\(\int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx\) [357]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {5 a^{5/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {a^3 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:

5*a^(5/2)*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^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^( 
1/2)+2*a^2*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.51 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {\sqrt {\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (7 (89+28 \cos (c+d x)+3 \cos (2 (c+d x))) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2},\frac {7}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )+24 (3+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {9}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2(c+d x)+6 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (\frac {3}{2},2,\frac {5}{2};1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{420 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2),x]
 

Output:

(Sqrt[Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[(c + d*x)/2]^4*Sqrt[S 
ec[c + d*x]]*(7*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Hypergeometric 
2F1[1/2, 3/2, 7/2, 2*Sin[(c + d*x)/2]^2] + 24*(3 + Cos[c + d*x])*Hypergeom 
etric2F1[3/2, 5/2, 9/2, 2*Sin[(c + d*x)/2]^2]*Sin[c + d*x]^2 + 6*Csc[(c + 
d*x)/2]^2*HypergeometricPFQ[{3/2, 2, 5/2}, {1, 9/2}, 2*Sin[(c + d*x)/2]^2] 
*Sin[c + d*x]^4)*Tan[(c + d*x)/2])/(420*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4710, 3042, 3241, 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.

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\)

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a \left (\frac {5}{2} a \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a \left (\frac {5}{2} a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a \left (-\frac {5 a \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}}+a \left (\frac {5 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin 
[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + a*((5*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 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]
 

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]
 

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 = 14.68 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (5 \cos \left (d x +c \right )^{2} \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\cos \left (d x +c \right ) \left (\cos \left (d x +c \right )+2\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) a^{2}}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(135\)

Input:

int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(3/2)/(cos(d*x+c)+1)/(cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*(5*cos(d*x+c)^2*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2))+cos(d*x+c)*(cos(d*x+c)+2)*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2))*a^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {5 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{d \cos \left (d x + c\right ) + d} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

-(5*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt( 
cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (a^2*cos(d*x + c) + 2*a^2)*sqrt(a* 
cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*sec(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 973 vs. \(2 (116) = 232\).

Time = 0.34 (sec) , antiderivative size = 973, normalized size of antiderivative = 7.26 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

1/4*(2*(a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d 
*x + c) - (a^2*cos(d*x + c) - a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c) + 1)))*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2* 
d*x + 2*c) + 1)*sqrt(a) + 5*(a^2*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^2*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*ar 
ctan2(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*arct 
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan 
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - a^2*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*arct 
an2(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^2*arctan2((cos(2*d*x + 2*c)^2 + si...
 

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2),x)
 

Output:

int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x +\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) \] Input:

int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(3/2),x)
 

Output:

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