\(\int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx\) [7]

Optimal result

Integrand size = 15, antiderivative size = 88 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=-\frac {a \cosh (c+d x)}{2 x^2}-\frac {b \cosh (c+d x)}{x}+\frac {1}{2} a d^2 \cosh (c) \text {Chi}(d x)+b d \text {Chi}(d x) \sinh (c)-\frac {a d \sinh (c+d x)}{2 x}+b d \cosh (c) \text {Shi}(d x)+\frac {1}{2} a d^2 \sinh (c) \text {Shi}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6874, 3378, 3384, 3379, 3382} \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\frac {1}{2} a d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} a d^2 \sinh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{2 x^2}-\frac {a d \sinh (c+d x)}{2 x}+b d \sinh (c) \text {Chi}(d x)+b d \cosh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{x} \]

[In]

Int[((a + b*x)*Cosh[c + d*x])/x^3,x]

[Out]

-1/2*(a*Cosh[c + d*x])/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 3378

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 3379

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 3382

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 3384

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\frac {-a \cosh (c+d x)-2 b x \cosh (c+d x)+d x^2 \text {Chi}(d x) (a d \cosh (c)+2 b \sinh (c))-a d x \sinh (c+d x)+d x^2 (2 b \cosh (c)+a d \sinh (c)) \text {Shi}(d x)}{2 x^2} \]

[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)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.56

method	result	size
risch	\(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a \,d^{2} x^{2}+{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a \,d^{2} x^{2}+2 \,{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) b d \,x^{2}-2 \,{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) b d \,x^{2}-{\mathrm e}^{-d x -c} a d x +{\mathrm e}^{d x +c} a d x +2 \,{\mathrm e}^{-d x -c} b x +2 \,{\mathrm e}^{d x +c} b x +{\mathrm e}^{-d x -c} a +a \,{\mathrm e}^{d x +c}}{4 x^{2}}\)	\(137\)
meijerg	\(\frac {i d b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {d b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Chi}\left (d x \right )-4 \ln \left (d x \right )-4 \gamma }{\sqrt {\pi }}\right )}{4}-\frac {a \cosh \left (c \right ) \sqrt {\pi }\, d^{2} \left (\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+2 \ln \left (i d \right )\right )}{\sqrt {\pi }}-\frac {4 \left (\frac {9 x^{2} d^{2}}{2}+3\right )}{3 \sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \cosh \left (d x \right )}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}-\frac {4 \left (\operatorname {Chi}\left (d x \right )-\ln \left (d x \right )-\gamma \right )}{\sqrt {\pi }}\right )}{8}+\frac {i a \sinh \left (c \right ) \sqrt {\pi }\, d^{2} \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}+\frac {4 i \sinh \left (d x \right )}{x^{2} d^{2} \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(278\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=-\frac {2 \, a d x \sinh \left (d x + c\right ) + 2 \, {\left (2 \, b x + a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{2} + 2 \, b d\right )} x^{2} {\rm Ei}\left (d x\right ) + {\left (a d^{2} - 2 \, b d\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - {\left ({\left (a d^{2} + 2 \, b d\right )} x^{2} {\rm Ei}\left (d x\right ) - {\left (a d^{2} - 2 \, b d\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, x^{2}} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\frac {1}{4} \, {\left (a d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + a d e^{c} \Gamma \left (-1, -d x\right ) - 2 \, b {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b {\rm Ei}\left (d x\right ) e^{c}\right )} d - \frac {{\left (2 \, b x + a\right )} \cosh \left (d x + c\right )}{2 \, x^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\frac {a d^{2} x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{2} x^{2} {\rm Ei}\left (d x\right ) e^{c} - 2 \, b d x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b d x^{2} {\rm Ei}\left (d x\right ) e^{c} - a d x e^{\left (d x + c\right )} + a d x e^{\left (-d x - c\right )} - 2 \, b x e^{\left (d x + c\right )} - 2 \, b x e^{\left (-d x - c\right )} - a e^{\left (d x + c\right )} - a e^{\left (-d x - c\right )}}{4 \, x^{2}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) \cosh (c+d x)}{x^3} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (a+b\,x\right )}{x^3} \,d x \]

[In]

int((cosh(c + d*x)*(a + b*x))/x^3,x)

[Out]

int((cosh(c + d*x)*(a + b*x))/x^3, x)