\(\int x^3 \text {Chi}(b x) \, dx\) [70]

Optimal result

Integrand size = 8, antiderivative size = 63 \[ \int x^3 \text {Chi}(b x) \, dx=\frac {3 \cosh (b x)}{2 b^4}+\frac {3 x^2 \cosh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {3 x \sinh (b x)}{2 b^3}-\frac {x^3 \sinh (b x)}{4 b} \]

[Out]

1/4*x^4*Chi(b*x)+3/2*cosh(b*x)/b^4+3/4*x^2*cosh(b*x)/b^2-3/2*x*sinh(b*x)/b^3-1/4*x^3*sinh(b*x)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6668, 12, 3377, 2718} \[ \int x^3 \text {Chi}(b x) \, dx=\frac {3 \cosh (b x)}{2 b^4}-\frac {3 x \sinh (b x)}{2 b^3}+\frac {3 x^2 \cosh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {x^3 \sinh (b x)}{4 b} \]

[In]

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

[Out]

(3*Cosh[b*x])/(2*b^4) + (3*x^2*Cosh[b*x])/(4*b^2) + (x^4*CoshIntegral[b*x])/4 - (3*x*Sinh[b*x])/(2*b^3) - (x^3
*Sinh[b*x])/(4*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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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*} \text {integral}& = \frac {1}{4} x^4 \text {Chi}(b x)-\frac {1}{4} b \int \frac {x^3 \cosh (b x)}{b} \, dx \\ & = \frac {1}{4} x^4 \text {Chi}(b x)-\frac {1}{4} \int x^3 \cosh (b x) \, dx \\ & = \frac {1}{4} x^4 \text {Chi}(b x)-\frac {x^3 \sinh (b x)}{4 b}+\frac {3 \int x^2 \sinh (b x) \, dx}{4 b} \\ & = \frac {3 x^2 \cosh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {x^3 \sinh (b x)}{4 b}-\frac {3 \int x \cosh (b x) \, dx}{2 b^2} \\ & = \frac {3 x^2 \cosh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {3 x \sinh (b x)}{2 b^3}-\frac {x^3 \sinh (b x)}{4 b}+\frac {3 \int \sinh (b x) \, dx}{2 b^3} \\ & = \frac {3 \cosh (b x)}{2 b^4}+\frac {3 x^2 \cosh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {3 x \sinh (b x)}{2 b^3}-\frac {x^3 \sinh (b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^3 \text {Chi}(b x) \, dx=\frac {3 \left (2+b^2 x^2\right ) \cosh (b x)}{4 b^4}+\frac {1}{4} x^4 \text {Chi}(b x)-\frac {x \left (6+b^2 x^2\right ) \sinh (b x)}{4 b^3} \]

[In]

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

[Out]

(3*(2 + b^2*x^2)*Cosh[b*x])/(4*b^4) + (x^4*CoshIntegral[b*x])/4 - (x*(6 + b^2*x^2)*Sinh[b*x])/(4*b^3)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86

method	result	size
parts	\(\frac {x^{4} \operatorname {Chi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sinh \left (b x \right )-3 b^{2} x^{2} \cosh \left (b x \right )+6 b x \sinh \left (b x \right )-6 \cosh \left (b x \right )}{4 b^{4}}\)	\(54\)
derivativedivides	\(\frac {\frac {b^{4} x^{4} \operatorname {Chi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sinh \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \cosh \left (b x \right )}{4}-\frac {3 b x \sinh \left (b x \right )}{2}+\frac {3 \cosh \left (b x \right )}{2}}{b^{4}}\)	\(56\)
default	\(\frac {\frac {b^{4} x^{4} \operatorname {Chi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sinh \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \cosh \left (b x \right )}{4}-\frac {3 b x \sinh \left (b x \right )}{2}+\frac {3 \cosh \left (b x \right )}{2}}{b^{4}}\)	\(56\)

[In]

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

[Out]

1/4*x^4*Chi(b*x)-1/4/b^4*(b^3*x^3*sinh(b*x)-3*b^2*x^2*cosh(b*x)+6*b*x*sinh(b*x)-6*cosh(b*x))

Fricas [F]

\[ \int x^3 \text {Chi}(b x) \, dx=\int { x^{3} {\rm Chi}\left (b x\right ) \,d x } \]

[In]

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

[Out]

integral(x^3*cosh_integral(b*x), x)

Sympy [A] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35 \[ \int x^3 \text {Chi}(b x) \, dx=- \frac {x^{4} \log {\left (b x \right )}}{4} + \frac {x^{4} \log {\left (b^{2} x^{2} \right )}}{8} + \frac {x^{4} \operatorname {Chi}\left (b x\right )}{4} - \frac {x^{3} \sinh {\left (b x \right )}}{4 b} + \frac {3 x^{2} \cosh {\left (b x \right )}}{4 b^{2}} - \frac {3 x \sinh {\left (b x \right )}}{2 b^{3}} + \frac {3 \cosh {\left (b x \right )}}{2 b^{4}} \]

[In]

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

[Out]

-x**4*log(b*x)/4 + x**4*log(b**2*x**2)/8 + x**4*Chi(b*x)/4 - x**3*sinh(b*x)/(4*b) + 3*x**2*cosh(b*x)/(4*b**2)
- 3*x*sinh(b*x)/(2*b**3) + 3*cosh(b*x)/(2*b**4)

Maxima [F]

\[ \int x^3 \text {Chi}(b x) \, dx=\int { x^{3} {\rm Chi}\left (b x\right ) \,d x } \]

[In]

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

[Out]

integrate(x^3*Chi(b*x), x)

Giac [F]

\[ \int x^3 \text {Chi}(b x) \, dx=\int { x^{3} {\rm Chi}\left (b x\right ) \,d x } \]

[In]

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

[Out]

integrate(x^3*Chi(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \text {Chi}(b x) \, dx=\int x^3\,\mathrm {coshint}\left (b\,x\right ) \,d x \]

[In]

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

[Out]

int(x^3*coshint(b*x), x)