3.7 \(\int \frac{(a+b x) \cosh (c+d x)}{x^3} \, dx\)

Optimal. Leaf size=88 \[ \frac{1}{2} a d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a d^2 \sinh (c) \text{Shi}(d x)-\frac{a \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{2 x}+b d \sinh (c) \text{Chi}(d x)+b d \cosh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.28109, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3298, 3301} \[ \frac{1}{2} a d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a d^2 \sinh (c) \text{Shi}(d x)-\frac{a \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{2 x}+b d \sinh (c) \text{Chi}(d x)+b d \cosh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Cosh[c + d*x])/x^3,x]

[Out]

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

Rule 6742

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.155352, size = 78, normalized size = 0.89 \[ \frac{d x^2 \text{Chi}(d x) (a d \cosh (c)+2 b \sinh (c))+d x^2 \text{Shi}(d x) (a d \sinh (c)+2 b \cosh (c))-a d x \sinh (c+d x)-a \cosh (c+d x)-2 b x \cosh (c+d x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Cosh[c + d*x])/x^3,x]

[Out]

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

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Maple [A]  time = 0.047, size = 139, normalized size = 1.6 \begin{align*}{\frac{da{{\rm e}^{-dx-c}}}{4\,x}}-{\frac{a{{\rm e}^{-dx-c}}}{4\,{x}^{2}}}-{\frac{{d}^{2}a{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{4}}-{\frac{b{{\rm e}^{-dx-c}}}{2\,x}}+{\frac{db{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}-{\frac{a{{\rm e}^{dx+c}}}{4\,{x}^{2}}}-{\frac{ad{{\rm e}^{dx+c}}}{4\,x}}-{\frac{{d}^{2}a{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{4}}-{\frac{b{{\rm e}^{dx+c}}}{2\,x}}-{\frac{db{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*cosh(d*x+c)/x^3,x)

[Out]

1/4*d*a*exp(-d*x-c)/x-1/4*a*exp(-d*x-c)/x^2-1/4*d^2*a*exp(-c)*Ei(1,d*x)-1/2*b*exp(-d*x-c)/x+1/2*d*b*exp(-c)*Ei
(1,d*x)-1/4*a/x^2*exp(d*x+c)-1/4*d*a/x*exp(d*x+c)-1/4*d^2*a*exp(c)*Ei(1,-d*x)-1/2*b/x*exp(d*x+c)-1/2*d*b*exp(c
)*Ei(1,-d*x)

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Maxima [A]  time = 1.417, size = 89, normalized size = 1.01 \begin{align*} \frac{1}{4} \,{\left (a d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + a d e^{c} \Gamma \left (-1, -d x\right ) - 2 \, b{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b{\rm Ei}\left (d x\right ) e^{c}\right )} d - \frac{{\left (2 \, b x + a\right )} \cosh \left (d x + c\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d*e^(-c)*gamma(-1, d*x) + a*d*e^c*gamma(-1, -d*x) - 2*b*Ei(-d*x)*e^(-c) + 2*b*Ei(d*x)*e^c)*d - 1/2*(2*b
*x + a)*cosh(d*x + c)/x^2

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Fricas [A]  time = 1.99403, size = 278, normalized size = 3.16 \begin{align*} -\frac{2 \, a d x \sinh \left (d x + c\right ) + 2 \,{\left (2 \, b x + a\right )} \cosh \left (d x + c\right ) -{\left ({\left (a d^{2} + 2 \, b d\right )} x^{2}{\rm Ei}\left (d x\right ) +{\left (a d^{2} - 2 \, b d\right )} x^{2}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) -{\left ({\left (a d^{2} + 2 \, b d\right )} x^{2}{\rm Ei}\left (d x\right ) -{\left (a d^{2} - 2 \, b d\right )} x^{2}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*d*x*sinh(d*x + c) + 2*(2*b*x + a)*cosh(d*x + c) - ((a*d^2 + 2*b*d)*x^2*Ei(d*x) + (a*d^2 - 2*b*d)*x^2
*Ei(-d*x))*cosh(c) - ((a*d^2 + 2*b*d)*x^2*Ei(d*x) - (a*d^2 - 2*b*d)*x^2*Ei(-d*x))*sinh(c))/x^2

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c)/x**3,x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.15187, size = 181, normalized size = 2.06 \begin{align*} \frac{a d^{2} x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{2} x^{2}{\rm Ei}\left (d x\right ) e^{c} - 2 \, b d x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b d x^{2}{\rm Ei}\left (d x\right ) e^{c} - a d x e^{\left (d x + c\right )} + a d x e^{\left (-d x - c\right )} - 2 \, b x e^{\left (d x + c\right )} - 2 \, b x e^{\left (-d x - c\right )} - a e^{\left (d x + c\right )} - a e^{\left (-d x - c\right )}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^2*x^2*Ei(-d*x)*e^(-c) + a*d^2*x^2*Ei(d*x)*e^c - 2*b*d*x^2*Ei(-d*x)*e^(-c) + 2*b*d*x^2*Ei(d*x)*e^c - a
*d*x*e^(d*x + c) + a*d*x*e^(-d*x - c) - 2*b*x*e^(d*x + c) - 2*b*x*e^(-d*x - c) - a*e^(d*x + c) - a*e^(-d*x - c
))/x^2