_integral(b*x)/x^3, x)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (39) = 78\).
time = 2.41, 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 {Chi}\left (b x\right )}{4} - \frac {b \sinh {\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 {\cosh {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Chi}\left (b x\right )}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**2*log(b*x)/4 + b**2*log(b**2*x**2)/8 + b**2*Chi(b*x)/4 - b*sinh(b*x)/(4*x) + log(b*x)/(2*x**2) - log(b**2*
x**2)/(4*x**2) - cosh(b*x)/(4*x**2) - Chi(b*x)/(2*x**2)

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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(Chi(b*x)/x^3,x, algorithm="giac")

[Out]

integrate(Chi(b*x)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*coshint(b*x))/4 - (coshint(b*x)/2 + cosh(b*x)/4 + (b*x*sinh(b*x))/4)/x^2

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