3.48 \(\int \frac{(a+b x^2) \cosh (c+d x)}{x^5} \, dx\)

Optimal. Leaf size=149 \[ \frac{1}{24} a d^4 \cosh (c) \text{Chi}(d x)+\frac{1}{24} a d^4 \sinh (c) \text{Shi}(d x)-\frac{a d^2 \cosh (c+d x)}{24 x^2}-\frac{a d^3 \sinh (c+d x)}{24 x}-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{a \cosh (c+d x)}{4 x^4}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{2 x^2}-\frac{b d \sinh (c+d x)}{2 x} \]

[Out]

-(a*Cosh[c + d*x])/(4*x^4) - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(24*x^2) + (b*d^2*Cosh[c]*CoshI
ntegral[d*x])/2 + (a*d^4*Cosh[c]*CoshIntegral[d*x])/24 - (a*d*Sinh[c + d*x])/(12*x^3) - (b*d*Sinh[c + d*x])/(2
*x) - (a*d^3*Sinh[c + d*x])/(24*x) + (b*d^2*Sinh[c]*SinhIntegral[d*x])/2 + (a*d^4*Sinh[c]*SinhIntegral[d*x])/2
4

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Rubi [A]  time = 0.290435, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {5287, 3297, 3303, 3298, 3301} \[ \frac{1}{24} a d^4 \cosh (c) \text{Chi}(d x)+\frac{1}{24} a d^4 \sinh (c) \text{Shi}(d x)-\frac{a d^2 \cosh (c+d x)}{24 x^2}-\frac{a d^3 \sinh (c+d x)}{24 x}-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{a \cosh (c+d x)}{4 x^4}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)-\frac{b \cosh (c+d x)}{2 x^2}-\frac{b d \sinh (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Cosh[c + d*x])/x^5,x]

[Out]

-(a*Cosh[c + d*x])/(4*x^4) - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(24*x^2) + (b*d^2*Cosh[c]*CoshI
ntegral[d*x])/2 + (a*d^4*Cosh[c]*CoshIntegral[d*x])/24 - (a*d*Sinh[c + d*x])/(12*x^3) - (b*d*Sinh[c + d*x])/(2
*x) - (a*d^3*Sinh[c + d*x])/(24*x) + (b*d^2*Sinh[c]*SinhIntegral[d*x])/2 + (a*d^4*Sinh[c]*SinhIntegral[d*x])/2
4

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 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^2\right ) \cosh (c+d x)}{x^5} \, dx &=\int \left (\frac{a \cosh (c+d x)}{x^5}+\frac{b \cosh (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac{\cosh (c+d x)}{x^5} \, dx+b \int \frac{\cosh (c+d x)}{x^3} \, dx\\ &=-\frac{a \cosh (c+d x)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}+\frac{1}{4} (a d) \int \frac{\sinh (c+d x)}{x^4} \, dx+\frac{1}{2} (b d) \int \frac{\sinh (c+d x)}{x^2} \, dx\\ &=-\frac{a \cosh (c+d x)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{12} \left (a d^2\right ) \int \frac{\cosh (c+d x)}{x^3} \, dx+\frac{1}{2} \left (b d^2\right ) \int \frac{\cosh (c+d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{24 x^2}-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{b d \sinh (c+d x)}{2 x}+\frac{1}{24} \left (a d^3\right ) \int \frac{\sinh (c+d x)}{x^2} \, 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)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{24 x^2}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{b d \sinh (c+d x)}{2 x}-\frac{a d^3 \sinh (c+d x)}{24 x}+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)+\frac{1}{24} \left (a d^4\right ) \int \frac{\cosh (c+d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{24 x^2}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{b d \sinh (c+d x)}{2 x}-\frac{a d^3 \sinh (c+d x)}{24 x}+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)+\frac{1}{24} \left (a d^4 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\frac{1}{24} \left (a d^4 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a \cosh (c+d x)}{4 x^4}-\frac{b \cosh (c+d x)}{2 x^2}-\frac{a d^2 \cosh (c+d x)}{24 x^2}+\frac{1}{2} b d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{24} a d^4 \cosh (c) \text{Chi}(d x)-\frac{a d \sinh (c+d x)}{12 x^3}-\frac{b d \sinh (c+d x)}{2 x}-\frac{a d^3 \sinh (c+d x)}{24 x}+\frac{1}{2} b d^2 \sinh (c) \text{Shi}(d x)+\frac{1}{24} a d^4 \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.274468, size = 127, normalized size = 0.85 \[ -\frac{-d^2 x^4 \cosh (c) \left (a d^2+12 b\right ) \text{Chi}(d x)-d^2 x^4 \sinh (c) \left (a d^2+12 b\right ) \text{Shi}(d x)+a d^3 x^3 \sinh (c+d x)+a d^2 x^2 \cosh (c+d x)+2 a d x \sinh (c+d x)+6 a \cosh (c+d x)+12 b d x^3 \sinh (c+d x)+12 b x^2 \cosh (c+d x)}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*Cosh[c + d*x])/x^5,x]

[Out]

-(6*a*Cosh[c + d*x] + 12*b*x^2*Cosh[c + d*x] + a*d^2*x^2*Cosh[c + d*x] - d^2*(12*b + a*d^2)*x^4*Cosh[c]*CoshIn
tegral[d*x] + 2*a*d*x*Sinh[c + d*x] + 12*b*d*x^3*Sinh[c + d*x] + a*d^3*x^3*Sinh[c + d*x] - d^2*(12*b + a*d^2)*
x^4*Sinh[c]*SinhIntegral[d*x])/(24*x^4)

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Maple [A]  time = 0.066, size = 238, normalized size = 1.6 \begin{align*}{\frac{a{d}^{3}{{\rm e}^{-dx-c}}}{48\,x}}-{\frac{a{d}^{2}{{\rm e}^{-dx-c}}}{48\,{x}^{2}}}+{\frac{da{{\rm e}^{-dx-c}}}{24\,{x}^{3}}}-{\frac{a{{\rm e}^{-dx-c}}}{8\,{x}^{4}}}+{\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{{d}^{4}a{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{48}}-{\frac{{d}^{4}a{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{48}}-{\frac{a{{\rm e}^{dx+c}}}{8\,{x}^{4}}}-{\frac{da{{\rm e}^{dx+c}}}{24\,{x}^{3}}}-{\frac{a{d}^{2}{{\rm e}^{dx+c}}}{48\,{x}^{2}}}-{\frac{a{d}^{3}{{\rm e}^{dx+c}}}{48\,x}}-{\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^2+a)*cosh(d*x+c)/x^5,x)

[Out]

1/48*d^3*a*exp(-d*x-c)/x-1/48*d^2*a*exp(-d*x-c)/x^2+1/24*d*a*exp(-d*x-c)/x^3-1/8*a*exp(-d*x-c)/x^4+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)-1/48*d^4*a*exp(-c)*Ei(1,d*x)-1/48*d^4*a*exp(c)*Ei
(1,-d*x)-1/8*a/x^4*exp(d*x+c)-1/24*d*a/x^3*exp(d*x+c)-1/48*d^2*a/x^2*exp(d*x+c)-1/48*d^3*a/x*exp(d*x+c)-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 = 1.21251, size = 103, normalized size = 0.69 \begin{align*} \frac{1}{8} \,{\left (a d^{3} e^{\left (-c\right )} \Gamma \left (-3, d x\right ) + a d^{3} e^{c} \Gamma \left (-3, -d x\right ) + 2 \, b d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 2 \, b d e^{c} \Gamma \left (-1, -d x\right )\right )} d - \frac{{\left (2 \, b x^{2} + a\right )} \cosh \left (d x + c\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="maxima")

[Out]

1/8*(a*d^3*e^(-c)*gamma(-3, d*x) + a*d^3*e^c*gamma(-3, -d*x) + 2*b*d*e^(-c)*gamma(-1, d*x) + 2*b*d*e^c*gamma(-
1, -d*x))*d - 1/4*(2*b*x^2 + a)*cosh(d*x + c)/x^4

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Fricas [A]  time = 2.00653, size = 352, normalized size = 2.36 \begin{align*} -\frac{2 \,{\left ({\left (a d^{2} + 12 \, b\right )} x^{2} + 6 \, a\right )} \cosh \left (d x + c\right ) -{\left ({\left (a d^{4} + 12 \, b d^{2}\right )} x^{4}{\rm Ei}\left (d x\right ) +{\left (a d^{4} + 12 \, b d^{2}\right )} x^{4}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \,{\left ({\left (a d^{3} + 12 \, b d\right )} x^{3} + 2 \, a d x\right )} \sinh \left (d x + c\right ) -{\left ({\left (a d^{4} + 12 \, b d^{2}\right )} x^{4}{\rm Ei}\left (d x\right ) -{\left (a d^{4} + 12 \, b d^{2}\right )} x^{4}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{48 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="fricas")

[Out]

-1/48*(2*((a*d^2 + 12*b)*x^2 + 6*a)*cosh(d*x + c) - ((a*d^4 + 12*b*d^2)*x^4*Ei(d*x) + (a*d^4 + 12*b*d^2)*x^4*E
i(-d*x))*cosh(c) + 2*((a*d^3 + 12*b*d)*x^3 + 2*a*d*x)*sinh(d*x + c) - ((a*d^4 + 12*b*d^2)*x^4*Ei(d*x) - (a*d^4
+ 12*b*d^2)*x^4*Ei(-d*x))*sinh(c))/x^4

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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**2+a)*cosh(d*x+c)/x**5,x)

[Out]

Timed out

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Giac [A]  time = 1.19852, size = 320, normalized size = 2.15 \begin{align*} \frac{a d^{4} x^{4}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{4} x^{4}{\rm Ei}\left (d x\right ) e^{c} + 12 \, b d^{2} x^{4}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 12 \, b d^{2} x^{4}{\rm Ei}\left (d x\right ) e^{c} - a d^{3} x^{3} e^{\left (d x + c\right )} + a d^{3} x^{3} e^{\left (-d x - c\right )} - a d^{2} x^{2} e^{\left (d x + c\right )} - 12 \, b d x^{3} e^{\left (d x + c\right )} - a d^{2} x^{2} e^{\left (-d x - c\right )} + 12 \, b d x^{3} e^{\left (-d x - c\right )} - 2 \, a d x e^{\left (d x + c\right )} - 12 \, b x^{2} e^{\left (d x + c\right )} + 2 \, a d x e^{\left (-d x - c\right )} - 12 \, b x^{2} e^{\left (-d x - c\right )} - 6 \, a e^{\left (d x + c\right )} - 6 \, a e^{\left (-d x - c\right )}}{48 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="giac")

[Out]

1/48*(a*d^4*x^4*Ei(-d*x)*e^(-c) + a*d^4*x^4*Ei(d*x)*e^c + 12*b*d^2*x^4*Ei(-d*x)*e^(-c) + 12*b*d^2*x^4*Ei(d*x)*
e^c - a*d^3*x^3*e^(d*x + c) + a*d^3*x^3*e^(-d*x - c) - a*d^2*x^2*e^(d*x + c) - 12*b*d*x^3*e^(d*x + c) - a*d^2*
x^2*e^(-d*x - c) + 12*b*d*x^3*e^(-d*x - c) - 2*a*d*x*e^(d*x + c) - 12*b*x^2*e^(d*x + c) + 2*a*d*x*e^(-d*x - c)
- 12*b*x^2*e^(-d*x - c) - 6*a*e^(d*x + c) - 6*a*e^(-d*x - c))/x^4