3.92 \(\int \frac {\text {Chi}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=111 \[ \frac {b^2 \text {Chi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a^2}-\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Chi}(a+b x)}{2 x^2}-\frac {b \cosh (a+b x)}{2 a x} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6533, 6742, 3297, 3303, 3298, 3301} \[ \frac {b^2 \text {Chi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a^2}-\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Chi}(a+b x)}{2 x^2}-\frac {b \cosh (a+b x)}{2 a x} \]

Antiderivative was successfully verified.

[In]

Int[CoshIntegral[a + b*x]/x^3,x]

[Out]

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

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3298

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 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]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 80, normalized size = 0.72 \[ \frac {\left (b^2 x^2-a^2\right ) \text {Chi}(a+b x)+b^2 x^2 (a \sinh (a)-\cosh (a)) \text {Chi}(b x)+b x (b x (a \cosh (a)-\sinh (a)) \text {Shi}(b x)-a \cosh (a+b x))}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[CoshIntegral[a + b*x]/x^3,x]

[Out]

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

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fricas [F]  time = 1.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {Chi}\left (b x + a\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cosh_integral(b*x + a)/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Chi}\left (b x + a\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x + a)/x^3, x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\Chi \left (b x +a \right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(Chi(b*x+a)/x^3,x)

[Out]

int(Chi(b*x+a)/x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Chi}\left (b x + a\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x + a)/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {coshint}\left (a+b\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coshint(a + b*x)/x^3,x)

[Out]

int(coshint(a + b*x)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Chi}\left (a + b x\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Chi(b*x+a)/x**3,x)

[Out]

Integral(Chi(a + b*x)/x**3, x)

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