\(\int (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \, dx\) [350]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \, dx=\frac {3 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a^2 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \, dx=\frac {a \sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \] Input:

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

Output:

(a*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec 
[c + d*x]]*(3*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d* 
x]]*Sin[(c + d*x)/2]))/(2*d)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4710, 3042, 3242, 27, 2011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((3*a^(3/2)*ArcSin[(Sqrt[a]*Sin[c + 
d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x]) 
/(d*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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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]
 

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 12.55 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+3 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )\right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) a}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(117\)

Input:

int((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+3*arctan(tan(d*x+c)*(cos 
(d*x+c)/(cos(d*x+c)+1))^(1/2)))*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(1/2)* 
cos(d*x+c)/(cos(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \, dx=\frac {\sqrt {a \cos \left (d x + c\right ) + a} a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{d \cos \left (d x + c\right ) + d} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

(sqrt(a*cos(d*x + c) + a)*a*sqrt(cos(d*x + c))*sin(d*x + c) - 3*(a*cos(d*x 
 + c) + a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqr 
t(a)*sin(d*x + c))))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 803 vs. \(2 (81) = 162\).

Time = 0.27 (sec) , antiderivative size = 803, normalized size of antiderivative = 8.45 \[ \int (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

1/4*(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*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(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)*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) + a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*...
 

Giac [F(-1)]

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

integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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