3.3.99 \(\int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx\) [299]

3.3.99.1 Optimal result
3.3.99.2 Mathematica [C] (verified)
3.3.99.3 Rubi [A] (verified)
3.3.99.4 Maple [A] (verified)
3.3.99.5 Fricas [C] (verification not implemented)
3.3.99.6 Sympy [F]
3.3.99.7 Maxima [F]
3.3.99.8 Giac [F(-1)]
3.3.99.9 Mupad [F(-1)]

3.3.99.1 Optimal result

Integrand size = 27, antiderivative size = 200 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a d}-\frac {e^{3/2} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a d (1+\cos (c+d x)+\sin (c+d x))}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a d (1+\cos (c+d x)+\sin (c+d x))} \]

output
e*(e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/a/d-e^(3/2)*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)/a/d/(1+c 
os(d*x+c)+sin(d*x+c))+e^(3/2)*arctan(sin(d*x+c)*e^(1/2)/(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)/a/d/( 
1+cos(d*x+c)+sin(d*x+c))
 
3.3.99.2 Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.38 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2\ 2^{3/4} (e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (1+\sin (c+d x))^{3/4} \sqrt {a (1+\sin (c+d x))}} \]

input
Integrate[(e*Cos[c + d*x])^(3/2)/Sqrt[a + a*Sin[c + d*x]],x]
 
output
(-2*2^(3/4)*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[1/4, 5/4, 9/4, (1 - S 
in[c + d*x])/2])/(5*d*e*(1 + Sin[c + d*x])^(3/4)*Sqrt[a*(1 + Sin[c + d*x]) 
])
 
3.3.99.3 Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3164, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3164

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3156

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

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

input
Int[(e*Cos[c + d*x])^(3/2)/Sqrt[a + a*Sin[c + d*x]],x]
 
output
(e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(a*d) + (e^2*((-2*ArcSin 
h[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]))))/(2*a)
 

3.3.99.3.1 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 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 3164
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)]], x_Symbol] :> Simp[g*Sqrt[g*Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x 
]]/(b*f)), x] + Simp[g^2/(2*a)   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[g*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]
 
3.3.99.4 Maple [A] (verified)

Time = 6.79 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.44

method result size
default \(\frac {\left (\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+\arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\cos ^{2}\left (d x +c \right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \sqrt {e \cos \left (d x +c \right )}\, e}{d \left (\cos \left (d x +c \right )-\sin \left (d x +c \right )+1\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}\) \(288\)

input
int((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
1/d*(arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)) 
*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+arctan((-cos(d*x+c)/(1+cos( 
d*x+c)))^(1/2))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+(-cos(d*x+c) 
/(1+cos(d*x+c)))^(1/2)*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+c 
os(d*x+c)))^(1/2))+(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((-cos(d*x+c)/ 
(1+cos(d*x+c)))^(1/2))+cos(d*x+c)^2+cos(d*x+c)*sin(d*x+c)+cos(d*x+c))*(e*c 
os(d*x+c))^(1/2)*e/(cos(d*x+c)-sin(d*x+c)+1)/(a*(1+sin(d*x+c)))^(1/2)
 
3.3.99.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 971, normalized size of antiderivative = 4.86 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Too large to display} \]

input
integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas 
")
 
output
-1/4*(I*a*d*(-e^6/(a^2*d^4))^(1/4)*log(-1/2*(2*(e^4*sin(d*x + c) + (a*d^2* 
e*cos(d*x + c) + a*d^2*e)*sqrt(-e^6/(a^2*d^4)))*sqrt(e*cos(d*x + c))*sqrt( 
a*sin(d*x + c) + a) + (I*a^2*d^3*cos(d*x + c) + I*a^2*d^3 + (2*I*a^2*d^3*c 
os(d*x + c) + I*a^2*d^3)*sin(d*x + c))*(-e^6/(a^2*d^4))^(3/4) - (2*I*a*d*e 
^3*cos(d*x + c)^2 + I*a*d*e^3*cos(d*x + c) - I*a*d*e^3*sin(d*x + c) - I*a* 
d*e^3)*(-e^6/(a^2*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - I*a*d* 
(-e^6/(a^2*d^4))^(1/4)*log(-1/2*(2*(e^4*sin(d*x + c) + (a*d^2*e*cos(d*x + 
c) + a*d^2*e)*sqrt(-e^6/(a^2*d^4)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + 
c) + a) + (-I*a^2*d^3*cos(d*x + c) - I*a^2*d^3 + (-2*I*a^2*d^3*cos(d*x + c 
) - I*a^2*d^3)*sin(d*x + c))*(-e^6/(a^2*d^4))^(3/4) - (-2*I*a*d*e^3*cos(d* 
x + c)^2 - I*a*d*e^3*cos(d*x + c) + I*a*d*e^3*sin(d*x + c) + I*a*d*e^3)*(- 
e^6/(a^2*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - a*d*(-e^6/(a^2* 
d^4))^(1/4)*log(-1/2*(2*(e^4*sin(d*x + c) - (a*d^2*e*cos(d*x + c) + a*d^2* 
e)*sqrt(-e^6/(a^2*d^4)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + ( 
a^2*d^3*cos(d*x + c) + a^2*d^3 + (2*a^2*d^3*cos(d*x + c) + a^2*d^3)*sin(d* 
x + c))*(-e^6/(a^2*d^4))^(3/4) + (2*a*d*e^3*cos(d*x + c)^2 + a*d*e^3*cos(d 
*x + c) - a*d*e^3*sin(d*x + c) - a*d*e^3)*(-e^6/(a^2*d^4))^(1/4))/(cos(d*x 
 + c) + sin(d*x + c) + 1)) + a*d*(-e^6/(a^2*d^4))^(1/4)*log(-1/2*(2*(e^4*s 
in(d*x + c) - (a*d^2*e*cos(d*x + c) + a*d^2*e)*sqrt(-e^6/(a^2*d^4)))*sqrt( 
e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (a^2*d^3*cos(d*x + c) + a^2*...
 
3.3.99.6 Sympy [F]

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

input
integrate((e*cos(d*x+c))**(3/2)/(a+a*sin(d*x+c))**(1/2),x)
 
output
Integral((e*cos(c + d*x))**(3/2)/sqrt(a*(sin(c + d*x) + 1)), x)
 
3.3.99.7 Maxima [F]

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

input
integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima 
")
 
output
integrate((e*cos(d*x + c))^(3/2)/sqrt(a*sin(d*x + c) + a), x)
 
3.3.99.8 Giac [F(-1)]

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

input
integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 
output
Timed out
 
3.3.99.9 Mupad [F(-1)]

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

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