3.87 \(\int x^2 \text{Chi}(a+b x) \, dx\)

Optimal. Leaf size=118 \[ \frac{a^3 \text{Chi}(a+b x)}{3 b^3}-\frac{a^2 \sinh (a+b x)}{3 b^3}+\frac{a x \sinh (a+b x)}{3 b^2}-\frac{2 \sinh (a+b x)}{3 b^3}-\frac{a \cosh (a+b x)}{3 b^3}+\frac{2 x \cosh (a+b x)}{3 b^2}+\frac{1}{3} x^3 \text{Chi}(a+b x)-\frac{x^2 \sinh (a+b x)}{3 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.280358, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6533, 6742, 2637, 3296, 2638, 3301} \[ \frac{a^3 \text{Chi}(a+b x)}{3 b^3}-\frac{a^2 \sinh (a+b x)}{3 b^3}+\frac{a x \sinh (a+b x)}{3 b^2}-\frac{2 \sinh (a+b x)}{3 b^3}-\frac{a \cosh (a+b x)}{3 b^3}+\frac{2 x \cosh (a+b x)}{3 b^2}+\frac{1}{3} x^3 \text{Chi}(a+b x)-\frac{x^2 \sinh (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*CoshIntegral[a + b*x],x]

[Out]

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

Rule 6533

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 6742

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

Rule 2637

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

Rule 3296

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 2638

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

Rule 3301

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]

Rubi steps

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

Mathematica [A]  time = 0.177024, size = 64, normalized size = 0.54 \[ -\frac{-\left (a^3+b^3 x^3\right ) \text{Chi}(a+b x)+\left (a^2-a b x+b^2 x^2+2\right ) \sinh (a+b x)+(a-2 b x) \cosh (a+b x)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.048, size = 101, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it Chi} \left ( bx+a \right ) }{3}}-{\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) }{3}}+{\frac{ \left ( 2\,bx+2\,a \right ) \cosh \left ( bx+a \right ) }{3}}-{\frac{2\,\sinh \left ( bx+a \right ) }{3}}+a \left ( \left ( bx+a \right ) \sinh \left ( bx+a \right ) -\cosh \left ( bx+a \right ) \right ) -{a}^{2}\sinh \left ( bx+a \right ) +{\frac{{a}^{3}{\it Chi} \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Chi(b*x+a),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Chi}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Chi(b*x+a),x, algorithm="maxima")

[Out]

integrate(x^2*Chi(b*x + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{Chi}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Chi(b*x+a),x, algorithm="fricas")

[Out]

integral(x^2*cosh_integral(b*x + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{Chi}\left (a + b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*Chi(a + b*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Chi}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Chi(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*Chi(b*x + a), x)