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

Optimal. Leaf size=54 \[ \frac{2 \sin \left (a+b \sqrt{c+d x}\right )}{b^2 d}-\frac{2 \sqrt{c+d x} \cos \left (a+b \sqrt{c+d x}\right )}{b d} \]

[Out]

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

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Rubi [A]  time = 0.0277695, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3361, 3296, 2637} \[ \frac{2 \sin \left (a+b \sqrt{c+d x}\right )}{b^2 d}-\frac{2 \sqrt{c+d x} \cos \left (a+b \sqrt{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*Sqrt[c + d*x]],x]

[Out]

(-2*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b*d) + (2*Sin[a + b*Sqrt[c + d*x]])/(b^2*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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0738688, size = 50, normalized size = 0.93 \[ \frac{2 \sin \left (a+b \sqrt{c+d x}\right )-2 b \sqrt{c+d x} \cos \left (a+b \sqrt{c+d x}\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*Sqrt[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.007, size = 61, normalized size = 1.1 \begin{align*} 2\,{\frac{\sin \left ( a+b\sqrt{dx+c} \right ) - \left ( a+b\sqrt{dx+c} \right ) \cos \left ( a+b\sqrt{dx+c} \right ) +a\cos \left ( a+b\sqrt{dx+c} \right ) }{d{b}^{2}}} \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*(sin(a+b*(d*x+c)^(1/2))-(a+b*(d*x+c)^(1/2))*cos(a+b*(d*x+c)^(1/2))+a*cos(a+b*(d*x+c)^(1/2)))

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Maxima [A]  time = 0.950191, size = 84, normalized size = 1.56 \begin{align*} -\frac{2 \,{\left ({\left (\sqrt{d x + c} b + a\right )} \cos \left (\sqrt{d x + c} b + a\right ) - a \cos \left (\sqrt{d x + c} b + a\right ) - \sin \left (\sqrt{d x + c} b + a\right )\right )}}{b^{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]

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

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Fricas [A]  time = 1.64205, size = 111, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (\sqrt{d x + c} b \cos \left (\sqrt{d x + c} b + a\right ) - \sin \left (\sqrt{d x + c} b + a\right )\right )}}{b^{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]

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

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Sympy [A]  time = 0.4912, size = 66, normalized size = 1.22 \begin{align*} \begin{cases} x \sin{\left (a \right )} & \text{for}\: b = 0 \wedge d = 0 \\x \sin{\left (a + b \sqrt{c} \right )} & \text{for}\: d = 0 \\x \sin{\left (a \right )} & \text{for}\: b = 0 \\- \frac{2 \sqrt{c + d x} \cos{\left (a + b \sqrt{c + d x} \right )}}{b d} + \frac{2 \sin{\left (a + b \sqrt{c + d x} \right )}}{b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sin(a), Eq(b, 0) & Eq(d, 0)), (x*sin(a + b*sqrt(c)), Eq(d, 0)), (x*sin(a), Eq(b, 0)), (-2*sqrt(c
+ d*x)*cos(a + b*sqrt(c + d*x))/(b*d) + 2*sin(a + b*sqrt(c + d*x))/(b**2*d), True))

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Giac [B]  time = 1.20166, size = 225, normalized size = 4.17 \begin{align*} -\frac{2 \,{\left ({\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )} \cos \left (-{\left (\sqrt{d x + c} b + a\right )} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) + a \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - a\right ) + \frac{b \sin \left (-{\left (\sqrt{d x + c} b + a\right )} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) + a \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - a\right )}{\mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )}\right )}}{b^{3} 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="giac")

[Out]

-2*(((sqrt(d*x + c)*b + a)*b - a*b)*cos(-(sqrt(d*x + c)*b + a)*sgn((sqrt(d*x + c)*b + a)*b - a*b) + a*sgn((sqr
t(d*x + c)*b + a)*b - a*b) - a) + b*sin(-(sqrt(d*x + c)*b + a)*sgn((sqrt(d*x + c)*b + a)*b - a*b) + a*sgn((sqr
t(d*x + c)*b + a)*b - a*b) - a)/sgn((sqrt(d*x + c)*b + a)*b - a*b))/(b^3*d)