\(\int \frac {\sinh (a+b \sqrt {c+d x})}{x^2} \, dx\) [97]

Optimal result

Integrand size = 18, antiderivative size = 184 \[ \int \frac {\sinh \left (a+b \sqrt {c+d x}\right )}{x^2} \, dx=-\frac {b d \cosh \left (a-b \sqrt {c}\right ) \text {Chi}\left (b \left (\sqrt {c}+\sqrt {c+d x}\right )\right )}{2 \sqrt {c}}+\frac {b d \cosh \left (a+b \sqrt {c}\right ) \text {Chi}\left (b \sqrt {c}-b \sqrt {c+d x}\right )}{2 \sqrt {c}}-\frac {\sinh \left (a+b \sqrt {c+d x}\right )}{x}-\frac {b d \sinh \left (a-b \sqrt {c}\right ) \text {Shi}\left (b \left (\sqrt {c}+\sqrt {c+d x}\right )\right )}{2 \sqrt {c}}-\frac {b d \sinh \left (a+b \sqrt {c}\right ) \text {Shi}\left (b \sqrt {c}-b \sqrt {c+d x}\right )}{2 \sqrt {c}} \] Output:

-1/2*b*d*cosh(a-b*c^(1/2))*Chi(b*(c^(1/2)+(d*x+c)^(1/2)))/c^(1/2)+1/2*b*d* 
cosh(a+b*c^(1/2))*Chi(b*c^(1/2)-b*(d*x+c)^(1/2))/c^(1/2)-sinh(a+b*(d*x+c)^ 
(1/2))/x-1/2*b*d*sinh(a-b*c^(1/2))*Shi(b*(c^(1/2)+(d*x+c)^(1/2)))/c^(1/2)- 
1/2*b*d*sinh(a+b*c^(1/2))*Shi(b*c^(1/2)-b*(d*x+c)^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 3.25 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.11 \[ \int \frac {\sinh \left (a+b \sqrt {c+d x}\right )}{x^2} \, dx=\frac {e^{-a} \left (2 \sqrt {c} e^{-b \sqrt {c+d x}}-2 \sqrt {c} e^{2 a+b \sqrt {c+d x}}+b d e^{-b \sqrt {c}} x \operatorname {ExpIntegralEi}\left (b \left (\sqrt {c}-\sqrt {c+d x}\right )\right )+b d e^{2 a+b \sqrt {c}} x \operatorname {ExpIntegralEi}\left (b \left (-\sqrt {c}+\sqrt {c+d x}\right )\right )-b d e^{b \sqrt {c}} x \operatorname {ExpIntegralEi}\left (-b \left (\sqrt {c}+\sqrt {c+d x}\right )\right )-b d e^{2 a-b \sqrt {c}} x \operatorname {ExpIntegralEi}\left (b \left (\sqrt {c}+\sqrt {c+d x}\right )\right )\right )}{4 \sqrt {c} x} \] Input:

Integrate[Sinh[a + b*Sqrt[c + d*x]]/x^2,x]
 

Output:

((2*Sqrt[c])/E^(b*Sqrt[c + d*x]) - 2*Sqrt[c]*E^(2*a + b*Sqrt[c + d*x]) + ( 
b*d*x*ExpIntegralEi[b*(Sqrt[c] - Sqrt[c + d*x])])/E^(b*Sqrt[c]) + b*d*E^(2 
*a + b*Sqrt[c])*x*ExpIntegralEi[b*(-Sqrt[c] + Sqrt[c + d*x])] - b*d*E^(b*S 
qrt[c])*x*ExpIntegralEi[-(b*(Sqrt[c] + Sqrt[c + d*x]))] - b*d*E^(2*a - b*S 
qrt[c])*x*ExpIntegralEi[b*(Sqrt[c] + Sqrt[c + d*x])])/(4*Sqrt[c]*E^a*x)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5887, 7267, 5811, 5804, 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.

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

\(\Big \downarrow \) 5887

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 5811

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

\(\Big \downarrow \) 5804

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 d \left (-\frac {1}{2} b \left (-\frac {\cosh \left (a+b \sqrt {c}\right ) \text {Chi}\left (b \sqrt {c}-b \sqrt {c+d x}\right )}{2 \sqrt {c}}+\frac {\cosh \left (a-b \sqrt {c}\right ) \text {Chi}\left (\sqrt {c} b+\sqrt {c+d x} b\right )}{2 \sqrt {c}}+\frac {\sinh \left (a+b \sqrt {c}\right ) \text {Shi}\left (b \sqrt {c}-b \sqrt {c+d x}\right )}{2 \sqrt {c}}+\frac {\sinh \left (a-b \sqrt {c}\right ) \text {Shi}\left (\sqrt {c} b+\sqrt {c+d x} b\right )}{2 \sqrt {c}}\right )-\frac {\sinh \left (a+b \sqrt {c+d x}\right )}{2 d x}\right )\)

Input:

Int[Sinh[a + b*Sqrt[c + d*x]]/x^2,x]
 

Output:

2*d*(-1/2*Sinh[a + b*Sqrt[c + d*x]]/(d*x) - (b*(-1/2*(Cosh[a + b*Sqrt[c]]* 
CoshIntegral[b*Sqrt[c] - b*Sqrt[c + d*x]])/Sqrt[c] + (Cosh[a - b*Sqrt[c]]* 
CoshIntegral[b*Sqrt[c] + b*Sqrt[c + d*x]])/(2*Sqrt[c]) + (Sinh[a + b*Sqrt[ 
c]]*SinhIntegral[b*Sqrt[c] - b*Sqrt[c + d*x]])/(2*Sqrt[c]) + (Sinh[a - b*S 
qrt[c]]*SinhIntegral[b*Sqrt[c] + b*Sqrt[c + d*x]])/(2*Sqrt[c])))/2)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5804

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

rule 5811

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*Sinh[(c_.) + (d_.)*(x_ 
)], x_Symbol] :> Simp[e^m*(a + b*x^n)^(p + 1)*(Sinh[c + d*x]/(b*n*(p + 1))) 
, x] - Simp[d*(e^m/(b*n*(p + 1)))   Int[(a + b*x^n)^(p + 1)*Cosh[c + d*x], 
x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IntegerQ[p] && EqQ[m - n + 1, 
0] && LtQ[p, -1] && (IntegerQ[n] || GtQ[e, 0])
 

rule 5887

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

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [F]

\[\int \frac {\sinh \left (a +b \sqrt {d x +c}\right )}{x^{2}}d x\]

Input:

int(sinh(a+b*(d*x+c)^(1/2))/x^2,x)
 

Output:

int(sinh(a+b*(d*x+c)^(1/2))/x^2,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (142) = 284\).

Time = 0.10 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.71 \[ \int \frac {\sinh \left (a+b \sqrt {c+d x}\right )}{x^2} \, dx=\frac {{\left (\sqrt {b^{2} c} d x {\rm Ei}\left (\sqrt {d x + c} b - \sqrt {b^{2} c}\right ) + \sqrt {b^{2} c} d x {\rm Ei}\left (-\sqrt {d x + c} b + \sqrt {b^{2} c}\right )\right )} \cosh \left (a + \sqrt {b^{2} c}\right ) - {\left (\sqrt {b^{2} c} d x {\rm Ei}\left (\sqrt {d x + c} b + \sqrt {b^{2} c}\right ) + \sqrt {b^{2} c} d x {\rm Ei}\left (-\sqrt {d x + c} b - \sqrt {b^{2} c}\right )\right )} \cosh \left (-a + \sqrt {b^{2} c}\right ) - 4 \, c \sinh \left (\sqrt {d x + c} b + a\right ) + {\left (\sqrt {b^{2} c} d x {\rm Ei}\left (\sqrt {d x + c} b - \sqrt {b^{2} c}\right ) - \sqrt {b^{2} c} d x {\rm Ei}\left (-\sqrt {d x + c} b + \sqrt {b^{2} c}\right )\right )} \sinh \left (a + \sqrt {b^{2} c}\right ) + {\left (\sqrt {b^{2} c} d x {\rm Ei}\left (\sqrt {d x + c} b + \sqrt {b^{2} c}\right ) - \sqrt {b^{2} c} d x {\rm Ei}\left (-\sqrt {d x + c} b - \sqrt {b^{2} c}\right )\right )} \sinh \left (-a + \sqrt {b^{2} c}\right )}{4 \, c x} \] Input:

integrate(sinh(a+b*(d*x+c)^(1/2))/x^2,x, algorithm="fricas")
 

Output:

1/4*((sqrt(b^2*c)*d*x*Ei(sqrt(d*x + c)*b - sqrt(b^2*c)) + sqrt(b^2*c)*d*x* 
Ei(-sqrt(d*x + c)*b + sqrt(b^2*c)))*cosh(a + sqrt(b^2*c)) - (sqrt(b^2*c)*d 
*x*Ei(sqrt(d*x + c)*b + sqrt(b^2*c)) + sqrt(b^2*c)*d*x*Ei(-sqrt(d*x + c)*b 
 - sqrt(b^2*c)))*cosh(-a + sqrt(b^2*c)) - 4*c*sinh(sqrt(d*x + c)*b + a) + 
(sqrt(b^2*c)*d*x*Ei(sqrt(d*x + c)*b - sqrt(b^2*c)) - sqrt(b^2*c)*d*x*Ei(-s 
qrt(d*x + c)*b + sqrt(b^2*c)))*sinh(a + sqrt(b^2*c)) + (sqrt(b^2*c)*d*x*Ei 
(sqrt(d*x + c)*b + sqrt(b^2*c)) - sqrt(b^2*c)*d*x*Ei(-sqrt(d*x + c)*b - sq 
rt(b^2*c)))*sinh(-a + sqrt(b^2*c)))/(c*x)
 

Sympy [F]

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

integrate(sinh(a+b*(d*x+c)**(1/2))/x**2,x)
 

Output:

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

Maxima [F]

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

integrate(sinh(a+b*(d*x+c)^(1/2))/x^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(sinh(a+b*(d*x+c)^(1/2))/x^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

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

Mupad [F(-1)]

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

int(sinh(a + b*(c + d*x)^(1/2))/x^2,x)
 

Output:

int(sinh(a + b*(c + d*x)^(1/2))/x^2, x)
 

Reduce [F]

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

int(sinh(a+b*(d*x+c)^(1/2))/x^2,x)
 

Output:

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