\(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx\) [293]

Optimal result

Integrand size = 27, antiderivative size = 204 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{5/2} (1+\cos (c+d x)+\sin (c+d x))}-\frac {2 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{5/2} (1+\cos (c+d x)+\sin (c+d x))}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}} \] Output:

2*a^2*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin( 
d*x+c))^(1/2)/d/e^(5/2)/(1+cos(d*x+c)+sin(d*x+c))-2*a^2*arctan(e^(1/2)*sin 
(d*x+c)/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)*(a 
+a*sin(d*x+c))^(1/2)/d/e^(5/2)/(1+cos(d*x+c)+sin(d*x+c))+4/3*a*(a+a*sin(d* 
x+c))^(3/2)/d/e/(e*cos(d*x+c))^(3/2)
 

Mathematica [C] (verified)

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

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.38 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{5/2}}{3 d e (e \cos (c+d x))^{3/2} (1+\sin (c+d x))^{7/4}} \] Input:

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

Output:

(4*2^(3/4)*Hypergeometric2F1[-3/4, -3/4, 1/4, (1 - Sin[c + d*x])/2]*(a*(1 
+ Sin[c + d*x]))^(5/2))/(3*d*e*(e*Cos[c + d*x])^(3/2)*(1 + Sin[c + d*x])^( 
7/4))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3155, 3042, 3156, 3042, 25, 3254, 216, 3312, 63, 222}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{(e \cos (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3155

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sqrt {e \cos (c+d x)}}dx}{e^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sqrt {e \cos (c+d x)}}dx}{e^2}\)

\(\Big \downarrow \) 3156

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}\right )}{e^2}\)

Input:

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

Output:

(4*a*(a + a*Sin[c + d*x])^(3/2))/(3*d*e*(e*Cos[c + d*x])^(3/2)) - (a^2*((- 
2*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a* 
Sin[c + d*x]])/(d*Sqrt[e]*(1 + Cos[c + d*x] + Sin[c + d*x])) + (2*ArcTan[( 
Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqrt[ 
1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*Sqrt[e]*(1 + Cos[c + d*x] + 
 Sin[c + d*x]))))/e^2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

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

rule 3155

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[-2*b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(p + 1))), x] + Simp[b^2*((2*m + p - 1)/(g^2*(p + 1))) 
 Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ 
[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && Int 
egersQ[2*m, 2*p]
 

rule 3156

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)] 
*(g_.)], x_Symbol] :> Simp[a*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x 
]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x]))   Int[Sqrt[1 + Cos[e + f*x]]/Sqrt 
[g*Cos[e + f*x]], x], x] + Simp[b*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e 
+ f*x]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x]))   Int[Sin[e + f*x]/(Sqrt[g*C 
os[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] & 
& EqQ[a^2 - b^2, 0]
 

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]
 

rule 3312

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 5.86 (sec) , antiderivative size = 1104, normalized size of antiderivative = 5.41

method	result	size
default	\(\text {Expression too large to display}\)	\(1104\)

Input:

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

Output:

1/3/d/(1+2^(1/2))/(2*2^(1/2)+3)^(1/2)/e^(11/2)*((((12*cos(1/2*d*x+1/2*c)^2 
+24*cos(1/2*d*x+1/2*c)+12)*sin(1/2*d*x+1/2*c)-12*cos(1/2*d*x+1/2*c)*(cos(1 
/2*d*x+1/2*c)^2+2*cos(1/2*d*x+1/2*c)+1))*2^(1/2)+(24*cos(1/2*d*x+1/2*c)^2+ 
48*cos(1/2*d*x+1/2*c)+24)*sin(1/2*d*x+1/2*c)-24*cos(1/2*d*x+1/2*c)*(cos(1/ 
2*d*x+1/2*c)^2+2*cos(1/2*d*x+1/2*c)+1))*(e*(2*cos(1/2*d*x+1/2*c)^2-1)/(cos 
(1/2*d*x+1/2*c)+1)^2)^(1/2)*e^(7/2)*((2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)+2 
*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*(-2*(2^(1/2)*cos(1/2* 
d*x+1/2*c)-2^(1/2)-2*cos(1/2*d*x+1/2*c)+1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*E 
llipticPi((2*2^(1/2)+3)^(1/2)*(csc(1/2*d*x+1/2*c)-cot(1/2*d*x+1/2*c)),-1/( 
2*2^(1/2)+3),(-2*2^(1/2)+3)^(1/2)/(2*2^(1/2)+3)^(1/2))+(((-18*cos(1/2*d*x+ 
1/2*c)^2+9)*sin(1/2*d*x+1/2*c)+9*cos(1/2*d*x+1/2*c)*(2*cos(1/2*d*x+1/2*c)^ 
2-1))*2^(1/2)+(-24*cos(1/2*d*x+1/2*c)^2+12)*sin(1/2*d*x+1/2*c)+12*cos(1/2* 
d*x+1/2*c)*(2*cos(1/2*d*x+1/2*c)^2-1))*arctanh(2^(1/2)*e^(1/2)*cos(1/2*d*x 
+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/(e*(2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+ 
1/2*c)+1)^2)^(1/2))*e^4+(((-6*cos(1/2*d*x+1/2*c)^2-12*cos(1/2*d*x+1/2*c)-6 
)*sin(1/2*d*x+1/2*c)+6*cos(1/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)^2+2*cos(1/2* 
d*x+1/2*c)+1))*2^(1/2)+(-12*cos(1/2*d*x+1/2*c)^2-24*cos(1/2*d*x+1/2*c)-12) 
*sin(1/2*d*x+1/2*c)+12*cos(1/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)^2+2*cos(1/2* 
d*x+1/2*c)+1))*e^(7/2)*(-2*(2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)-2*cos(1/2*d 
*x+1/2*c)+1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*EllipticF((1+2^(1/2))*(csc(1...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (176) = 352\).

Time = 0.17 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.05 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {16 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} a^{2} + 12 \, \sqrt {2} {\left (a^{2} e \sin \left (d x + c\right ) - a^{2} e\right )} \sqrt {\frac {a}{e}} \arctan \left (\frac {\sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right )}\right ) - 3 \, \sqrt {2} {\left (a^{2} e \sin \left (d x + c\right ) - a^{2} e\right )} \sqrt {\frac {a}{e}} \log \left (\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} {\left (\cos \left (d x + c\right ) + 1\right )} + 2 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right ) + 3 \, \sqrt {2} {\left (a^{2} e \sin \left (d x + c\right ) - a^{2} e\right )} \sqrt {\frac {a}{e}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} {\left (\cos \left (d x + c\right ) + 1\right )} - 2 \, a \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right )}{12 \, {\left (d e^{3} \sin \left (d x + c\right ) - d e^{3}\right )}} \] Input:

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

Output:

-1/12*(16*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*a^2 + 12*sqrt(2)*( 
a^2*e*sin(d*x + c) - a^2*e)*sqrt(a/e)*arctan(sqrt(2)*sqrt(e*cos(d*x + c))* 
sqrt(a*sin(d*x + c) + a)*sqrt(a/e)*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos( 
d*x + c)*sin(d*x + c) + a*cos(d*x + c))) - 3*sqrt(2)*(a^2*e*sin(d*x + c) - 
 a^2*e)*sqrt(a/e)*log((2*sqrt(2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) 
+ a)*sqrt(a/e)*(cos(d*x + c) + 1) + 2*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) 
+ (2*a*cos(d*x + c) + a)*sin(d*x + c) + a)/(cos(d*x + c) + sin(d*x + c) + 
1)) + 3*sqrt(2)*(a^2*e*sin(d*x + c) - a^2*e)*sqrt(a/e)*log(-(2*sqrt(2)*sqr 
t(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt(a/e)*(cos(d*x + c) + 1) - 
2*a*cos(d*x + c)^2 - 3*a*cos(d*x + c) - (2*a*cos(d*x + c) + a)*sin(d*x + c 
) - a)/(cos(d*x + c) + sin(d*x + c) + 1)))/(d*e^3*sin(d*x + c) - d*e^3)
 

Sympy [F(-1)]

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

integrate((a+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(5/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

integrate((a*sin(d*x + c) + a)^(5/2)/(e*cos(d*x + c))^(5/2), x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(5/2),x)
 

Output:

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(5/2), x)
 

Reduce [F]

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

int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(5/2),x)
 

Output:

(sqrt(e)*sqrt(a)*a**2*(int((sqrt(sin(c + d*x) + 1)*sqrt(cos(c + d*x))*sin( 
c + d*x)**2)/cos(c + d*x)**3,x) + 2*int((sqrt(sin(c + d*x) + 1)*sqrt(cos(c 
 + d*x))*sin(c + d*x))/cos(c + d*x)**3,x) + int((sqrt(sin(c + d*x) + 1)*sq 
rt(cos(c + d*x)))/cos(c + d*x)**3,x)))/e**3