3.96 \(\int \frac{\sinh (a+b \sqrt{c+d x})}{x} \, dx\)

Optimal. Leaf size=124 \[ \sinh \left (a-b \sqrt{c}\right ) \text{Chi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right )+\sinh \left (a+b \sqrt{c}\right ) \text{Chi}\left (b \left (\sqrt{c}-\sqrt{c+d x}\right )\right )-\cosh \left (a+b \sqrt{c}\right ) \text{Shi}\left (b \left (\sqrt{c}-\sqrt{c+d x}\right )\right )+\cosh \left (a-b \sqrt{c}\right ) \text{Shi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right ) \]

[Out]

CoshIntegral[b*(Sqrt[c] + Sqrt[c + d*x])]*Sinh[a - b*Sqrt[c]] + CoshIntegral[b*(Sqrt[c] - Sqrt[c + d*x])]*Sinh
[a + b*Sqrt[c]] - Cosh[a + b*Sqrt[c]]*SinhIntegral[b*(Sqrt[c] - Sqrt[c + d*x])] + Cosh[a - b*Sqrt[c]]*SinhInte
gral[b*(Sqrt[c] + Sqrt[c + d*x])]

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Rubi [A]  time = 0.286707, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5364, 5292, 3303, 3298, 3301} \[ \sinh \left (a-b \sqrt{c}\right ) \text{Chi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right )+\sinh \left (a+b \sqrt{c}\right ) \text{Chi}\left (b \left (\sqrt{c}-\sqrt{c+d x}\right )\right )-\cosh \left (a+b \sqrt{c}\right ) \text{Shi}\left (b \left (\sqrt{c}-\sqrt{c+d x}\right )\right )+\cosh \left (a-b \sqrt{c}\right ) \text{Shi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Sqrt[c + d*x]]/x,x]

[Out]

CoshIntegral[b*(Sqrt[c] + Sqrt[c + d*x])]*Sinh[a - b*Sqrt[c]] + CoshIntegral[b*(Sqrt[c] - Sqrt[c + d*x])]*Sinh
[a + b*Sqrt[c]] - Cosh[a + b*Sqrt[c]]*SinhIntegral[b*(Sqrt[c] - Sqrt[c + d*x])] + Cosh[a - b*Sqrt[c]]*SinhInte
gral[b*(Sqrt[c] + Sqrt[c + d*x])]

Rule 5364

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Dist[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 5292

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[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 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> 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 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> 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]

Rubi steps

\begin{align*} \int \frac{\sinh \left (a+b \sqrt{c+d x}\right )}{x} \, dx &=\operatorname{Subst}\left (\int \frac{\sinh \left (a+b \sqrt{x}\right )}{-c+x} \, dx,x,c+d x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x \sinh (a+b x)}{-c+x^2} \, dx,x,\sqrt{c+d x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (-\frac{\sinh (a+b x)}{2 \left (\sqrt{c}-x\right )}+\frac{\sinh (a+b x)}{2 \left (\sqrt{c}+x\right )}\right ) \, dx,x,\sqrt{c+d x}\right )\\ &=-\operatorname{Subst}\left (\int \frac{\sinh (a+b x)}{\sqrt{c}-x} \, dx,x,\sqrt{c+d x}\right )+\operatorname{Subst}\left (\int \frac{\sinh (a+b x)}{\sqrt{c}+x} \, dx,x,\sqrt{c+d x}\right )\\ &=\cosh \left (a-b \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (b \sqrt{c}+b x\right )}{\sqrt{c}+x} \, dx,x,\sqrt{c+d x}\right )+\cosh \left (a+b \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (b \sqrt{c}-b x\right )}{\sqrt{c}-x} \, dx,x,\sqrt{c+d x}\right )+\sinh \left (a-b \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (b \sqrt{c}+b x\right )}{\sqrt{c}+x} \, dx,x,\sqrt{c+d x}\right )-\sinh \left (a+b \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (b \sqrt{c}-b x\right )}{\sqrt{c}-x} \, dx,x,\sqrt{c+d x}\right )\\ &=\text{Chi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right ) \sinh \left (a-b \sqrt{c}\right )+\text{Chi}\left (b \sqrt{c}-b \sqrt{c+d x}\right ) \sinh \left (a+b \sqrt{c}\right )+\cosh \left (a-b \sqrt{c}\right ) \text{Shi}\left (b \left (\sqrt{c}+\sqrt{c+d x}\right )\right )-\cosh \left (a+b \sqrt{c}\right ) \text{Shi}\left (b \sqrt{c}-b \sqrt{c+d x}\right )\\ \end{align*}

Mathematica [A]  time = 0.963794, size = 130, normalized size = 1.05 \[ \frac{1}{2} e^{-a-b \sqrt{c}} \left (e^{2 \left (a+b \sqrt{c}\right )} \text{ExpIntegralEi}\left (b \left (\sqrt{c+d x}-\sqrt{c}\right )\right )+e^{2 a} \text{ExpIntegralEi}\left (b \left (\sqrt{c+d x}+\sqrt{c}\right )\right )-\text{ExpIntegralEi}\left (b \left (\sqrt{c}-\sqrt{c+d x}\right )\right )-e^{2 b \sqrt{c}} \text{ExpIntegralEi}\left (-b \left (\sqrt{c+d x}+\sqrt{c}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-a - b*Sqrt[c])*(-ExpIntegralEi[b*(Sqrt[c] - Sqrt[c + d*x])] + E^(2*(a + b*Sqrt[c]))*ExpIntegralEi[b*(-Sqr
t[c] + Sqrt[c + d*x])] - E^(2*b*Sqrt[c])*ExpIntegralEi[-(b*(Sqrt[c] + Sqrt[c + d*x]))] + E^(2*a)*ExpIntegralEi
[b*(Sqrt[c] + Sqrt[c + d*x])]))/2

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sinh \left ( a+b\sqrt{dx+c} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (\sqrt{d x + c} b + a\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.1562, size = 547, normalized size = 4.41 \begin{align*} \frac{1}{2} \,{\left ({\rm Ei}\left (\sqrt{d x + c} b - \sqrt{b^{2} c}\right ) -{\rm Ei}\left (-\sqrt{d x + c} b + \sqrt{b^{2} c}\right )\right )} \cosh \left (a + \sqrt{b^{2} c}\right ) + \frac{1}{2} \,{\left ({\rm Ei}\left (\sqrt{d x + c} b + \sqrt{b^{2} c}\right ) -{\rm Ei}\left (-\sqrt{d x + c} b - \sqrt{b^{2} c}\right )\right )} \cosh \left (-a + \sqrt{b^{2} c}\right ) + \frac{1}{2} \,{\left ({\rm Ei}\left (\sqrt{d x + c} b - \sqrt{b^{2} c}\right ) +{\rm Ei}\left (-\sqrt{d x + c} b + \sqrt{b^{2} c}\right )\right )} \sinh \left (a + \sqrt{b^{2} c}\right ) - \frac{1}{2} \,{\left ({\rm Ei}\left (\sqrt{d x + c} b + \sqrt{b^{2} c}\right ) +{\rm Ei}\left (-\sqrt{d x + c} b - \sqrt{b^{2} c}\right )\right )} \sinh \left (-a + \sqrt{b^{2} c}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b \sqrt{c + d x} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (\sqrt{d x + c} b + a\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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