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

Optimal. Leaf size=69 \[ \frac{1}{2} a d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a d^2 \sinh (c) \text{Shi}(d x)-\frac{a \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{2 x}+\frac{b \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.138576, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {5287, 2637, 3297, 3303, 3298, 3301} \[ \frac{1}{2} a d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a d^2 \sinh (c) \text{Shi}(d x)-\frac{a \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{2 x}+\frac{b \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5287

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A]  time = 0.141124, size = 86, normalized size = 1.25 \[ \frac{1}{2} a d^2 (\cosh (c) \text{Chi}(d x)+\sinh (c) \text{Shi}(d x))-\frac{a \cosh (d x) (d x \sinh (c)+\cosh (c))}{2 x^2}-\frac{a \sinh (d x) (d x \cosh (c)+\sinh (c))}{2 x^2}+\frac{b \sinh (c) \cosh (d x)}{d}+\frac{b \cosh (c) \sinh (d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 114, normalized size = 1.7 \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\,d}}-{\frac{a{{\rm e}^{dx+c}}}{4\,{x}^{2}}}-{\frac{da{{\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\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+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/d*exp(-d*x-c)-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/d*exp(d*x+c)

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Maxima [A]  time = 1.17684, size = 117, normalized size = 1.7 \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 ) - \frac{2 \,{\left (d x e^{c} - e^{c}\right )} b e^{\left (d x\right )}}{d^{2}} - \frac{2 \,{\left (d x + 1\right )} b e^{\left (-d x - c\right )}}{d^{2}}\right )} d + \frac{1}{2} \,{\left (2 \, b x - \frac{a}{x^{2}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.83341, size = 238, normalized size = 3.45 \begin{align*} -\frac{2 \, a d \cosh \left (d x + c\right ) -{\left (a d^{3} x^{2}{\rm Ei}\left (d x\right ) + a d^{3} x^{2}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \,{\left (a d^{2} x - 2 \, b x^{2}\right )} \sinh \left (d x + c\right ) -{\left (a d^{3} x^{2}{\rm Ei}\left (d x\right ) - a d^{3} x^{2}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, d x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*d*cosh(d*x + c) - (a*d^3*x^2*Ei(d*x) + a*d^3*x^2*Ei(-d*x))*cosh(c) + 2*(a*d^2*x - 2*b*x^2)*sinh(d*x
+ c) - (a*d^3*x^2*Ei(d*x) - a*d^3*x^2*Ei(-d*x))*sinh(c))/(d*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**3+a)*cosh(d*x+c)/x**3,x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.25569, size = 159, normalized size = 2.3 \begin{align*} \frac{a d^{3} x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{3} x^{2}{\rm Ei}\left (d x\right ) e^{c} - a d^{2} x e^{\left (d x + c\right )} + a d^{2} x e^{\left (-d x - c\right )} + 2 \, b x^{2} e^{\left (d x + c\right )} - 2 \, b x^{2} e^{\left (-d x - c\right )} - a d e^{\left (d x + c\right )} - a d e^{\left (-d x - c\right )}}{4 \, d x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^3*x^2*Ei(-d*x)*e^(-c) + a*d^3*x^2*Ei(d*x)*e^c - a*d^2*x*e^(d*x + c) + a*d^2*x*e^(-d*x - c) + 2*b*x^2*
e^(d*x + c) - 2*b*x^2*e^(-d*x - c) - a*d*e^(d*x + c) - a*d*e^(-d*x - c))/(d*x^2)