3.1.87 \(\int x^2 \text {Chi}(a+b x) \, dx\) [87]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6668, 6874, 2717, 3377, 2718, 3382} \begin {gather*} \frac {a^3 \text {Chi}(a+b x)}{3 b^3}-\frac {a^2 \sinh (a+b x)}{3 b^3}-\frac {2 \sinh (a+b x)}{3 b^3}-\frac {a \cosh (a+b x)}{3 b^3}+\frac {a x \sinh (a+b x)}{3 b^2}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*CoshIntegral[a + b*x],x]

[Out]

-1/3*(a*Cosh[a + b*x])/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 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^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*}

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Mathematica [A]
time = 0.13, size = 64, normalized size = 0.54 \begin {gather*} -\frac {(a-2 b x) \cosh (a+b x)-\left (a^3+b^3 x^3\right ) \text {Chi}(a+b x)+\left (2+a^2-a b x+b^2 x^2\right ) \sinh (a+b x)}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((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])/b^3

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Maple [A]
time = 0.25, size = 101, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\hyperbolicCosineIntegral \left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \hyperbolicCosineIntegral \left (b x +a \right )}{3}-a^{2} \sinh \left (b x +a \right )+a \left (\left (b x +a \right ) \sinh \left (b x +a \right )-\cosh \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right )}{3}+\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )}{3}-\frac {2 \sinh \left (b x +a \right )}{3}}{b^{3}}\)	\(101\)
default	\(\frac {\frac {\hyperbolicCosineIntegral \left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \hyperbolicCosineIntegral \left (b x +a \right )}{3}-a^{2} \sinh \left (b x +a \right )+a \left (\left (b x +a \right ) \sinh \left (b x +a \right )-\cosh \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right )}{3}+\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )}{3}-\frac {2 \sinh \left (b x +a \right )}{3}}{b^{3}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2*Chi(b*x+a),x, algorithm="maxima")

[Out]

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

[Out]

integral(x^2*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^{2} \operatorname {Chi}\left (a + b x\right )\, dx \end {gather*}

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)

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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^2*Chi(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*Chi(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coshint(a + b*x),x)

[Out]

int(x^2*coshint(a + b*x), x)

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