3.199 \(\int \cosh ^2(a+b x) \log (x) \, dx\)

Optimal. Leaf size=66 \[ -\frac {\sinh (2 a) \text {Chi}(2 b x)}{4 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b}+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac {x}{2}+\frac {1}{2} x \log (x) \]

[Out]

-1/2*x+1/2*x*ln(x)-1/4*cosh(2*a)*Shi(2*b*x)/b-1/4*Chi(2*b*x)*sinh(2*a)/b+1/2*cosh(b*x+a)*ln(x)*sinh(b*x+a)/b

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Rubi [A]  time = 0.13, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {2635, 8, 2554, 12, 5274, 3303, 3298, 3301} \[ -\frac {\sinh (2 a) \text {Chi}(2 b x)}{4 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b}+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac {x}{2}+\frac {1}{2} x \log (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^2*Log[x],x]

[Out]

-x/2 + (x*Log[x])/2 - (CoshIntegral[2*b*x]*Sinh[2*a])/(4*b) + (Cosh[a + b*x]*Log[x]*Sinh[a + b*x])/(2*b) - (Co
sh[2*a]*SinhIntegral[2*b*x])/(4*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

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 5274

Int[(u_)^(m_.)*((a_.) + (b_.)*Sinh[v_])^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*(a + b*Sinh[ExpandToSum[v,
x]])^n, x] /; FreeQ[{a, b, m, n}, x] && LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {align*} \int \cosh ^2(a+b x) \log (x) \, dx &=\frac {1}{2} x \log (x)+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\int \frac {1}{4} \left (2+\frac {\sinh (2 (a+b x))}{b x}\right ) \, dx\\ &=\frac {1}{2} x \log (x)+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {1}{4} \int \left (2+\frac {\sinh (2 (a+b x))}{b x}\right ) \, dx\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\int \frac {\sinh (2 (a+b x))}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\int \frac {\sinh (2 a+2 b x)}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\cosh (2 a) \int \frac {\sinh (2 b x)}{x} \, dx}{4 b}-\frac {\sinh (2 a) \int \frac {\cosh (2 b x)}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)-\frac {\text {Chi}(2 b x) \sinh (2 a)}{4 b}+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 50, normalized size = 0.76 \[ -\frac {\sinh (2 a) \text {Chi}(2 b x)+\cosh (2 a) \text {Shi}(2 b x)-\log (x) \sinh (2 (a+b x))+2 b x-2 b x \log (x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^2*Log[x],x]

[Out]

-1/4*(2*b*x - 2*b*x*Log[x] + CoshIntegral[2*b*x]*Sinh[2*a] - Log[x]*Sinh[2*(a + b*x)] + Cosh[2*a]*SinhIntegral
[2*b*x])/b

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fricas [B]  time = 0.48, size = 305, normalized size = 4.62 \[ \frac {4 \, \cosh \left (b x + a\right ) \log \relax (x) \sinh \left (b x + a\right )^{3} + \log \relax (x) \sinh \left (b x + a\right )^{4} - {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right )^{2} \sinh \left (2 \, a\right ) - {\left (4 \, b x + {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right )^{2} - {\left (4 \, b x + {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, {\left (2 \, b x + 3 \, \cosh \left (b x + a\right )^{2}\right )} \log \relax (x) + {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )\right )} \sinh \left (b x + a\right )^{2} + {\left (4 \, b x \cosh \left (b x + a\right )^{2} + \cosh \left (b x + a\right )^{4} - 1\right )} \log \relax (x) - 2 \, {\left ({\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right ) \sinh \left (2 \, a\right ) + {\left (4 \, b x + {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right ) - 2 \, {\left (2 \, b x \cosh \left (b x + a\right ) + \cosh \left (b x + a\right )^{3}\right )} \log \relax (x)\right )} \sinh \left (b x + a\right )}{8 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*cosh(b*x + a)*log(x)*sinh(b*x + a)^3 + log(x)*sinh(b*x + a)^4 - (Ei(2*b*x) + Ei(-2*b*x))*cosh(b*x + a)^
2*sinh(2*a) - (4*b*x + (Ei(2*b*x) - Ei(-2*b*x))*cosh(2*a))*cosh(b*x + a)^2 - (4*b*x + (Ei(2*b*x) - Ei(-2*b*x))
*cosh(2*a) - 2*(2*b*x + 3*cosh(b*x + a)^2)*log(x) + (Ei(2*b*x) + Ei(-2*b*x))*sinh(2*a))*sinh(b*x + a)^2 + (4*b
*x*cosh(b*x + a)^2 + cosh(b*x + a)^4 - 1)*log(x) - 2*((Ei(2*b*x) + Ei(-2*b*x))*cosh(b*x + a)*sinh(2*a) + (4*b*
x + (Ei(2*b*x) - Ei(-2*b*x))*cosh(2*a))*cosh(b*x + a) - 2*(2*b*x*cosh(b*x + a) + cosh(b*x + a)^3)*log(x))*sinh
(b*x + a))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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giac [A]  time = 0.23, size = 67, normalized size = 1.02 \[ \frac {1}{8} \, {\left (4 \, x + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \relax (x) - \frac {4 \, b x + {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x + e^(2*b*x + 2*a)/b - e^(-2*b*x - 2*a)/b)*log(x) - 1/8*(4*b*x + Ei(2*b*x)*e^(2*a) - Ei(-2*b*x)*e^(-2*
a))/b

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maple [A]  time = 1.24, size = 97, normalized size = 1.47 \[ \frac {a \ln \left (b x \right )}{2 b}-\frac {a \ln \left (-b x \right )}{2 b}+\frac {\Ei \left (1, -2 b x \right ) {\mathrm e}^{2 a}}{8 b}-\frac {\Ei \left (1, 2 b x \right ) {\mathrm e}^{-2 a}}{8 b}-\frac {x}{2}+\left (\frac {x}{2}-\frac {{\mathrm e}^{-2 b x -2 a}}{8 b}+\frac {{\mathrm e}^{2 b x +2 a}}{8 b}\right ) \ln \relax (x )-\frac {a}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^2*ln(x),x)

[Out]

(1/2*x+1/8/b*exp(2*b*x+2*a)-1/8/b*exp(-2*b*x-2*a))*ln(x)+1/8/b*exp(2*a)*Ei(1,-2*b*x)+1/2*a/b*ln(b*x)-1/2*a/b*l
n(-b*x)-1/2*x-1/2*a/b-1/8/b*exp(-2*a)*Ei(1,2*b*x)

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maxima [A]  time = 0.68, size = 67, normalized size = 1.02 \[ \frac {1}{8} \, {\left (4 \, x + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \relax (x) - \frac {1}{2} \, x - \frac {{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )}}{8 \, b} + \frac {{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x + e^(2*b*x + 2*a)/b - e^(-2*b*x - 2*a)/b)*log(x) - 1/2*x - 1/8*Ei(2*b*x)*e^(2*a)/b + 1/8*Ei(-2*b*x)*e
^(-2*a)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {cosh}\left (a+b\,x\right )}^2\,\ln \relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^2*log(x),x)

[Out]

int(cosh(a + b*x)^2*log(x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\relax (x )} \cosh ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2*ln(x),x)

[Out]

Integral(log(x)*cosh(a + b*x)**2, x)

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