\(\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\) [344]
Optimal result
Integrand size = 25, antiderivative size = 92 \[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}
\]
[Out]
a*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+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 = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4307, 2849, 2853, 222}
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}
\]
[In]
Int[Sqrt[a + a*Cos[c + d*x]]/Sqrt[Sec[c + d*x]],x]
[Out]
(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]])/d + (a
*Sin[c + d*x])/(d*Sqrt[a + a*Cos[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 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {a \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {1}{2} \left (\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)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (\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 )}{d} \\ & = \frac {\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)}}{d}+\frac {a \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}
\]
[In]
Integrate[Sqrt[a + a*Cos[c + d*x]]/Sqrt[Sec[c + d*x]],x]
[Out]
(Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(Sqrt[2]*ArcSin[Sqrt[2]*Sin
[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*Sin[(c + d*x)/2]))/(2*d)
Maple [A] (verified)
Time = 14.16 (sec) , antiderivative size = 108, normalized size of antiderivative =
1.17
| | |
method | result | size |
| | |
default |
\(\frac {\left (\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) |
\(108\) |
| | |
|
|
|
[In]
int((a+cos(d*x+c)*a)^(1/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/d*(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*(a*(1+
cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
Fricas [A] (verification not implemented)
none
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d}
\]
[In]
integrate((a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
-(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)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)
Sympy [F]
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx
\]
[In]
integrate((a+a*cos(d*x+c))**(1/2)/sec(d*x+c)**(1/2),x)
[Out]
Integral(sqrt(a*(cos(c + d*x) + 1))/sqrt(sec(c + d*x)), x)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (78) = 156\).
Time = 0.41 (sec) , antiderivative size = 791, normalized size of antiderivative = 8.60
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display}
\]
[In]
integrate((a+a*cos(d*x+c))^(1/2)/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*arct
an2(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*arctan
2(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) + arc
tan2((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*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1)))/d
Giac [F]
\[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {\sqrt {a \cos \left (d x + c\right ) + a}}{\sqrt {\sec \left (d x + c\right )}} \,d x }
\]
[In]
integrate((a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(a*cos(d*x + c) + a)/sqrt(sec(d*x + c)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x
\]
[In]
int((a + a*cos(c + d*x))^(1/2)/(1/cos(c + d*x))^(1/2),x)
[Out]
int((a + a*cos(c + d*x))^(1/2)/(1/cos(c + d*x))^(1/2), x)