Integrand size = 8, antiderivative size = 25 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=b \text {Chi}(b x)-\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x} \] Output:
b*Chi(b*x)-sinh(b*x)/x-Shi(b*x)/x
Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=b \text {Chi}(b x)-\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x} \] Input:
Integrate[SinhIntegral[b*x]/x^2,x]
Output:
b*CoshIntegral[b*x] - Sinh[b*x]/x - SinhIntegral[b*x]/x
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {7086, 27, 3042, 26, 3778, 3042, 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.
\(\displaystyle \int \frac {\text {Shi}(b x)}{x^2} \, dx\) |
\(\Big \downarrow \) 7086 |
\(\displaystyle b \int \frac {\sinh (b x)}{b x^2}dx-\frac {\text {Shi}(b x)}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\sinh (b x)}{x^2}dx-\frac {\text {Shi}(b x)}{x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\text {Shi}(b x)}{x}+\int -\frac {i \sin (i b x)}{x^2}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {\text {Shi}(b x)}{x}-i \int \frac {\sin (i b x)}{x^2}dx\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle -\frac {\text {Shi}(b x)}{x}-i \left (i b \int \frac {\cosh (b x)}{x}dx-\frac {i \sinh (b x)}{x}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\text {Shi}(b x)}{x}-i \left (i b \int \frac {\sin \left (i b x+\frac {\pi }{2}\right )}{x}dx-\frac {i \sinh (b x)}{x}\right )\) |
\(\Big \downarrow \) 3782 |
\(\displaystyle -\frac {\text {Shi}(b x)}{x}-i \left (i b \text {Chi}(b x)-\frac {i \sinh (b x)}{x}\right )\) |
Input:
Int[SinhIntegral[b*x]/x^2,x]
Output:
(-I)*(I*b*CoshIntegral[b*x] - (I*Sinh[b*x])/x) - SinhIntegral[b*x]/x
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[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[((c_.) + (d_.)*(x_))^(m_.)*SinhIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinhIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/ (d*(m + 1)) Int[(c + d*x)^(m + 1)*(Sinh[a + b*x]/(a + b*x)), x], x] /; Fr eeQ[{a, b, c, d, m}, x] && NeQ[m, -1]
Time = 0.56 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
parts | \(-\frac {\operatorname {Shi}\left (b x \right )}{x}+b \left (-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\) | \(30\) |
derivativedivides | \(b \left (-\frac {\operatorname {Shi}\left (b x \right )}{b x}-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\) | \(32\) |
default | \(b \left (-\frac {\operatorname {Shi}\left (b x \right )}{b x}-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\) | \(32\) |
meijerg | \(\frac {\sqrt {\pi }\, b \left (\frac {8 \gamma -16+8 \ln \left (x \right )+8 \ln \left (i b \right )}{\sqrt {\pi }}+\frac {16}{\sqrt {\pi }}-\frac {4 \,{\mathrm e}^{b x}}{\sqrt {\pi }\, b x}+\frac {4 \,{\mathrm e}^{-b x}}{\sqrt {\pi }\, b x}-\frac {4 \left (-9 b x +9\right ) \left (-\gamma -\ln \left (-b x \right )-\operatorname {expIntegral}_{1}\left (-b x \right )\right )}{9 \sqrt {\pi }\, b x}+\frac {4 \left (9 b x +9\right ) \left (-\gamma -\ln \left (b x \right )-\operatorname {expIntegral}_{1}\left (b x \right )\right )}{9 \sqrt {\pi }\, b x}\right )}{8}\) | \(135\) |
Input:
int(Shi(b*x)/x^2,x,method=_RETURNVERBOSE)
Output:
-Shi(b*x)/x+b*(-sinh(b*x)/b/x+Chi(b*x))
\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \] Input:
integrate(Shi(b*x)/x^2,x, algorithm="fricas")
Output:
integral(sinh_integral(b*x)/x^2, x)
Time = 0.61 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\frac {b^{3} x^{2} {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {3}{2} \\ 2, 2, \frac {5}{2}, \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{36} + \frac {b \log {\left (b^{2} x^{2} \right )}}{2} \] Input:
integrate(Shi(b*x)/x**2,x)
Output:
b**3*x**2*hyper((1, 1, 3/2), (2, 2, 5/2, 5/2), b**2*x**2/4)/36 + b*log(b** 2*x**2)/2
\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \] Input:
integrate(Shi(b*x)/x^2,x, algorithm="maxima")
Output:
integrate(Shi(b*x)/x^2, x)
\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \] Input:
integrate(Shi(b*x)/x^2,x, algorithm="giac")
Output:
integrate(Shi(b*x)/x^2, x)
Timed out. \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int \frac {\mathrm {sinhint}\left (b\,x\right )}{x^2} \,d x \] Input:
int(sinhint(b*x)/x^2,x)
Output:
int(sinhint(b*x)/x^2, x)
\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int \frac {\mathit {shi} \left (b x \right )}{x^{2}}d x \] Input:
int(Shi(b*x)/x^2,x)
Output:
int(shi(b*x)/x**2,x)