3.8 \(\int \frac{(a+b x) \cosh (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=132 \[ \frac{1}{6} a d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a d^3 \cosh (c) \text{Shi}(d x)-\frac{a d^2 \cosh (c+d x)}{6 x}-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{a \cosh (c+d x)}{3 x^3}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{2 x^2}-\frac{b d \sinh (c+d x)}{2 x} \]

[Out]

-(a*Cosh[c + d*x])/(3*x^3) - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(6*x) + (b*d^2*Cosh[c]*CoshInte
gral[d*x])/2 + (a*d^3*CoshIntegral[d*x]*Sinh[c])/6 - (a*d*Sinh[c + d*x])/(6*x^2) - (b*d*Sinh[c + d*x])/(2*x) +
(a*d^3*Cosh[c]*SinhIntegral[d*x])/6 + (b*d^2*Sinh[c]*SinhIntegral[d*x])/2

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Rubi [A]  time = 0.339799, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ \frac{1}{6} a d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a d^3 \cosh (c) \text{Shi}(d x)-\frac{a d^2 \cosh (c+d x)}{6 x}-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{a \cosh (c+d x)}{3 x^3}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{2 x^2}-\frac{b d \sinh (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Cosh[c + d*x])/x^4,x]

[Out]

-(a*Cosh[c + d*x])/(3*x^3) - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(6*x) + (b*d^2*Cosh[c]*CoshInte
gral[d*x])/2 + (a*d^3*CoshIntegral[d*x]*Sinh[c])/6 - (a*d*Sinh[c + d*x])/(6*x^2) - (b*d*Sinh[c + d*x])/(2*x) +
(a*d^3*Cosh[c]*SinhIntegral[d*x])/6 + (b*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^4} \, dx &=\int \left (\frac{a \cosh (c+d x)}{x^4}+\frac{b \cosh (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac{\cosh (c+d x)}{x^4} \, dx+b \int \frac{\cosh (c+d x)}{x^3} \, dx\\ &=-\frac{a \cosh (c+d x)}{3 x^3}-\frac{b \cosh (c+d x)}{2 x^2}+\frac{1}{3} (a d) \int \frac{\sinh (c+d x)}{x^3} \, dx+\frac{1}{2} (b d) \int \frac{\sinh (c+d x)}{x^2} \, dx\\ &=-\frac{a \cosh (c+d x)}{3 x^3}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{6} \left (a d^2\right ) \int \frac{\cosh (c+d x)}{x^2} \, dx+\frac{1}{2} \left (b d^2\right ) \int \frac{\cosh (c+d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{3 x^3}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{6 x}-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{6} \left (a d^3\right ) \int \frac{\sinh (c+d x)}{x} \, dx+\frac{1}{2} \left (b d^2 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\frac{1}{2} \left (b d^2 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{3 x^3}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{6 x}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)+\frac{1}{6} \left (a d^3 \cosh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx+\frac{1}{6} \left (a d^3 \sinh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{3 x^3}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{6 x}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{6} a d^3 \text{Chi}(d x) \sinh (c)-\frac{a d \sinh (c+d x)}{6 x^2}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{6} a d^3 \cosh (c) \text{Shi}(d x)+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.2723, size = 110, normalized size = 0.83 \[ -\frac{-d^2 x^3 \text{Chi}(d x) (a d \sinh (c)+3 b \cosh (c))-d^2 x^3 \text{Shi}(d x) (a d \cosh (c)+3 b \sinh (c))+a d^2 x^2 \cosh (c+d x)+a d x \sinh (c+d x)+2 a \cosh (c+d x)+3 b d x^2 \sinh (c+d x)+3 b x \cosh (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Cosh[c + d*x])/x^4,x]

[Out]

-(2*a*Cosh[c + d*x] + 3*b*x*Cosh[c + d*x] + a*d^2*x^2*Cosh[c + d*x] - d^2*x^3*CoshIntegral[d*x]*(3*b*Cosh[c] +
a*d*Sinh[c]) + a*d*x*Sinh[c + d*x] + 3*b*d*x^2*Sinh[c + d*x] - d^2*x^3*(a*d*Cosh[c] + 3*b*Sinh[c])*SinhIntegr
al[d*x])/(6*x^3)

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Maple [A]  time = 0.053, size = 205, normalized size = 1.6 \begin{align*} -{\frac{a{d}^{2}{{\rm e}^{-dx-c}}}{12\,x}}+{\frac{da{{\rm e}^{-dx-c}}}{12\,{x}^{2}}}-{\frac{a{{\rm e}^{-dx-c}}}{6\,{x}^{3}}}+{\frac{{d}^{3}a{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{12}}+{\frac{bd{{\rm e}^{-dx-c}}}{4\,x}}-{\frac{b{{\rm e}^{-dx-c}}}{4\,{x}^{2}}}-{\frac{{d}^{2}b{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{4}}-{\frac{a{{\rm e}^{dx+c}}}{6\,{x}^{3}}}-{\frac{ad{{\rm e}^{dx+c}}}{12\,{x}^{2}}}-{\frac{a{d}^{2}{{\rm e}^{dx+c}}}{12\,x}}-{\frac{{d}^{3}a{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{12}}-{\frac{b{{\rm e}^{dx+c}}}{4\,{x}^{2}}}-{\frac{bd{{\rm e}^{dx+c}}}{4\,x}}-{\frac{{d}^{2}b{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*cosh(d*x+c)/x^4,x)

[Out]

-1/12*d^2*a*exp(-d*x-c)/x+1/12*d*a*exp(-d*x-c)/x^2-1/6*a*exp(-d*x-c)/x^3+1/12*d^3*a*exp(-c)*Ei(1,d*x)+1/4*d*b*
exp(-d*x-c)/x-1/4*b*exp(-d*x-c)/x^2-1/4*d^2*b*exp(-c)*Ei(1,d*x)-1/6*a/x^3*exp(d*x+c)-1/12*d*a/x^2*exp(d*x+c)-1
/12*d^2*a/x*exp(d*x+c)-1/12*d^3*a*exp(c)*Ei(1,-d*x)-1/4*b/x^2*exp(d*x+c)-1/4*d*b/x*exp(d*x+c)-1/4*d^2*b*exp(c)
*Ei(1,-d*x)

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Maxima [A]  time = 1.39474, size = 105, normalized size = 0.8 \begin{align*} \frac{1}{12} \,{\left (2 \, a d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - 2 \, a d^{2} e^{c} \Gamma \left (-2, -d x\right ) + 3 \, b d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 3 \, b d e^{c} \Gamma \left (-1, -d x\right )\right )} d - \frac{{\left (3 \, b x + 2 \, a\right )} \cosh \left (d x + c\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c)/x^4,x, algorithm="maxima")

[Out]

1/12*(2*a*d^2*e^(-c)*gamma(-2, d*x) - 2*a*d^2*e^c*gamma(-2, -d*x) + 3*b*d*e^(-c)*gamma(-1, d*x) + 3*b*d*e^c*ga
mma(-1, -d*x))*d - 1/6*(3*b*x + 2*a)*cosh(d*x + c)/x^3

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Fricas [A]  time = 1.98033, size = 328, normalized size = 2.48 \begin{align*} -\frac{2 \,{\left (a d^{2} x^{2} + 3 \, b x + 2 \, a\right )} \cosh \left (d x + c\right ) -{\left ({\left (a d^{3} + 3 \, b d^{2}\right )} x^{3}{\rm Ei}\left (d x\right ) -{\left (a d^{3} - 3 \, b d^{2}\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \,{\left (3 \, b d x^{2} + a d x\right )} \sinh \left (d x + c\right ) -{\left ({\left (a d^{3} + 3 \, b d^{2}\right )} x^{3}{\rm Ei}\left (d x\right ) +{\left (a d^{3} - 3 \, b d^{2}\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c)/x^4,x, algorithm="fricas")

[Out]

-1/12*(2*(a*d^2*x^2 + 3*b*x + 2*a)*cosh(d*x + c) - ((a*d^3 + 3*b*d^2)*x^3*Ei(d*x) - (a*d^3 - 3*b*d^2)*x^3*Ei(-
d*x))*cosh(c) + 2*(3*b*d*x^2 + a*d*x)*sinh(d*x + c) - ((a*d^3 + 3*b*d^2)*x^3*Ei(d*x) + (a*d^3 - 3*b*d^2)*x^3*E
i(-d*x))*sinh(c))/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c)/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.14764, size = 269, normalized size = 2.04 \begin{align*} -\frac{a d^{3} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x^{3}{\rm Ei}\left (d x\right ) e^{c} - 3 \, b d^{2} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 3 \, b d^{2} x^{3}{\rm Ei}\left (d x\right ) e^{c} + a d^{2} x^{2} e^{\left (d x + c\right )} + a d^{2} x^{2} e^{\left (-d x - c\right )} + 3 \, b d x^{2} e^{\left (d x + c\right )} - 3 \, b d x^{2} e^{\left (-d x - c\right )} + a d x e^{\left (d x + c\right )} - a d x e^{\left (-d x - c\right )} + 3 \, b x e^{\left (d x + c\right )} + 3 \, b x e^{\left (-d x - c\right )} + 2 \, a e^{\left (d x + c\right )} + 2 \, a e^{\left (-d x - c\right )}}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c)/x^4,x, algorithm="giac")

[Out]

-1/12*(a*d^3*x^3*Ei(-d*x)*e^(-c) - a*d^3*x^3*Ei(d*x)*e^c - 3*b*d^2*x^3*Ei(-d*x)*e^(-c) - 3*b*d^2*x^3*Ei(d*x)*e
^c + a*d^2*x^2*e^(d*x + c) + a*d^2*x^2*e^(-d*x - c) + 3*b*d*x^2*e^(d*x + c) - 3*b*d*x^2*e^(-d*x - c) + a*d*x*e
^(d*x + c) - a*d*x*e^(-d*x - c) + 3*b*x*e^(d*x + c) + 3*b*x*e^(-d*x - c) + 2*a*e^(d*x + c) + 2*a*e^(-d*x - c))
/x^3