3.92 \(\int \cos (a+b \sqrt{c+d x}) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0274606, 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 = {3362, 3296, 2638} \[ \frac{2 \cos \left (a+b \sqrt{c+d x}\right )}{b^2 d}+\frac{2 \sqrt{c+d x} \sin \left (a+b \sqrt{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*Sqrt[c + d*x]],x]

[Out]

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

Rule 3362

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Cos[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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.0734567, size = 48, normalized size = 0.89 \[ \frac{2 \left (b \sqrt{c+d x} \sin \left (a+b \sqrt{c+d x}\right )+\cos \left (a+b \sqrt{c+d x}\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*Sqrt[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.033, size = 61, normalized size = 1.1 \begin{align*} 2\,{\frac{\cos \left ( a+b\sqrt{dx+c} \right ) + \left ( a+b\sqrt{dx+c} \right ) \sin \left ( a+b\sqrt{dx+c} \right ) -a\sin \left ( a+b\sqrt{dx+c} \right ) }{{b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15409, size = 81, normalized size = 1.5 \begin{align*} \frac{2 \,{\left ({\left (\sqrt{d x + c} b + a\right )} \sin \left (\sqrt{d x + c} b + a\right ) - a \sin \left (\sqrt{d x + c} b + a\right ) + \cos \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(cos(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57493, size = 109, normalized size = 2.02 \begin{align*} \frac{2 \,{\left (\sqrt{d x + c} b \sin \left (\sqrt{d x + c} b + a\right ) + \cos \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(cos(a+b*(d*x+c)^(1/2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.535975, size = 66, normalized size = 1.22 \begin{align*} \begin{cases} x \cos{\left (a \right )} & \text{for}\: b = 0 \wedge d = 0 \\x \cos{\left (a + b \sqrt{c} \right )} & \text{for}\: d = 0 \\x \cos{\left (a \right )} & \text{for}\: b = 0 \\\frac{2 \sqrt{c + d x} \sin{\left (a + b \sqrt{c + d x} \right )}}{b d} + \frac{2 \cos{\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(cos(a+b*(d*x+c)**(1/2)),x)

[Out]

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

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Giac [B]  time = 1.18209, size = 227, normalized size = 4.2 \begin{align*} -\frac{2 \,{\left ({\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )} \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 ) - \frac{b \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 )}{\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(cos(a+b*(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

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