3.4.82 \(\int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx\) [382]

3.4.82.1 Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=b \cosh (2 a) \text {Chi}(2 b x)-\frac {\sinh (2 a+2 b x)}{2 x}+b \sinh (2 a) \text {Shi}(2 b x)-\text {Int}\left (\frac {\tanh (a+b x)}{x^2},x\right ) \]

output

b*Chi(2*b*x)*cosh(2*a)+b*Shi(2*b*x)*sinh(2*a)-1/2*sinh(2*b*x+2*a)/x-Uninte 
grable(tanh(b*x+a)/x^2,x)
 

3.4.82.2 Mathematica [N/A]

Not integrable

Time = 10.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx \]

input

Integrate[(Sinh[a + b*x]^2*Tanh[a + b*x])/x^2,x]
 

output

Integrate[(Sinh[a + b*x]^2*Tanh[a + b*x])/x^2, x]
 

3.4.82.3 Rubi [N/A]

Not integrable

Time = 0.63 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {5972, 3042, 26, 4222, 5971, 27, 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.

\(\displaystyle \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx\)

\(\Big \downarrow \) 5972

\(\displaystyle \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^2}dx-\int \frac {\tanh (a+b x)}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^2}dx-\int -\frac {i \tan (i a+i b x)}{x^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^2}dx+i \int \frac {\tan (i a+i b x)}{x^2}dx\)

\(\Big \downarrow \) 4222

\(\displaystyle \int \frac {\cosh (a+b x) \sinh (a+b x)}{x^2}dx-\int \frac {\tanh (a+b x)}{x^2}dx\)

\(\Big \downarrow \) 5971

\(\displaystyle \int \frac {\sinh (2 a+2 b x)}{2 x^2}dx-\int \frac {\tanh (a+b x)}{x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {\sinh (2 a+2 b x)}{x^2}dx-\int \frac {\tanh (a+b x)}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx+\frac {1}{2} \int -\frac {i \sin (2 i a+2 i b x)}{x^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \int \frac {\sin (2 i a+2 i b x)}{x^2}dx\)

\(\Big \downarrow \) 3778

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \int \frac {\cosh (2 a+2 b x)}{x}dx-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \int \frac {\sin \left (2 i a+2 i b x+\frac {\pi }{2}\right )}{x}dx-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 3784

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \left (\cosh (2 a) \int \frac {\cosh (2 b x)}{x}dx-i \sinh (2 a) \int \frac {i \sinh (2 b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \left (\sinh (2 a) \int \frac {\sinh (2 b x)}{x}dx+\cosh (2 a) \int \frac {\cosh (2 b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \left (\sinh (2 a) \int -\frac {i \sin (2 i b x)}{x}dx+\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \left (\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx-i \sinh (2 a) \int \frac {\sin (2 i b x)}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 3779

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b \left (\sinh (2 a) \text {Shi}(2 b x)+\cosh (2 a) \int \frac {\sin \left (2 i b x+\frac {\pi }{2}\right )}{x}dx\right )-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

\(\Big \downarrow \) 3782

\(\displaystyle -\int \frac {\tanh (a+b x)}{x^2}dx-\frac {1}{2} i \left (2 i b (\cosh (2 a) \text {Chi}(2 b x)+\sinh (2 a) \text {Shi}(2 b x))-\frac {i \sinh (2 a+2 b x)}{x}\right )\)

input

Int[(Sinh[a + b*x]^2*Tanh[a + b*x])/x^2,x]
 

output

$Aborted
 

3.4.82.3.1 Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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]
 

rule 3779

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]
 

rule 3782

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]
 

rule 3784

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]
 

rule 4222

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> 
Simp[If[MatchQ[f, (f1_.)*(Complex[0, j_])], If[MatchQ[e, (e1_.) + Pi/2], I^ 
n*Unintegrable[(c + d*x)^m*Coth[(-I)*(e - Pi/2) - I*f*x]^n, x], I^n*Uninteg 
rable[(c + d*x)^m*Tanh[(-I)*e - I*f*x]^n, x]], If[MatchQ[e, (e1_.) + Pi/2], 
 (-1)^n*Unintegrable[(c + d*x)^m*Cot[e - Pi/2 + f*x]^n, x], Unintegrable[(c 
 + d*x)^m*Tan[e + f*x]^n, x]]], x] /; FreeQ[{c, d, e, f, m, n}, x] && Integ 
erQ[n]
 

rule 5971

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]
 

rule 5972

Int[((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + 
(b_.)*(x_)]^(p_.), x_Symbol] :> Int[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b* 
x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p, x] 
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
 

3.4.82.4 Maple [N/A] (verified)

Not integrable

Time = 0.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

\[\int \frac {\operatorname {sech}\left (b x +a \right ) \sinh \left (b x +a \right )^{3}}{x^{2}}d x\]

input

int(sech(b*x+a)*sinh(b*x+a)^3/x^2,x)
 

output

int(sech(b*x+a)*sinh(b*x+a)^3/x^2,x)
 

3.4.82.5 Fricas [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int { \frac {\operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(sech(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="fricas")
 

output

integral(sech(b*x + a)*sinh(b*x + a)^3/x^2, x)
 

3.4.82.6 Sympy [N/A]

Not integrable

Time = 3.44 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}}{x^{2}}\, dx \]

input

integrate(sech(b*x+a)*sinh(b*x+a)**3/x**2,x)
 

output

Integral(sinh(a + b*x)**3*sech(a + b*x)/x**2, x)
 

3.4.82.7 Maxima [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.94 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int { \frac {\operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(sech(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="maxima")
 

output

1/2*b*e^(-2*a)*gamma(-1, 2*b*x) + 1/2*b*e^(2*a)*gamma(-1, -2*b*x) + 1/x + 
2*integrate(1/(x^2*e^(2*b*x + 2*a) + x^2), x)
 

3.4.82.8 Giac [N/A]

Not integrable

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int { \frac {\operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(sech(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="giac")
 

output

integrate(sech(b*x + a)*sinh(b*x + a)^3/x^2, x)
 

3.4.82.9 Mupad [N/A]

Not integrable

Time = 2.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {\sinh ^2(a+b x) \tanh (a+b x)}{x^2} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^2\,\mathrm {cosh}\left (a+b\,x\right )} \,d x \]

input

int(sinh(a + b*x)^3/(x^2*cosh(a + b*x)),x)
 

output

int(sinh(a + b*x)^3/(x^2*cosh(a + b*x)), x)