\(\int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx\) [6]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=-\frac {a \cosh (c+d x)}{x}+b \cosh (c) \text {Chi}(d x)+a d \text {Chi}(d x) \sinh (c)+a d \cosh (c) \text {Shi}(d x)+b \sinh (c) \text {Shi}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6874, 3378, 3384, 3379, 3382} \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}+b \cosh (c) \text {Chi}(d x)+b \sinh (c) \text {Shi}(d x) \]

[In]

Int[((a + b*x)*Cosh[c + d*x])/x^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=-\frac {a \cosh (c) \cosh (d x)}{x}+b \cosh (c) \text {Chi}(d x)-\frac {a \sinh (c) \sinh (d x)}{x}+b \sinh (c) \text {Shi}(d x)+a d (\text {Chi}(d x) \sinh (c)+\cosh (c) \text {Shi}(d x)) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60

method	result	size
risch	\(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a d x -{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a d x +{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) b x +{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) b x +{\mathrm e}^{-d x -c} a +a \,{\mathrm e}^{d x +c}}{2 x}\)	\(75\)
meijerg	\(\frac {b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {2 \gamma +2 \ln \left (x \right )+2 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Chi}\left (d x \right )-2 \ln \left (d x \right )-2 \gamma }{\sqrt {\pi }}\right )}{2}+b \,\operatorname {Shi}\left (d x \right ) \sinh \left (c \right )+\frac {i a \cosh \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sinh \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Chi}\left (d x \right )-4 \ln \left (d x \right )-4 \gamma }{\sqrt {\pi }}\right )}{4}\)	\(164\)

[In]

int((b*x+a)*cosh(d*x+c)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=-\frac {2 \, a \cosh \left (d x + c\right ) - {\left ({\left (a d + b\right )} x {\rm Ei}\left (d x\right ) - {\left (a d - b\right )} x {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - {\left ({\left (a d + b\right )} x {\rm Ei}\left (d x\right ) + {\left (a d - b\right )} x {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{2 \, x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=\int \frac {\left (a + b x\right ) \cosh {\left (c + d x \right )}}{x^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.74 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=-\frac {1}{2} \, {\left ({\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - {\rm Ei}\left (d x\right ) e^{c}\right )} a + \frac {2 \, b \cosh \left (d x + c\right ) \log \left (x\right )}{d} - \frac {{\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} b}{d}\right )} d + {\left (b \log \left (x\right ) - \frac {a}{x}\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=-\frac {a d x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d x {\rm Ei}\left (d x\right ) e^{c} - b x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - b x {\rm Ei}\left (d x\right ) e^{c} + a e^{\left (d x + c\right )} + a e^{\left (-d x - c\right )}}{2 \, x} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) \cosh (c+d x)}{x^2} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (a+b\,x\right )}{x^2} \,d x \]

[In]

int((cosh(c + d*x)*(a + b*x))/x^2,x)

[Out]

int((cosh(c + d*x)*(a + b*x))/x^2, x)