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

Optimal. Leaf size=94 \[ \frac{b^2 \sin (a) \text{CosIntegral}\left (\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b^2 \cos (a) \text{Si}\left (\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d} \]

[Out]

(b*Sqrt[c + d*x]*Cos[a + b/Sqrt[c + d*x]])/d + (b^2*CosIntegral[b/Sqrt[c + d*x]]*Sin[a])/d + ((c + d*x)*Sin[a
+ b/Sqrt[c + d*x]])/d + (b^2*Cos[a]*SinIntegral[b/Sqrt[c + d*x]])/d

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Rubi [A]  time = 0.118033, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3361, 3297, 3303, 3299, 3302} \[ \frac{b^2 \sin (a) \text{CosIntegral}\left (\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b^2 \cos (a) \text{Si}\left (\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/Sqrt[c + d*x]],x]

[Out]

(b*Sqrt[c + d*x]*Cos[a + b/Sqrt[c + d*x]])/d + (b^2*CosIntegral[b/Sqrt[c + d*x]]*Sin[a])/d + ((c + d*x)*Sin[a
+ b/Sqrt[c + d*x]])/d + (b^2*Cos[a]*SinIntegral[b/Sqrt[c + d*x]])/d

Rule 3361

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

Rule 3297

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] - Dist[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 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 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \sin \left (a+\frac{b}{\sqrt{c+d x}}\right ) \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^3} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^2} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{\left (b^2 \cos (a)\right ) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}+\frac{\left (b^2 \sin (a)\right ) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b^2 \text{Ci}\left (\frac{b}{\sqrt{c+d x}}\right ) \sin (a)}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d}+\frac{b^2 \cos (a) \text{Si}\left (\frac{b}{\sqrt{c+d x}}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0762403, size = 99, normalized size = 1.05 \[ \frac{b^2 \sin (a) \text{CosIntegral}\left (\frac{b}{\sqrt{c+d x}}\right )+b^2 \cos (a) \text{Si}\left (\frac{b}{\sqrt{c+d x}}\right )+c \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )+d x \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )+b \sqrt{c+d x} \cos \left (a+\frac{b}{\sqrt{c+d x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/Sqrt[c + d*x]],x]

[Out]

(b*Sqrt[c + d*x]*Cos[a + b/Sqrt[c + d*x]] + b^2*CosIntegral[b/Sqrt[c + d*x]]*Sin[a] + c*Sin[a + b/Sqrt[c + d*x
]] + d*x*Sin[a + b/Sqrt[c + d*x]] + b^2*Cos[a]*SinIntegral[b/Sqrt[c + d*x]])/d

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Maple [A]  time = 0.015, size = 84, normalized size = 0.9 \begin{align*} -2\,{\frac{{b}^{2}}{d} \left ( -1/2\,{\frac{dx+c}{{b}^{2}}\sin \left ( a+{\frac{b}{\sqrt{dx+c}}} \right ) }-1/2\,{\frac{\sqrt{dx+c}}{b}\cos \left ( a+{\frac{b}{\sqrt{dx+c}}} \right ) }-1/2\,{\it Si} \left ({\frac{b}{\sqrt{dx+c}}} \right ) \cos \left ( a \right ) -1/2\,{\it Ci} \left ({\frac{b}{\sqrt{dx+c}}} \right ) \sin \left ( a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^2*(-1/2*sin(a+b/(d*x+c)^(1/2))*(d*x+c)/b^2-1/2*cos(a+b/(d*x+c)^(1/2))*(d*x+c)^(1/2)/b-1/2*Si(b/(d*x+c)^
(1/2))*cos(a)-1/2*Ci(b/(d*x+c)^(1/2))*sin(a))

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Maxima [C]  time = 1.20623, size = 167, normalized size = 1.78 \begin{align*} \frac{{\left ({\left (-i \,{\rm Ei}\left (\frac{i \, b}{\sqrt{d x + c}}\right ) + i \,{\rm Ei}\left (-\frac{i \, b}{\sqrt{d x + c}}\right )\right )} \cos \left (a\right ) +{\left ({\rm Ei}\left (\frac{i \, b}{\sqrt{d x + c}}\right ) +{\rm Ei}\left (-\frac{i \, b}{\sqrt{d x + c}}\right )\right )} \sin \left (a\right )\right )} b^{2} + 2 \, \sqrt{d x + c} b \cos \left (\frac{\sqrt{d x + c} a + b}{\sqrt{d x + c}}\right ) + 2 \,{\left (d x + c\right )} \sin \left (\frac{\sqrt{d x + c} a + b}{\sqrt{d x + c}}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((-I*Ei(I*b/sqrt(d*x + c)) + I*Ei(-I*b/sqrt(d*x + c)))*cos(a) + (Ei(I*b/sqrt(d*x + c)) + Ei(-I*b/sqrt(d*x
+ c)))*sin(a))*b^2 + 2*sqrt(d*x + c)*b*cos((sqrt(d*x + c)*a + b)/sqrt(d*x + c)) + 2*(d*x + c)*sin((sqrt(d*x +
c)*a + b)/sqrt(d*x + c)))/d

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Fricas [A]  time = 2.16723, size = 360, normalized size = 3.83 \begin{align*} \frac{b^{2} \operatorname{Ci}\left (\frac{b}{\sqrt{d x + c}}\right ) \sin \left (a\right ) + b^{2} \operatorname{Ci}\left (-\frac{b}{\sqrt{d x + c}}\right ) \sin \left (a\right ) + 2 \, b^{2} \cos \left (a\right ) \operatorname{Si}\left (\frac{b}{\sqrt{d x + c}}\right ) + 2 \, \sqrt{d x + c} b \cos \left (\frac{a d x + a c + \sqrt{d x + c} b}{d x + c}\right ) + 2 \,{\left (d x + c\right )} \sin \left (\frac{a d x + a c + \sqrt{d x + c} b}{d x + c}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*cos_integral(b/sqrt(d*x + c))*sin(a) + b^2*cos_integral(-b/sqrt(d*x + c))*sin(a) + 2*b^2*cos(a)*sin_i
ntegral(b/sqrt(d*x + c)) + 2*sqrt(d*x + c)*b*cos((a*d*x + a*c + sqrt(d*x + c)*b)/(d*x + c)) + 2*(d*x + c)*sin(
(a*d*x + a*c + sqrt(d*x + c)*b)/(d*x + c)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/sqrt(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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