\(\int \log (x) \sinh ^2(a+b x) \, dx\) [196]

Optimal result

Integrand size = 11, antiderivative size = 66 \[ \int \log (x) \sinh ^2(a+b x) \, dx=\frac {x}{2}-\frac {1}{2} x \log (x)-\frac {\text {Chi}(2 b x) \sinh (2 a)}{4 b}+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b} \]

[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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {2715, 8, 2634, 12, 5382, 3384, 3379, 3382} \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {\sinh (2 a) \text {Chi}(2 b x)}{4 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b}+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x}{2}-\frac {1}{2} x \log (x) \]

[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 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 2634

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 2715

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] && Integ
erQ[2*n]

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 5382

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*} \text {integral}& = -\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] (verified)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {-2 b x+2 b x \log (x)+\text {Chi}(2 b x) \sinh (2 a)-\log (x) \sinh (2 (a+b x))+\cosh (2 a) \text {Shi}(2 b x)}{4 b} \]

[In]

Integrate[Log[x]*Sinh[a + b*x]^2,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]*SinhIntegra
l[2*b*x])/b

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50

method	result	size
risch	\(-\frac {\ln \left (x \right ) x}{2}+\frac {{\mathrm e}^{2 b x +2 a} \ln \left (x \right )}{8 b}-\frac {{\mathrm e}^{-2 b x -2 a} \ln \left (x \right )}{8 b}+\frac {{\mathrm e}^{2 a} \operatorname {Ei}_{1}\left (-2 b x \right )}{8 b}-\frac {a \ln \left (b x \right )}{2 b}+\frac {a \ln \left (-b x \right )}{2 b}-\frac {{\mathrm e}^{-2 a} \operatorname {Ei}_{1}\left (2 b x \right )}{8 b}+\frac {x}{2}+\frac {a}{2 b}\)	\(99\)

[In]

int(ln(x)*sinh(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (56) = 112\).

Time = 0.34 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.74 \[ \int \log (x) \sinh ^2(a+b x) \, dx=\frac {4 \, \cosh \left (b x + a\right ) \log \left (x\right ) \sinh \left (b x + a\right )^{3} + \log \left (x\right ) \sinh \left (b x + a\right )^{4} - {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right )^{2} \sinh \left (2 \, a\right ) + {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right )^{2} + {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, {\left (2 \, b x - 3 \, \cosh \left (b x + a\right )^{2}\right )} \log \left (x\right ) - {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )\right )} \sinh \left (b x + a\right )^{2} - {\left (4 \, b x \cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right )^{4} + 1\right )} \log \left (x\right ) - 2 \, {\left ({\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right ) \sinh \left (2 \, a\right ) - {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right ) + 2 \, {\left (2 \, b x \cosh \left (b x + a\right ) - \cosh \left (b x + a\right )^{3}\right )} \log \left (x\right )\right )} \sinh \left (b x + a\right )}{8 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]

[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)

Sympy [F]

\[ \int \log (x) \sinh ^2(a+b x) \, dx=\int \log {\left (x \right )} \sinh ^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral(log(x)*sinh(a + b*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {1}{8} \, {\left (4 \, x - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \left (x\right ) + \frac {1}{2} \, x - \frac {{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )}}{8 \, b} + \frac {{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )}}{8 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {1}{8} \, {\left (4 \, x - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \left (x\right ) + \frac {4 \, b x - {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )}}{8 \, b} \]

[In]

integrate(log(x)*sinh(b*x+a)^2,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

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x)^2*log(x),x)

[Out]

int(sinh(a + b*x)^2*log(x), x)