\(\int \cos ^{\frac {3}{2}}(a+b x) \, dx\) [79]

Optimal result

Integrand size = 10, antiderivative size = 42 \[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b} \]

[Out]

2/3*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))/b+2/3*sin(b*x+a)*cos
(b*x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2715, 2720} \[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b}+\frac {2 \sin (a+b x) \sqrt {\cos (a+b x)}}{3 b} \]

[In]

Int[Cos[a + b*x]^(3/2),x]

[Out]

(2*EllipticF[(a + b*x)/2, 2])/(3*b) + (2*Sqrt[Cos[a + b*x]]*Sin[a + b*x])/(3*b)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b}+\frac {1}{3} \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx \\ & = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \left (\operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+\sqrt {\cos (a+b x)} \sin (a+b x)\right )}{3 b} \]

[In]

Integrate[Cos[a + b*x]^(3/2),x]

[Out]

(2*(EllipticF[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*Sin[a + b*x]))/(3*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(62)=124\).

Time = 2.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 4.26

method	result	size
default	\(-\frac {2 \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, b}\)	\(179\)

[In]

int(cos(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*((-1+2*cos(1/2*b*x+1/2*a)^2)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)-2*sin
(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a)+(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(c
os(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 = 68, normalized size of antiderivative = 1.62 \[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \, \sqrt {\cos \left (b x + a\right )} \sin \left (b x + a\right ) - i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{3 \, b} \]

[In]

integrate(cos(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

1/3*(2*sqrt(cos(b*x + a))*sin(b*x + a) - I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) +
I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)))/b

Sympy [F]

\[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\int \cos ^{\frac {3}{2}}{\left (a + b x \right )}\, dx \]

[In]

integrate(cos(b*x+a)**(3/2),x)

[Out]

Integral(cos(a + b*x)**(3/2), x)

Maxima [F]

\[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\int { \cos \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(3/2), x)

Giac [F]

\[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\int { \cos \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \cos ^{\frac {3}{2}}(a+b x) \, dx=\frac {2\,\mathrm {F}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{3\,b}+\frac {2\,\sqrt {\cos \left (a+b\,x\right )}\,\sin \left (a+b\,x\right )}{3\,b} \]

[In]

int(cos(a + b*x)^(3/2),x)

[Out]

(2*ellipticF(a/2 + (b*x)/2, 2))/(3*b) + (2*cos(a + b*x)^(1/2)*sin(a + b*x))/(3*b)