\(\int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx\) [118]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}} \]

[Out]

-arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))*a^(1/2)/d+a*sin(d*x+c)/d/(a-a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3890, 3859, 209} \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\frac {a \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d} \]

[In]

Int[Cos[c + d*x]*Sqrt[a - a*Sec[c + d*x]],x]

[Out]

-((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/d) + (a*Sin[c + d*x])/(d*Sqrt[a - a*Sec[c
+ d*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3890

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e
+ f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[a*((2*n + 1)/(2*b*d*n)), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {a \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {1}{2} \int \sqrt {a-a \sec (c+d x)} \, dx \\ & = \frac {a \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {a \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d} \\ & = -\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \sqrt {a-a \sec (c+d x)} \left (-\text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\cos (c+d x) \sqrt {1+\sec (c+d x)}\right )}{d \sqrt {1+\sec (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]*Sqrt[a - a*Sec[c + d*x]],x]

[Out]

-((Cot[(c + d*x)/2]*Sqrt[a - a*Sec[c + d*x]]*(-ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 + Sec[c +
d*x]]))/(d*Sqrt[1 + Sec[c + d*x]]))

Maple [A] (verified)

Time = 22.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (\arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\cos \left (d x +c \right )\right ) \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}\, \left (\cos \left (d x +c \right )+1\right ) \csc \left (d x +c \right )}{d}\)	\(82\)

[In]

int(cos(d*x+c)*(a-a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/d*(arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+cos(d*x+c))*(-a*(sec(d*x+c
)-1))^(1/2)*(cos(d*x+c)+1)*csc(d*x+c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (57) = 114\).

Time = 0.32 (sec) , antiderivative size = 294, normalized size of antiderivative = 4.52 \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\left [\frac {\sqrt {-a} \log \left (\frac {4 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, d \sin \left (d x + c\right )}, \frac {\sqrt {a} \arctan \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \, d \sin \left (d x + c\right )}\right ] \]

[In]

integrate(cos(d*x+c)*(a-a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(-a)*log((4*(2*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/
cos(d*x + c)) - (8*a*cos(d*x + c)^2 + 8*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*(cos(
d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(d*sin(d*x + c)), 1/2*(sqrt(a)*arctan(2*(c
os(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))/((2*a*cos(d*x + c) + a)*sin(d*x
+ c)))*sin(d*x + c) - 2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(d*sin(d*x +
c))]

Sympy [F]

\[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\int \sqrt {- a \left (\sec {\left (c + d x \right )} - 1\right )} \cos {\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)*(a-a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(-a*(sec(c + d*x) - 1))*cos(c + d*x), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (57) = 114\).

Time = 0.48 (sec) , antiderivative size = 791, normalized size of antiderivative = 12.17 \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)*(a-a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/4*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) + 1)))*sqrt(a) + sqrt(a)*(arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4
)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*
(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c
) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) + arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2
*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - ar
ctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (57) = 114\).

Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.06 \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {2 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{2 \, d} \]

[In]

integrate(cos(d*x+c)*(a-a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))*sgn(tan(1/2*d*x +
1/2*c)^3 + tan(1/2*d*x + 1/2*c)) - 2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a*sgn(tan(1/2*d*x + 1/2*c)^3 + tan(1/2
*d*x + 1/2*c))/(a*tan(1/2*d*x + 1/2*c)^2 + a))*sgn(cos(d*x + c))/d

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) \sqrt {a-a \sec (c+d x)} \, dx=\int \cos \left (c+d\,x\right )\,\sqrt {a-\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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