\(\int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [176]

Optimal result

Integrand size = 35, antiderivative size = 133 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^{3/2} (12 A+7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}+\frac {a^2 (4 A+5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {a B \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*a^(3/2)*(12*A+7*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d 
+1/4*a^2*(4*A+5*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/ 
2*a*B*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.76 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} (12 A+7 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (4 A+7 B+2 B \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*(12*A + 7*B)*ArcSi 
n[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(4*A + 7*B + 2*B*Cos[c 
+ d*x])*Sin[(c + d*x)/2]))/(8*d)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 3455, 27, 3042, 3460, 3042, 3253, 223}

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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} (a (4 A+B)+a (4 A+5 B) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} (a (4 A+B)+a (4 A+5 B) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (4 A+B)+a (4 A+5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (12 A+7 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (12 A+7 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {1}{4} \left (\frac {a^2 (4 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {a (12 A+7 B) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{4} \left (\frac {a^{3/2} (12 A+7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^2 (4 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\)

Input:

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

Output:

(a*B*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ((a 
^(3/2)*(12*A + 7*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]] 
])/d + (a^2*(4*A + 5*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos 
[c + d*x]]))/4
 

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 223

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 3042

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

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3455

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3460

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] + Simp[(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]
 

Maple [A] (verified)

Time = 21.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\left (12 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+7 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+4 A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (2 d x +2 c \right )+7 \sin \left (d x +c \right )\right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, a}{4 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right )}\)	\(184\)
parts	\(\frac {A \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\frac {\sin \left (2 d x +2 c \right )}{2}+3 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )\right ) a}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (7 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (2 d x +2 c \right )+7 \sin \left (d x +c \right )\right )\right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sqrt {\cos \left (d x +c \right )}\, a}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(230\)

Input:

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/4/d*(12*A*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+7*B*arcta 
n(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4*A*(cos(d*x+c)/(cos(d*x+c 
)+1))^(1/2)*sin(d*x+c)+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(sin(2*d*x+2*c)+7 
*sin(d*x+c))*B)*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)/(cos 
(d*x+c)+1))^(1/2)/(cos(d*x+c)+1)*a
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {{\left (2 \, B a \cos \left (d x + c\right ) + {\left (4 \, A + 7 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (12 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + {\left (12 \, A + 7 \, B\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algo 
rithm="fricas")
 

Output:

1/4*((2*B*a*cos(d*x + c) + (4*A + 7*B)*a)*sqrt(a*cos(d*x + c) + a)*sqrt(co 
s(d*x + c))*sin(d*x + c) + ((12*A + 7*B)*a*cos(d*x + c) + (12*A + 7*B)*a)* 
sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x 
 + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1884 vs. \(2 (113) = 226\).

Time = 0.34 (sec) , antiderivative size = 1884, normalized size of antiderivative = 14.17 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algo 
rithm="maxima")
 

Output:

1/16*(4*(2*(a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin 
(d*x + c) - (a*cos(d*x + c) - a)*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)*sqrt(a) + 3*(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) - 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*arctan 
2(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(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - 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*arctan2(si 
n(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), c 
os(2*d*x + 2*c) + 1)) + 1) + a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x ...
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\sqrt {a}\, a \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}d x \right ) b \right ) \] Input:

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

Output:

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