3.1.81 \(\int \text {Chi}(b x)^2 \, dx\) [81]

Optimal. Leaf size=31 \[ x \text {Chi}(b x)^2-\frac {2 \text {Chi}(b x) \sinh (b x)}{b}+\frac {\text {Shi}(2 b x)}{b} \]

[Out]

x*Chi(b*x)^2+Shi(2*b*x)/b-2*Chi(b*x)*sinh(b*x)/b

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Rubi [A]
time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6670, 6676, 12, 5556, 3379} \begin {gather*} x \text {Chi}(b x)^2-\frac {2 \text {Chi}(b x) \sinh (b x)}{b}+\frac {\text {Shi}(2 b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*CoshIntegral[b*x]^2 - (2*CoshIntegral[b*x]*Sinh[b*x])/b + SinhIntegral[2*b*x]/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 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 6670

Int[CoshIntegral[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(CoshIntegral[a + b*x]^2/b), x] - Dist[2,
Int[Cosh[a + b*x]*CoshIntegral[a + b*x], x], x] /; FreeQ[{a, b}, x]

Rule 6676

Int[Cosh[(a_.) + (b_.)*(x_)]*CoshIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sinh[a + b*x]*(CoshIntegral[c
+ d*x]/b), x] - Dist[d/b, Int[Sinh[a + b*x]*(Cosh[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 31, normalized size = 1.00 \begin {gather*} x \text {Chi}(b x)^2-\frac {2 \text {Chi}(b x) \sinh (b x)}{b}+\frac {\text {Shi}(2 b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 30, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {\hyperbolicCosineIntegral \left (b x \right )^{2} b x -2 \hyperbolicCosineIntegral \left (b x \right ) \sinh \left (b x \right )+\hyperbolicSineIntegral \left (2 b x \right )}{b}\)	\(30\)
default	\(\frac {\hyperbolicCosineIntegral \left (b x \right )^{2} b x -2 \hyperbolicCosineIntegral \left (b x \right ) \sinh \left (b x \right )+\hyperbolicSineIntegral \left (2 b x \right )}{b}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(Chi(b*x)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral(cosh_integral(b*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {Chi}^{2}\left (b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Chi(b*x)**2, 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(Chi(b*x)^2,x, algorithm="giac")

[Out]

integrate(Chi(b*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int {\mathrm {coshint}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(coshint(b*x)^2, x)

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