\(\int \sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)} \, dx\) [1209]

Optimal result

Integrand size = 37, antiderivative size = 144 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\frac {\sqrt {a} (8 A+3 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a C \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}} \]

[Out]

1/4*a*C*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/2*C*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x
+c)^(1/2)+1/4*(8*A+3*C)*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 = 0.49 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4306, 3125, 3060, 2853, 222} \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\frac {\sqrt {a} (8 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}+\frac {C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d \sqrt {\sec (c+d x)}}+\frac {a C \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(Sqrt[a]*(8*A + 3*C)*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*C*Sin[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (C*Sqrt[a + a*Cos[c + d*x
]]*Sin[c + d*x])/(2*d*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 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 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 3125

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(
n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Si
mp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^
(-1)] && NeQ[m + n + 2, 0]

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[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 \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (4 A+C)+\frac {1}{2} a C \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{2 a} \\ & = \frac {a C \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {1}{8} \left ((8 A+3 C) \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 C \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}-\frac {\left ((8 A+3 C) \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 {\sqrt {a} (8 A+3 C) \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 C \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} (8 A+3 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 C \sqrt {\cos (c+d x)} \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{8 d} \]

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*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]*(8*A + 3*C)*ArcSin
[Sqrt[2]*Sin[(c + d*x)/2]] + 2*C*Sqrt[Cos[c + d*x]]*(2*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(8*d)

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {\left (2 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+8 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+3 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(184\)
parts	\(\frac {2 A \left (\sqrt {\sec }\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right )}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \left (2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+3 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(241\)

[In]

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

[Out]

1/4/d*(2*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)+8*A*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+3*C*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ta
n(d*x+c)))*sec(d*x+c)^(1/2)*((1+cos(d*x+c))*a)^(1/2)*cos(d*x+c)/(1+cos(d*x+c))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=-\frac {{\left ({\left (8 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

-1/4*(((8*A + 3*C)*cos(d*x + c) + 8*A + 3*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(
a)*sin(d*x + c))) - (2*C*cos(d*x + c)^2 + 3*C*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x
+ c)))/(d*cos(d*x + c) + d)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + C*cos(c + d*x)**2)*sqrt(sec(c + d*x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1207 vs. \(2 (120) = 240\).

Time = 0.75 (sec) , antiderivative size = 1207, normalized size of antiderivative = 8.38 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/16*(16*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)*sin(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co
s(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(d*x + c)) + (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))) + s
in(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*ar
ctan2(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)*(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)*(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*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)*(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*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)*(co
s(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 + 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)))*C)/d

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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