\(\int \sqrt {\cos (a+b x)} \, dx\) [76]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \sqrt {\cos (a+b x)} \, dx=\frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b} \]

[Out]

2*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2719} \[ \int \sqrt {\cos (a+b x)} \, dx=\frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b} \]

[In]

Int[Sqrt[Cos[a + b*x]],x]

[Out]

(2*EllipticE[(a + b*x)/2, 2])/b

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \sqrt {\cos (a+b x)} \, dx=\frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b} \]

[In]

Integrate[Sqrt[Cos[a + b*x]],x]

[Out]

(2*EllipticE[(a + b*x)/2, 2])/b

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(132\) vs. \(2(42)=84\).

Time = 1.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 8.31

method	result	size
default	\(\frac {2 \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, b}\)	\(133\)
risch	\(-\frac {i \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) {\mathrm e}^{-i \left (b x +a \right )}}}{b}-\frac {i \left (-\frac {2 \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{\sqrt {\left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) {\mathrm e}^{i \left (b x +a \right )}}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (b x +a \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}}\right ) \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) {\mathrm e}^{-i \left (b x +a \right )}}\, \sqrt {\left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) {\mathrm e}^{i \left (b x +a \right )}}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}\)	\(285\)

[In]

int(cos(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*((-1+2*cos(1/2*b*x+1/2*a)^2)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)
^2+1)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))/(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)/sin(1/2
*b*x+1/2*a)/(-1+2*cos(1/2*b*x+1/2*a)^2)^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 3.56 \[ \int \sqrt {\cos (a+b x)} \, dx=\frac {i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) - i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{b} \]

[In]

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

[Out]

(I*sqrt(2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) - I*sqrt(2)*weier
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))))/b

Sympy [F]

\[ \int \sqrt {\cos (a+b x)} \, dx=\int \sqrt {\cos {\left (a + b x \right )}}\, dx \]

[In]

integrate(cos(b*x+a)**(1/2),x)

[Out]

Integral(sqrt(cos(a + b*x)), x)

Maxima [F]

\[ \int \sqrt {\cos (a+b x)} \, dx=\int { \sqrt {\cos \left (b x + a\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(cos(b*x + a)), x)

Giac [F]

\[ \int \sqrt {\cos (a+b x)} \, dx=\int { \sqrt {\cos \left (b x + a\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(cos(b*x + a)), x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \sqrt {\cos (a+b x)} \, dx=\frac {2\,\mathrm {E}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{b} \]

[In]

int(cos(a + b*x)^(1/2),x)

[Out]

(2*ellipticE(a/2 + (b*x)/2, 2))/b