3.1.71 \(\int x^2 \text {Chi}(b x) \, dx\) [71]

Optimal. Leaf size=49 \[ \frac {2 x \cosh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Chi}(b x)-\frac {2 \sinh (b x)}{3 b^3}-\frac {x^2 \sinh (b x)}{3 b} \]

[Out]

1/3*x^3*Chi(b*x)+2/3*x*cosh(b*x)/b^2-2/3*sinh(b*x)/b^3-1/3*x^2*sinh(b*x)/b

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 49, 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, 3377, 2717} \begin {gather*} -\frac {2 \sinh (b x)}{3 b^3}+\frac {2 x \cosh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Chi}(b x)-\frac {x^2 \sinh (b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*CoshIntegral[b*x],x]

[Out]

(2*x*Cosh[b*x])/(3*b^2) + (x^3*CoshIntegral[b*x])/3 - (2*Sinh[b*x])/(3*b^3) - (x^2*Sinh[b*x])/(3*b)

Rule 12

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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 x^2 \text {Chi}(b x) \, dx &=\frac {1}{3} x^3 \text {Chi}(b x)-\frac {1}{3} b \int \frac {x^2 \cosh (b x)}{b} \, dx\\ &=\frac {1}{3} x^3 \text {Chi}(b x)-\frac {1}{3} \int x^2 \cosh (b x) \, dx\\ &=\frac {1}{3} x^3 \text {Chi}(b x)-\frac {x^2 \sinh (b x)}{3 b}+\frac {2 \int x \sinh (b x) \, dx}{3 b}\\ &=\frac {2 x \cosh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Chi}(b x)-\frac {x^2 \sinh (b x)}{3 b}-\frac {2 \int \cosh (b x) \, dx}{3 b^2}\\ &=\frac {2 x \cosh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Chi}(b x)-\frac {2 \sinh (b x)}{3 b^3}-\frac {x^2 \sinh (b x)}{3 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 44, normalized size = 0.90 \begin {gather*} \frac {2 x \cosh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Chi}(b x)-\frac {\left (2+b^2 x^2\right ) \sinh (b x)}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*CoshIntegral[b*x],x]

[Out]

(2*x*Cosh[b*x])/(3*b^2) + (x^3*CoshIntegral[b*x])/3 - ((2 + b^2*x^2)*Sinh[b*x])/(3*b^3)

________________________________________________________________________________________

Maple [A]
time = 0.26, size = 44, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\frac {b^{3} x^{3} \hyperbolicCosineIntegral \left (b x \right )}{3}-\frac {b^{2} x^{2} \sinh \left (b x \right )}{3}+\frac {2 b x \cosh \left (b x \right )}{3}-\frac {2 \sinh \left (b x \right )}{3}}{b^{3}}\)	\(44\)
default	\(\frac {\frac {b^{3} x^{3} \hyperbolicCosineIntegral \left (b x \right )}{3}-\frac {b^{2} x^{2} \sinh \left (b x \right )}{3}+\frac {2 b x \cosh \left (b x \right )}{3}-\frac {2 \sinh \left (b x \right )}{3}}{b^{3}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*b^3*x^3*Chi(b*x)-1/3*b^2*x^2*sinh(b*x)+2/3*b*x*cosh(b*x)-2/3*sinh(b*x))

________________________________________________________________________________________

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(x^2*Chi(b*x),x, algorithm="maxima")

[Out]

integrate(x^2*Chi(b*x), x)

________________________________________________________________________________________

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(x^2*Chi(b*x),x, algorithm="fricas")

[Out]

integral(x^2*cosh_integral(b*x), x)

________________________________________________________________________________________

Sympy [A]
time = 1.22, size = 70, normalized size = 1.43 \begin {gather*} - \frac {x^{3} \log {\left (b x \right )}}{3} + \frac {x^{3} \log {\left (b^{2} x^{2} \right )}}{6} + \frac {x^{3} \operatorname {Chi}\left (b x\right )}{3} - \frac {x^{2} \sinh {\left (b x \right )}}{3 b} + \frac {2 x \cosh {\left (b x \right )}}{3 b^{2}} - \frac {2 \sinh {\left (b x \right )}}{3 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Chi(b*x),x)

[Out]

-x**3*log(b*x)/3 + x**3*log(b**2*x**2)/6 + x**3*Chi(b*x)/3 - x**2*sinh(b*x)/(3*b) + 2*x*cosh(b*x)/(3*b**2) - 2
*sinh(b*x)/(3*b**3)

________________________________________________________________________________________

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(x^2*Chi(b*x),x, algorithm="giac")

[Out]

integrate(x^2*Chi(b*x), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \frac {x^3\,\mathrm {coshint}\left (b\,x\right )}{3}-\frac {\frac {2\,\mathrm {sinh}\left (b\,x\right )}{3}+\frac {b^2\,x^2\,\mathrm {sinh}\left (b\,x\right )}{3}-\frac {2\,b\,x\,\mathrm {cosh}\left (b\,x\right )}{3}}{b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coshint(b*x),x)

[Out]

(x^3*coshint(b*x))/3 - ((2*sinh(b*x))/3 + (b^2*x^2*sinh(b*x))/3 - (2*b*x*cosh(b*x))/3)/b^3

________________________________________________________________________________________