Integrand size = 11, antiderivative size = 66 \[ \int \log (x) \sinh ^2(a+b x) \, dx=\frac {x}{2}-\frac {1}{2} x \log (x)-\frac {\text {Chi}(2 b x) \sinh (2 a)}{4 b}+\frac {\cosh (a+b x) \log (x) \sinh (a+b x)}{2 b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{4 b} \] Output:
1/2*x-1/2*x*ln(x)-1/4*Chi(2*b*x)*sinh(2*a)/b+1/2*cosh(b*x+a)*ln(x)*sinh(b* x+a)/b-1/4*cosh(2*a)*Shi(2*b*x)/b
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {-2 b x+2 b x \log (x)+\text {Chi}(2 b x) \sinh (2 a)-\log (x) \sinh (2 (a+b x))+\cosh (2 a) \text {Shi}(2 b x)}{4 b} \] Input:
Integrate[Log[x]*Sinh[a + b*x]^2,x]
Output:
-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
Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3034, 27, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \log (x) \sinh ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3034 |
\(\displaystyle -\int \frac {1}{4} \left (\frac {\sinh (2 (a+b x))}{b x}-2\right )dx+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac {1}{2} x \log (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {\sinh (2 (a+b x))}{b x}-2\right )dx+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac {1}{2} x \log (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {\sinh (2 a) \text {Chi}(2 b x)}{b}-\frac {\cosh (2 a) \text {Shi}(2 b x)}{b}+2 x\right )+\frac {\log (x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac {1}{2} x \log (x)\) |
Input:
Int[Log[x]*Sinh[a + b*x]^2,x]
Output:
-1/2*(x*Log[x]) + (Cosh[a + b*x]*Log[x]*Sinh[a + b*x])/(2*b) + (2*x - (Cos hIntegral[2*b*x]*Sinh[2*a])/b - (Cosh[2*a]*SinhIntegral[2*b*x])/b)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[Log[u] w, x ] - Int[SimplifyIntegrand[w*(D[u, x]/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]
Time = 1.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50
method | result | size |
risch | \(-\frac {x \ln \left (x \right )}{2}+\frac {{\mathrm e}^{2 b x +2 a} \ln \left (x \right )}{8 b}-\frac {{\mathrm e}^{-2 b x -2 a} \ln \left (x \right )}{8 b}+\frac {{\mathrm e}^{2 a} \operatorname {expIntegral}_{1}\left (-2 b x \right )}{8 b}-\frac {a \ln \left (b x \right )}{2 b}+\frac {\ln \left (-b x \right ) a}{2 b}-\frac {{\mathrm e}^{-2 a} \operatorname {expIntegral}_{1}\left (2 b x \right )}{8 b}+\frac {x}{2}+\frac {a}{2 b}\) | \(99\) |
Input:
int(ln(x)*sinh(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*x*ln(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*ln(-b*x)*a-1/8/b*exp(-2*a)*Ei( 1,2*b*x)+1/2*x+1/2/b*a
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (56) = 112\).
Time = 0.08 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.74 \[ \int \log (x) \sinh ^2(a+b x) \, dx=\frac {4 \, \cosh \left (b x + a\right ) \log \left (x\right ) \sinh \left (b x + a\right )^{3} + \log \left (x\right ) \sinh \left (b x + a\right )^{4} - {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right )^{2} \sinh \left (2 \, a\right ) + {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right )^{2} + {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, {\left (2 \, b x - 3 \, \cosh \left (b x + a\right )^{2}\right )} \log \left (x\right ) - {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )\right )} \sinh \left (b x + a\right )^{2} - {\left (4 \, b x \cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right )^{4} + 1\right )} \log \left (x\right ) - 2 \, {\left ({\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (b x + a\right ) \sinh \left (2 \, a\right ) - {\left (4 \, b x - {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right )\right )} \cosh \left (b x + a\right ) + 2 \, {\left (2 \, b x \cosh \left (b x + a\right ) - \cosh \left (b x + a\right )^{3}\right )} \log \left (x\right )\right )} \sinh \left (b x + a\right )}{8 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \] Input:
integrate(log(x)*sinh(b*x+a)^2,x, algorithm="fricas")
Output:
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))*c osh(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)
\[ \int \log (x) \sinh ^2(a+b x) \, dx=\int \log {\left (x \right )} \sinh ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(ln(x)*sinh(b*x+a)**2,x)
Output:
Integral(log(x)*sinh(a + b*x)**2, x)
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {1}{8} \, {\left (4 \, x - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \left (x\right ) + \frac {1}{2} \, x - \frac {{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )}}{8 \, b} + \frac {{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )}}{8 \, b} \] Input:
integrate(log(x)*sinh(b*x+a)^2,x, algorithm="maxima")
Output:
-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*E i(2*b*x)*e^(2*a)/b + 1/8*Ei(-2*b*x)*e^(-2*a)/b
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \log (x) \sinh ^2(a+b x) \, dx=-\frac {1}{8} \, {\left (4 \, x - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \log \left (x\right ) + \frac {{\left (4 \, b x e^{\left (2 \, a\right )} - {\rm Ei}\left (2 \, b x\right ) e^{\left (4 \, a\right )} + {\rm Ei}\left (-2 \, b x\right )\right )} e^{\left (-2 \, a\right )}}{8 \, b} \] Input:
integrate(log(x)*sinh(b*x+a)^2,x, algorithm="giac")
Output:
-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*e^ (2*a) - Ei(2*b*x)*e^(4*a) + Ei(-2*b*x))*e^(-2*a)/b
Timed out. \[ \int \log (x) \sinh ^2(a+b x) \, dx=\int {\mathrm {sinh}\left (a+b\,x\right )}^2\,\ln \left (x\right ) \,d x \] Input:
int(sinh(a + b*x)^2*log(x),x)
Output:
int(sinh(a + b*x)^2*log(x), x)
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \log (x) \sinh ^2(a+b x) \, dx=\frac {e^{2 b x} \mathit {ei} \left (-2 b x \right )-e^{2 b x +4 a} \mathit {ei} \left (2 b x \right )+e^{4 b x +4 a} \mathrm {log}\left (x \right )-4 e^{2 b x +2 a} \mathrm {log}\left (x \right ) b x +4 e^{2 b x +2 a} b x -\mathrm {log}\left (x \right )}{8 e^{2 b x +2 a} b} \] Input:
int(log(x)*sinh(b*x+a)^2,x)
Output:
(e**(2*b*x)*ei( - 2*b*x) - e**(4*a + 2*b*x)*ei(2*b*x) + e**(4*a + 4*b*x)*l og(x) - 4*e**(2*a + 2*b*x)*log(x)*b*x + 4*e**(2*a + 2*b*x)*b*x - log(x))/( 8*e**(2*a + 2*b*x)*b)