3.198 \(\int \cosh (a+b x) \log (x) \, dx\)

Optimal. Leaf size=35 \[ -\frac{\sinh (a) \text{Chi}(b x)}{b}-\frac{\cosh (a) \text{Shi}(b x)}{b}+\frac{\log (x) \sinh (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.070019, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {2637, 2554, 12, 3303, 3298, 3301} \[ -\frac{\sinh (a) \text{Chi}(b x)}{b}-\frac{\cosh (a) \text{Shi}(b x)}{b}+\frac{\log (x) \sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]*Log[x],x]

[Out]

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

Rule 2637

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

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0379113, size = 30, normalized size = 0.86 \[ -\frac{\sinh (a) \text{Chi}(b x)+\cosh (a) \text{Shi}(b x)-\log (x) \sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]*Log[x],x]

[Out]

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

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Maple [A]  time = 0.019, size = 58, normalized size = 1.7 \begin{align*} \left ({\frac{{{\rm e}^{bx+a}}}{2\,b}}-{\frac{{{\rm e}^{-bx-a}}}{2\,b}} \right ) \ln \left ( x \right ) +{\frac{{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{2\,b}}-{\frac{{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*ln(x),x)

[Out]

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

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Maxima [A]  time = 1.189, size = 50, normalized size = 1.43 \begin{align*} \frac{\log \left (x\right ) \sinh \left (b x + a\right )}{b} + \frac{{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} -{\rm Ei}\left (b x\right ) e^{a}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(x),x, algorithm="maxima")

[Out]

log(x)*sinh(b*x + a)/b + 1/2*(Ei(-b*x)*e^(-a) - Ei(b*x)*e^a)/b

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Fricas [B]  time = 1.78763, size = 393, normalized size = 11.23 \begin{align*} -\frac{{\left ({\rm Ei}\left (b x\right ) -{\rm Ei}\left (-b x\right )\right )} \cosh \left (b x + a\right ) \cosh \left (a\right ) - \log \left (x\right ) \sinh \left (b x + a\right )^{2} +{\left ({\rm Ei}\left (b x\right ) +{\rm Ei}\left (-b x\right )\right )} \cosh \left (b x + a\right ) \sinh \left (a\right ) -{\left (\cosh \left (b x + a\right )^{2} - 1\right )} \log \left (x\right ) +{\left ({\left ({\rm Ei}\left (b x\right ) -{\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) - 2 \, \cosh \left (b x + a\right ) \log \left (x\right ) +{\left ({\rm Ei}\left (b x\right ) +{\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )\right )} \sinh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(x),x, algorithm="fricas")

[Out]

-1/2*((Ei(b*x) - Ei(-b*x))*cosh(b*x + a)*cosh(a) - log(x)*sinh(b*x + a)^2 + (Ei(b*x) + Ei(-b*x))*cosh(b*x + a)
*sinh(a) - (cosh(b*x + a)^2 - 1)*log(x) + ((Ei(b*x) - Ei(-b*x))*cosh(a) - 2*cosh(b*x + a)*log(x) + (Ei(b*x) +
Ei(-b*x))*sinh(a))*sinh(b*x + a))/(b*cosh(b*x + a) + b*sinh(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (x \right )} \cosh{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*ln(x),x)

[Out]

Integral(log(x)*cosh(a + b*x), x)

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Giac [A]  time = 1.26758, size = 73, normalized size = 2.09 \begin{align*} \frac{1}{2} \,{\left (\frac{e^{\left (b x + a\right )}}{b} - \frac{e^{\left (-b x - a\right )}}{b}\right )} \log \left (x\right ) + \frac{{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} -{\rm Ei}\left (b x\right ) e^{a}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(x),x, algorithm="giac")

[Out]

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