\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [105]

Optimal result

Integrand size = 27, antiderivative size = 109 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 C \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.58 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (3 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 C \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )}{3 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]*(3*(A + C)*ArcTanh[Sin[(c + d*x)/2]] - 4*C*Sin[(c + d* 
x)/2]^3))/(3*d*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3503, 27, 3042, 3230, 3042, 3128, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 3503

\(\displaystyle \frac {2 \int \frac {a (3 A+C)-2 a C \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (3 A+C)-2 a C \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (3 A+C)-2 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {3 a (A+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {6 a (A+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \sqrt {2} \sqrt {a} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

Input:

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

Output:

(2*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) + ((3*Sqrt[2]*Sqrt[a]* 
(A + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])] 
)/d - (4*a*C*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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

rule 3230

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

rule 3503

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + 
 (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ 
(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^ 
m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a 
, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-4 C \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +3 C \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(173\)
parts	\(\frac {A \sqrt {2}\, \operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, 1\right )}{d \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-4 \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(193\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.30 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, {\left (C \cos \left (d x + c\right ) - C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {3 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/6*(4*(C*cos(d*x + c) - C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 3*sqrt 
(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)* 
sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d 
*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19528 vs. \(2 (92) = 184\).

Time = 0.74 (sec) , antiderivative size = 19528, normalized size of antiderivative = 179.16 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/60*(30*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* 
sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d 
*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*A/sqrt(a) + (20*(cos(d*x + c) 
 + 1)*sin(5/2*d*x + 5/2*c)^3 + 8*(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos( 
d*x + c) + 1)*sin(3/2*d*x + 3/2*c)^3 - 20*cos(5/2*d*x + 5/2*c)^3*sin(d*x + 
 c) + 2*(15*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1 
/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 
 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + 15*(log(cos(1/2*d*x + 1/2 
*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2 
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*si 
n(d*x + c)^2 + 30*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2 
*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/ 
2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c) + 4*(cos(d*x + c)^2 + s 
in(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 20*cos(3/2*d*x 
+ 3/2*c)*sin(d*x + c) + 15*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2* 
c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 15*log(cos(1/2*d*x + 1/2*c)^2 + sin(1 
/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(5/2*d*x + 5/2*c)^2 + 
30*((log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin 
(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + (log(cos(1/2*d*x + 1/2*c)^2 + ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2\,C\,\sin \left (c+d\,x\right )\,\left (a+a\,\cos \left (c+d\,x\right )\right )+3\,\sqrt {2}\,A\,a\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{a}}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-4\,\sqrt {2}\,C\,a\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{a}}\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )+3\,\sqrt {2}\,C\,a\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{a}}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )}{3\,a\,d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \] Input:

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

Output:

(2*C*sin(c + d*x)*(a + a*cos(c + d*x)) + 3*2^(1/2)*A*a*((a + a*cos(c + d*x 
))/a)^(1/2)*ellipticF(c/2 + (d*x)/2, 1) - 4*2^(1/2)*C*a*((a + a*cos(c + d* 
x))/a)^(1/2)*ellipticE(c/2 + (d*x)/2, 1) + 3*2^(1/2)*C*a*((a + a*cos(c + d 
*x))/a)^(1/2)*ellipticF(c/2 + (d*x)/2, 1))/(3*a*d*(a + a*cos(c + d*x))^(1/ 
2))
 

Reduce [F]

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

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

Output:

(sqrt(a)*(int(sqrt(cos(c + d*x) + 1)/(cos(c + d*x) + 1),x)*a + int((sqrt(c 
os(c + d*x) + 1)*cos(c + d*x)**2)/(cos(c + d*x) + 1),x)*c))/a