\(\int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [345]

Optimal result

Integrand size = 25, antiderivative size = 136 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 \sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

1/2*a*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+3/4*a*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c
)^(1/2)+3/4*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4307, 2849, 2853, 222} \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 \sqrt {a} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}+\frac {a \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {3 a \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[Sqrt[a + a*Cos[c + d*x]]/Sec[c + d*x]^(3/2),x]

[Out]

(3*Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(4*d
) + (a*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) + (3*a*Sin[c + d*x])/(4*d*Sqrt[a + a*Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])

Rule 222

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

Rule 2849

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[2*n*((b*c + a*d)
/(b*(2*n + 1))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 4307

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {a \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{4} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {a \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {1}{8} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {a \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d} \\ & = \frac {3 \sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{8 d} \]

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]/Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(3*Sqrt[2]*ArcSin[Sqrt[2]*S
in[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(2*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(8*d)

Maple [A] (verified)

Time = 14.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \sec \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(145\)

[In]

int((a+cos(d*x+c)*a)^(1/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/d*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(2*sin(d*x+c)
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*sec(d*x+c)*arctan(tan(d*x+
c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {\sqrt {a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/4*(3*sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))
- sqrt(a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c
) + d)

Sympy [F]

\[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+a*cos(d*x+c))**(1/2)/sec(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))/sec(c + d*x)**(3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1059 vs. \(2 (112) = 224\).

Time = 0.43 (sec) , antiderivative size = 1059, normalized size of antiderivative = 7.79 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/16*(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)))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c) -
2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))) - cos(2*d*x + 2*c) + 2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) +
3*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)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (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))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 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)))*sin(1/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (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))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c)))) - 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) + arctan2((cos(2*d*x + 2*c)^2 + si
n(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)), (co
s(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)))/d

Giac [F]

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

[In]

integrate((a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*cos(d*x + c) + a)/sec(d*x + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((a + a*cos(c + d*x))^(1/2)/(1/cos(c + d*x))^(3/2),x)

[Out]

int((a + a*cos(c + d*x))^(1/2)/(1/cos(c + d*x))^(3/2), x)