3.1.88 \(\int x \text {Chi}(a+b x) \, dx\) [88]

Optimal. Leaf size=71 \[ \frac {\cosh (a+b x)}{2 b^2}-\frac {a^2 \text {Chi}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {Chi}(a+b x)+\frac {a \sinh (a+b x)}{2 b^2}-\frac {x \sinh (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6668, 6874, 2717, 3377, 2718, 3382} \begin {gather*} -\frac {a^2 \text {Chi}(a+b x)}{2 b^2}+\frac {a \sinh (a+b x)}{2 b^2}+\frac {\cosh (a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {Chi}(a+b x)-\frac {x \sinh (a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*CoshIntegral[a + b*x],x]

[Out]

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

Rule 2717

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 47, normalized size = 0.66 \begin {gather*} \frac {\cosh (a+b x)+\left (-a^2+b^2 x^2\right ) \text {Chi}(a+b x)+(a-b x) \sinh (a+b x)}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*CoshIntegral[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.27, size = 60, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\hyperbolicCosineIntegral \left (b x +a \right ) \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )+a \sinh \left (b x +a \right )-\frac {\left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\cosh \left (b x +a \right )}{2}}{b^{2}}\)	\(60\)
default	\(\frac {\hyperbolicCosineIntegral \left (b x +a \right ) \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )+a \sinh \left (b x +a \right )-\frac {\left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\cosh \left (b x +a \right )}{2}}{b^{2}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Chi(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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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(x*Chi(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*Chi(b*x + a), 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(x*Chi(b*x+a),x, algorithm="fricas")

[Out]

integral(x*cosh_integral(b*x + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {Chi}\left (a + b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Chi(b*x+a),x)

[Out]

Integral(x*Chi(a + b*x), x)

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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(x*Chi(b*x+a),x, algorithm="giac")

[Out]

integrate(x*Chi(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \frac {x^2\,\mathrm {coshint}\left (a+b\,x\right )}{2}+\frac {\frac {{\mathrm {e}}^{-a-b\,x}\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-a+a\,{\mathrm {e}}^{2\,a+2\,b\,x}-2\,a^2\,\mathrm {coshint}\left (a+b\,x\right )\,{\mathrm {e}}^{a+b\,x}+1\right )}{4}+\frac {b\,{\mathrm {e}}^{-a-b\,x}\,\left (x-x\,{\mathrm {e}}^{2\,a+2\,b\,x}\right )}{4}}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coshint(a + b*x),x)

[Out]

(x^2*coshint(a + b*x))/2 + ((exp(- a - b*x)*(exp(2*a + 2*b*x) - a + a*exp(2*a + 2*b*x) - 2*a^2*coshint(a + b*x
)*exp(a + b*x) + 1))/4 + (b*exp(- a - b*x)*(x - x*exp(2*a + 2*b*x)))/4)/b^2

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