\(\int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [209]

Optimal result

Integrand size = 25, antiderivative size = 76 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {\cos (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3241, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

\(\displaystyle \frac {2 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\)

Input:

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

Output:

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

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 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

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]
 

Maple [A] (verified)

Time = 5.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {2 \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sin \left (d x +c \right )\right ) a}{d \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right )}\)	\(99\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.67 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \, {\left (\sqrt {a \cos \left (d x + c\right ) + a} a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )\right )}}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \] Input:

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

Output:

2*(sqrt(a*cos(d*x + c) + a)*a*sqrt(cos(d*x + c))*sin(d*x + c) + (a*cos(d*x 
 + c)^2 + a*cos(d*x + c))*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*c 
os(d*x + c)^2 + d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 997 vs. \(2 (66) = 132\).

Time = 0.23 (sec) , antiderivative size = 997, normalized size of antiderivative = 13.12 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((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*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(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c)))) + 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)))*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(s 
in(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), co 
s(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) - 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(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),...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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