3.196 \(\int \log (x) \sinh ^2(a+b x) \, dx\)

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]

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) - (Cos
h[2*a]*SinhIntegral[2*b*x])/(4*b)

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Rubi [A]  time = 0.142901, antiderivative size = 66, normalized size of antiderivative = 1., 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[Log[x]*Sinh[a + b*x]^2,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) - (Cos
h[2*a]*SinhIntegral[2*b*x])/(4*b)

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 8

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

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 \log (x) \sinh ^2(a+b 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*}

Mathematica [A]  time = 0.09926, 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[Log[x]*Sinh[a + b*x]^2,x]

[Out]

-(-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])/(4*b)

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Maple [A]  time = 0.036, size = 97, normalized size = 1.5 \begin{align*} \left ( -{\frac{x}{2}}+{\frac{{{\rm e}^{2\,bx+2\,a}}}{8\,b}}-{\frac{{{\rm e}^{-2\,bx-2\,a}}}{8\,b}} \right ) \ln \left ( x \right ) +{\frac{{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{8\,b}}-{\frac{a\ln \left ( bx \right ) }{2\,b}}+{\frac{a\ln \left ( -bx \right ) }{2\,b}}+{\frac{x}{2}}+{\frac{a}{2\,b}}-{\frac{{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)*sinh(b*x+a)^2,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/b*a*ln(b*x)+1/2/b*a*
ln(-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 = 1.15357, size = 90, normalized size = 1.36 \begin{align*} -\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 \left (x\right ) + \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sinh(b*x+a)^2,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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Fricas [B]  time = 1.85988, size = 859, normalized size = 13.02 \begin{align*} \frac{4 \, \cosh \left (b x + a\right ) \log \left (x\right ) \sinh \left (b x + a\right )^{3} + \log \left (x\right ) \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 \left (x\right ) -{\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 \left (x\right ) - 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 \left (x\right )\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sinh(b*x+a)^2,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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (x \right )} \sinh ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(x)*sinh(a + b*x)**2, x)

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Giac [A]  time = 1.37262, size = 123, normalized size = 1.86 \begin{align*} -\frac{{\left (4 \, b x -{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 4 \, a - e^{\left (2 \, b x + 2 \, a\right )}\right )} \log \left (x\right )}{8 \, b} + \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 )} + 4 \, a \log \left (x\right ) - 2 \, \log \left (x\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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