Integrand size = 16, antiderivative size = 85 \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=-\frac {b \cosh (2 a+2 b x)}{6 x^2}+\frac {2}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)-\frac {\sinh (2 a+2 b x)}{6 x^3}-\frac {b^2 \sinh (2 a+2 b x)}{3 x}+\frac {2}{3} b^3 \sinh (2 a) \text {Shi}(2 b x) \]
2/3*b^3*Chi(2*b*x)*cosh(2*a)-1/6*b*cosh(2*b*x+2*a)/x^2+2/3*b^3*Shi(2*b*x)* sinh(2*a)-1/6*sinh(2*b*x+2*a)/x^3-1/3*b^2*sinh(2*b*x+2*a)/x
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=-\frac {b x \cosh (2 (a+b x))-4 b^3 x^3 \cosh (2 a) \text {Chi}(2 b x)+\sinh (2 (a+b x))+2 b^2 x^2 \sinh (2 (a+b x))-4 b^3 x^3 \sinh (2 a) \text {Shi}(2 b x)}{6 x^3} \]
-1/6*(b*x*Cosh[2*(a + b*x)] - 4*b^3*x^3*Cosh[2*a]*CoshIntegral[2*b*x] + Si nh[2*(a + b*x)] + 2*b^2*x^2*Sinh[2*(a + b*x)] - 4*b^3*x^3*Sinh[2*a]*SinhIn tegral[2*b*x])/x^3
Result contains complex when optimal does not.
Time = 0.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {5971, 27, 3042, 26, 3778, 3042, 3778, 26, 3042, 26, 3778, 3042, 3784, 26, 3042, 26, 3779, 3782}
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.
\(\displaystyle \int \frac {\sinh (a+b x) \cosh (a+b x)}{x^4} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \frac {\sinh (2 a+2 b x)}{2 x^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {\sinh (2 a+2 b x)}{x^4}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int -\frac {i \sin (2 i a+2 i b x)}{x^4}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {1}{2} i \int \frac {\sin (2 i a+2 i b x)}{x^4}dx\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \int \frac {\cosh (2 a+2 b x)}{x^3}dx-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \int \frac {\sin \left (2 i a+2 i b x+\frac {\pi }{2}\right )}{x^3}dx-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}+i b \int -\frac {i \sinh (2 a+2 b x)}{x^2}dx\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (b \int \frac {\sinh (2 a+2 b x)}{x^2}dx-\frac {\cosh (2 a+2 b x)}{2 x^2}\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}+b \int -\frac {i \sin (2 i a+2 i b x)}{x^2}dx\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \int \frac {\sin (2 i a+2 i b x)}{x^2}dx\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \int \frac {\cosh (2 a+2 b x)}{x}dx-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \int \frac {\sin \left (2 i a+2 i b x+\frac {\pi }{2}\right )}{x}dx-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \left (\cosh (2 a) \int \frac {\cosh (2 b x)}{x}dx-i \sinh (2 a) \int \frac {i \sinh (2 b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \left (\sinh (2 a) \int \frac {\sinh (2 b x)}{x}dx+\cosh (2 a) \int \frac {\cosh (2 b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \left (\sinh (2 a) \int -\frac {i \sin (2 i b x)}{x}dx+\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \left (\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx-i \sinh (2 a) \int \frac {\sin (2 i b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3779 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b \left (\sinh (2 a) \text {Shi}(2 b x)+\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
\(\Big \downarrow \) 3782 |
\(\displaystyle -\frac {1}{2} i \left (\frac {2}{3} i b \left (-\frac {\cosh (2 a+2 b x)}{2 x^2}-i b \left (2 i b (\cosh (2 a) \text {Chi}(2 b x)+\sinh (2 a) \text {Shi}(2 b x))-\frac {i \sinh (2 a+2 b x)}{x}\right )\right )-\frac {i \sinh (2 a+2 b x)}{3 x^3}\right )\) |
(-1/2*I)*(((-1/3*I)*Sinh[2*a + 2*b*x])/x^3 + ((2*I)/3)*b*(-1/2*Cosh[2*a + 2*b*x]/x^2 - I*b*(((-I)*Sinh[2*a + 2*b*x])/x + (2*I)*b*(Cosh[2*a]*CoshInte gral[2*b*x] + Sinh[2*a]*SinhIntegral[2*b*x]))))
3.3.58.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> 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]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> 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]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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]
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Time = 1.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.42
method | result | size |
risch | \(-\frac {4 \,{\mathrm e}^{2 a} \operatorname {Ei}_{1}\left (-2 b x \right ) x^{3} b^{3}+4 \,{\mathrm e}^{-2 a} \operatorname {Ei}_{1}\left (2 b x \right ) x^{3} b^{3}+2 \,{\mathrm e}^{2 b x +2 a} b^{2} x^{2}-2 \,{\mathrm e}^{-2 b x -2 a} b^{2} x^{2}+{\mathrm e}^{2 b x +2 a} b x +{\mathrm e}^{-2 b x -2 a} b x +{\mathrm e}^{2 b x +2 a}-{\mathrm e}^{-2 b x -2 a}}{12 x^{3}}\) | \(121\) |
-1/12*(4*exp(2*a)*Ei(1,-2*b*x)*x^3*b^3+4*exp(-2*a)*Ei(1,2*b*x)*x^3*b^3+2*e xp(2*b*x+2*a)*b^2*x^2-2*exp(-2*b*x-2*a)*b^2*x^2+exp(2*b*x+2*a)*b*x+exp(-2* b*x-2*a)*b*x+exp(2*b*x+2*a)-exp(-2*b*x-2*a))/x^3
Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.35 \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=-\frac {b x \cosh \left (b x + a\right )^{2} + b x \sinh \left (b x + a\right )^{2} + 2 \, {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{6 \, x^{3}} \]
-1/6*(b*x*cosh(b*x + a)^2 + b*x*sinh(b*x + a)^2 + 2*(2*b^2*x^2 + 1)*cosh(b *x + a)*sinh(b*x + a) - 2*(b^3*x^3*Ei(2*b*x) + b^3*x^3*Ei(-2*b*x))*cosh(2* a) - 2*(b^3*x^3*Ei(2*b*x) - b^3*x^3*Ei(-2*b*x))*sinh(2*a))/x^3
\[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=\int \frac {\sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{4}}\, dx \]
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.36 \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=2 \, b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) + 2 \, b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) \]
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.41 \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=\frac {4 \, b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 4 \, b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 2 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} - b x e^{\left (2 \, b x + 2 \, a\right )} - b x e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{12 \, x^{3}} \]
1/12*(4*b^3*x^3*Ei(2*b*x)*e^(2*a) + 4*b^3*x^3*Ei(-2*b*x)*e^(-2*a) - 2*b^2* x^2*e^(2*b*x + 2*a) + 2*b^2*x^2*e^(-2*b*x - 2*a) - b*x*e^(2*b*x + 2*a) - b *x*e^(-2*b*x - 2*a) - e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))/x^3
Timed out. \[ \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx=\int \frac {\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )}{x^4} \,d x \]