\(\int \sinh ^2(\frac {a}{c+d x}) \, dx\) [290]

Optimal result

Integrand size = 12, antiderivative size = 39 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}-\frac {a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \] Output:

(d*x+c)*sinh(a/(d*x+c))^2/d-a*Shi(2*a/(d*x+c))/d
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )-a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \] Input:

Integrate[Sinh[a/(c + d*x)]^2,x]
 

Output:

((c + d*x)*Sinh[a/(c + d*x)]^2 - a*SinhIntegral[(2*a)/(c + d*x)])/d
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5833, 5825, 3042, 25, 3794, 27, 3042, 26, 3779}

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.

\(\displaystyle \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx\)

\(\Big \downarrow \) 5833

\(\displaystyle \frac {\int \sinh ^2\left (\frac {a}{c+d x}\right )d(c+d x)}{d}\)

\(\Big \downarrow \) 5825

\(\displaystyle -\frac {\int (c+d x)^2 \sinh ^2\left (\frac {a}{c+d x}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int -(c+d x)^2 \sin \left (\frac {i a}{c+d x}\right )^2d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int (c+d x)^2 \sin \left (\frac {i a}{c+d x}\right )^2d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3794

\(\displaystyle -\frac {-\left ((c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )\right )-2 i a \int \frac {1}{2} i (c+d x) \sinh \left (\frac {2 a}{c+d x}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a \int (c+d x) \sinh \left (\frac {2 a}{c+d x}\right )d\frac {1}{c+d x}-(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )+a \int -i (c+d x) \sin \left (\frac {2 i a}{c+d x}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {-\left ((c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )\right )-i a \int (c+d x) \sin \left (\frac {2 i a}{c+d x}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3779

\(\displaystyle -\frac {a \text {Shi}\left (\frac {2 a}{c+d x}\right )-(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}\)

Input:

Int[Sinh[a/(c + d*x)]^2,x]
 

Output:

-((-((c + d*x)*Sinh[a/(c + d*x)]^2) + a*SinhIntegral[(2*a)/(c + d*x)])/d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

rule 5825

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subs 
t[Int[(a + b*Sinh[c + d/x^n])^p/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d}, x] 
&& ILtQ[n, 0] && IntegerQ[p]
 

rule 5833

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Simp[ 
1/Coefficient[u, x, 1]   Subst[Int[(a + b*Sinh[c + d*x^n])^p, x], x, u], x] 
 /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u, x]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(-\frac {a \left (\frac {d x +c}{2 a}-\frac {\left (d x +c \right ) \cosh \left (\frac {2 a}{d x +c}\right )}{2 a}+\operatorname {Shi}\left (\frac {2 a}{d x +c}\right )\right )}{d}\)	\(50\)
default	\(-\frac {a \left (\frac {d x +c}{2 a}-\frac {\left (d x +c \right ) \cosh \left (\frac {2 a}{d x +c}\right )}{2 a}+\operatorname {Shi}\left (\frac {2 a}{d x +c}\right )\right )}{d}\)	\(50\)
risch	\(-\frac {x}{2}+\frac {{\mathrm e}^{-\frac {2 a}{d x +c}} x}{4}+\frac {{\mathrm e}^{-\frac {2 a}{d x +c}} c}{4 d}-\frac {a \,\operatorname {expIntegral}_{1}\left (\frac {2 a}{d x +c}\right )}{2 d}+\frac {{\mathrm e}^{\frac {2 a}{d x +c}} x}{4}+\frac {{\mathrm e}^{\frac {2 a}{d x +c}} c}{4 d}+\frac {a \,\operatorname {expIntegral}_{1}\left (-\frac {2 a}{d x +c}\right )}{2 d}\)	\(103\)

Input:

int(sinh(a/(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/d*a*(1/2/a*(d*x+c)-1/2/a*(d*x+c)*cosh(2*a/(d*x+c))+Shi(2*a/(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.87 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {{\left (d x + c\right )} \cosh \left (\frac {a}{d x + c}\right )^{2} + {\left (d x + c\right )} \sinh \left (\frac {a}{d x + c}\right )^{2} - d x - a {\rm Ei}\left (\frac {2 \, a}{d x + c}\right ) + a {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right )}{2 \, d} \] Input:

integrate(sinh(a/(d*x+c))^2,x, algorithm="fricas")
 

Output:

1/2*((d*x + c)*cosh(a/(d*x + c))^2 + (d*x + c)*sinh(a/(d*x + c))^2 - d*x - 
 a*Ei(2*a/(d*x + c)) + a*Ei(-2*a/(d*x + c)))/d
 

Sympy [F]

\[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\int \sinh ^{2}{\left (\frac {a}{c + d x} \right )}\, dx \] Input:

integrate(sinh(a/(d*x+c))**2,x)
 

Output:

Integral(sinh(a/(c + d*x))**2, x)
 

Maxima [F]

\[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\int { \sinh \left (\frac {a}{d x + c}\right )^{2} \,d x } \] Input:

integrate(sinh(a/(d*x+c))^2,x, algorithm="maxima")
 

Output:

1/2*a*d*integrate(x*e^(2*a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) - 1/2* 
a*d*integrate(x*e^(-2*a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) + 1/4*x*e 
^(2*a/(d*x + c)) + 1/4*x*e^(-2*a/(d*x + c)) - 1/2*x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (39) = 78\).

Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 3.33 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=-\frac {{\left (\frac {2 \, a^{3} {\rm Ei}\left (\frac {2 \, a}{d x + c}\right ) e^{\left (\frac {2 \, a}{d x + c}\right )}}{d x + c} - \frac {2 \, a^{3} {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right ) e^{\left (\frac {2 \, a}{d x + c}\right )}}{d x + c} - a^{2} e^{\left (\frac {4 \, a}{d x + c}\right )} + 2 \, a^{2} e^{\left (\frac {2 \, a}{d x + c}\right )} - a^{2}\right )} {\left (d x + c\right )} e^{\left (-\frac {2 \, a}{d x + c}\right )}}{4 \, a^{2} d} \] Input:

integrate(sinh(a/(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/4*(2*a^3*Ei(2*a/(d*x + c))*e^(2*a/(d*x + c))/(d*x + c) - 2*a^3*Ei(-2*a/ 
(d*x + c))*e^(2*a/(d*x + c))/(d*x + c) - a^2*e^(4*a/(d*x + c)) + 2*a^2*e^( 
2*a/(d*x + c)) - a^2)*(d*x + c)*e^(-2*a/(d*x + c))/(a^2*d)
 

Mupad [F(-1)]

Timed out. \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\int {\mathrm {sinh}\left (\frac {a}{c+d\,x}\right )}^2 \,d x \] Input:

int(sinh(a/(c + d*x))^2,x)
 

Output:

int(sinh(a/(c + d*x))^2, x)
 

Reduce [F]

\[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {2 e^{\frac {4 a}{d x +c}} a \,d^{2} x^{2}-e^{\frac {4 a}{d x +c}} c^{3}-e^{\frac {4 a}{d x +c}} c^{2} d x -4 e^{\frac {2 a}{d x +c}} \left (\int \frac {x^{2}}{e^{\frac {2 a}{d x +c}} c^{3}+3 e^{\frac {2 a}{d x +c}} c^{2} d x +3 e^{\frac {2 a}{d x +c}} c \,d^{2} x^{2}+e^{\frac {2 a}{d x +c}} d^{3} x^{3}}d x \right ) a^{2} c \,d^{3}-4 e^{\frac {2 a}{d x +c}} \left (\int \frac {x^{2}}{e^{\frac {2 a}{d x +c}} c^{3}+3 e^{\frac {2 a}{d x +c}} c^{2} d x +3 e^{\frac {2 a}{d x +c}} c \,d^{2} x^{2}+e^{\frac {2 a}{d x +c}} d^{3} x^{3}}d x \right ) a^{2} d^{4} x +4 e^{\frac {2 a}{d x +c}} \left (\int \frac {e^{\frac {2 a}{d x +c}} x^{2}}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \right ) a^{2} c \,d^{3}+4 e^{\frac {2 a}{d x +c}} \left (\int \frac {e^{\frac {2 a}{d x +c}} x^{2}}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \right ) a^{2} d^{4} x -4 e^{\frac {2 a}{d x +c}} a c d x -4 e^{\frac {2 a}{d x +c}} a \,d^{2} x^{2}+2 a \,d^{2} x^{2}+c^{3}+c^{2} d x}{8 e^{\frac {2 a}{d x +c}} a d \left (d x +c \right )} \] Input:

int(sinh(a/(d*x+c))^2,x)
 

Output:

(2*e**((4*a)/(c + d*x))*a*d**2*x**2 - e**((4*a)/(c + d*x))*c**3 - e**((4*a 
)/(c + d*x))*c**2*d*x - 4*e**((2*a)/(c + d*x))*int(x**2/(e**((2*a)/(c + d* 
x))*c**3 + 3*e**((2*a)/(c + d*x))*c**2*d*x + 3*e**((2*a)/(c + d*x))*c*d**2 
*x**2 + e**((2*a)/(c + d*x))*d**3*x**3),x)*a**2*c*d**3 - 4*e**((2*a)/(c + 
d*x))*int(x**2/(e**((2*a)/(c + d*x))*c**3 + 3*e**((2*a)/(c + d*x))*c**2*d* 
x + 3*e**((2*a)/(c + d*x))*c*d**2*x**2 + e**((2*a)/(c + d*x))*d**3*x**3),x 
)*a**2*d**4*x + 4*e**((2*a)/(c + d*x))*int((e**((2*a)/(c + d*x))*x**2)/(c* 
*3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3),x)*a**2*c*d**3 + 4*e**((2*a)/ 
(c + d*x))*int((e**((2*a)/(c + d*x))*x**2)/(c**3 + 3*c**2*d*x + 3*c*d**2*x 
**2 + d**3*x**3),x)*a**2*d**4*x - 4*e**((2*a)/(c + d*x))*a*c*d*x - 4*e**(( 
2*a)/(c + d*x))*a*d**2*x**2 + 2*a*d**2*x**2 + c**3 + c**2*d*x)/(8*e**((2*a 
)/(c + d*x))*a*d*(c + d*x))